electric field potential

bull11

Electric Chargebull Types

ndash Positivebull Glass rubbed with silk bull Missing electrons

ndash Negativebull RubberPlastic rubbed

with furbull Extra electrons

bull Arbitrary choice ndash convention attributed to

bull Units amount of charge is measured in

[Coulombs]bull Empirical Observations

ndash Like charges repelndash Unlike charges attract

Electrostaticrdquo ForceElectrostaticrdquo Force

bullElectrostaticrdquo ForceElectrostaticrdquo Force

bullCoulombs law states that the force between two electric charges is proportional to the product of the charges and inversely proportional to their separation

1 22E c

q qF k

r

bullq charge Coulomb C

bullr distance between charges m

bullFE Electric Force Newton N VECTOR

bullkc coulomb constant 899x109Nm2C2

Coulombs LawCoulombs Lawbull The force between charges is

directly proportional to the magnitude or amount of each charge

bull Doubling one charge doubles the force

bull Doubling both charges quadruples the forcebull The force between charges is inversely proportional to the square of the distance between them

bull Doubling the distance reduces the force by a factor of 22 = (4) decreasing the force to one-fourth its original value (14)

bull bull7

bullNotesNotes

bullk is a constant (9109) Q is in Coulombs r in meters

bullOne unit of charge (proton) has Q = 1610-19 Coulombs

bullLooks a lot like Newtonrsquos gravitation in form

bullElectron and proton attract each other 1040 times stronger electrically than gravitationally

bullTwo charges Q1 and Q2 separated by distance r exert a force on each other

F = (kmiddotQ1middotQ2) r2

bullGood thing charge is usually balanced

bullCoulomb Law Coulomb Law IllustratedIllustratedbullLike charges repel

bullUnlike charges attract

bullr

bullndash bull+

bullndashbullndashbullIf charges are of same magnitude (and same separation)

bullall the forces will be the same magnitude with different

bulldirections

bullCoulomb Force Law Coulomb Force Law QualitativelyQualitatively

bullDouble one of the charges

bullforce doubles

bullChange sign of one of the charges

bullforce changes direction

bullChange sign of both charges

bullforce stays the same

bullDouble the distance between charges

bullforce four times weaker

bullDouble both charges

bullforce four times stronger

bull11

bullAn object A with +825 x 10-6 C charge has two other charges nearby Object B -35 x 10-6 C is 0030 m to the right Object C +250 x 10-6 C is 0050 m below What is the net force and the angle on AV

X

Y

bull12

bullThree point charges lie along the x axis as shown in Figure The positive charge q1 150 C is at x 200 mthe positive charge q2 600 C is at the origin and the resultant force acting on q3 is zero What is the x coordinate of q3

bullSolution Because q3 is negative and q1 and q2 are positive

bullFor the resultant force on q3 to be zero F23 must be equal in magnitude and opposite in direction to F13 Setting the magnitudes of the two forces equal

bullif the charges have opposite signs the force is negative

bullAttractiveAttractive

bullIf the charges have the same sign the force is positive

bullRepulsiveRepulsive

Electric Force VectorElectric force in vector

form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1

Superposition PrincipleSuperposition PrincipleThe net force acting on any charge is the

vector sum of the forces due to the remaining charges in the distribution

znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF

bull1102006 bull184 Lecture 3 bull15

Learning Goals - we will learnbull How to use Coulombrsquos Law (and vector

addition) to calculate the force between electric chargesbull How to calculate the electric field caused by discrete electric chargesbull How to calculate the electric field caused by a continuous distribution of electric charge

bull17

bullExampleExample

bullWhat is the force between two charges of 1 C separated by 1 meter

bullAnswer 899 x 109 N

bullie huge

Example - Equilibrium PositionExample - Equilibrium PositionConsider two charges located on the x

axis

The charges are described byq1 = 015 C x1 = 00 m

q2 = 035 C x2 = 040 m

Where do we need to put a third charge for that charge to be at an equilibrium pointAt the equilibrium point the forces from the

two charges will cancel bull18

x1 x2

bullThe equilibrium point must be along the x-axis

bullThree regions along the x-axis where we might place our third charge

x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2

bullthird charge to be at an equilibrium point when

22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample

bullTwo electrostatic point charges of +200 μC and ndash300 μC exert attractive

bullforces on each other of ndash145 N What is the distance between the two

bullcharges

bullq1 = 200 times C q2 = minus300 times C

bullFelectric = minus145 N kC = 899 times 10 Nbullm2C2

bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample

bullSuppose two charges having equal but opposite charge are separated by

bull64 times 10 m If the magnitude of the electric force between the charges

bullis 562 times10ndash14 N what is the value of q

bull-8

example

bullHow will the force between two spheres change if the distance between two sphere doubles

(a) double (b) one half (c) quadruple (d) one forth (e) same

bullexamplebullThere are two balls Ball A has a charge of +18 x 10-8 C and Ball B has a charge of +15 x 10-8 C and is 0002 m away from A a What is the force acting on A

bullexampleexample

bullCharged spheres A and B are fixed in position as shown and have charges of +79 x 10-6 C and -23 x 10-6 C respectively Calculate the net force on sphere

C whose charge is +58 x 10-6 C

])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261

Example - The Helium NucleusPart 1 The nucleus of a helium atom has two protons and two neutrons What is the magnitude of the electric force between the two protons in the helium nucleus R of nucleus =

bull27

bullAnswer 58 NbullPart 2 What if the distance is doubled how will the force change

bullAnswer 145 N

bullInverse square law If the distance is doubled then

the force is reduced by a factor of 4

Example - Charged Pendulums

Consider two identical charged balls hanging from

the ceiling by strings of equal length 15 m (in equilibrium) Each ball has a charge of 25

C The balls hang at an angle = 25 with respect to

the vertical What is the mass of the balls

Step 1 Three forces act on

each ball Coulomb force gravity and the tension of the

stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully

Example - Charged Pendulums (2)

tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29

bullStep 2Step 2 The balls are in equilibrium positions That means the sum of all

forces acting on the ball is zero

bullAnswer m = 076 kg

bullA similar analysis applies to the ball on the right

bulld=2 l sin

Example - Four ChargesExample - Four Charges

Consider four charges placed at the corners of a square with sides of length 125 m as shown on the right What is the magnitude of the electric force on q4 resulting from the electric force from the remaining three charges

bull30

bullSet up an xy-coordinate system with its origin at q2

Three Charges in a Line

bull31

Question Suppose that three point charges qa qb and qc are arranged at the vertices of a right-angled triangle as shown in the diagram What is the magnitude and direction of the electrostatic force acting on the third charge and

bullmbullmSolution The magnitude of the force exerted by charge on charge is given by

Three Charges in a Plane

bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE

bullElectric Field E is defined as the force acting on a test particle divided by the charge of that test particle

bullThus Electric Field from a single charge is

rr

F

E

bullA charged body creates an electric fieldCoulomb force of repulsion between two charged bodies at A and B (having charges Q and qo respectively) has magnitude F = k |Q qo |r2 = qo [ k Qr2 ]

where we have factored out the small charge qo We can write the force in terms of an electric field E Therefore we can write for

bullthe electric fieldE = [ k Q r2 ]

bullElectric Force and Field Force

bullWhat -- Action on a distance

bullHow ndash Electric Field

bullWhy ndash Field Force

bullWhere ndash in the space surrounding charges

bullElectric FieldbullElectric field is said to exist in the

region of space around a charged object the source charge

bullConcept of test charge

bullSmall and positive

bullDoes not affect charge distribution

bullElectric field

bullExistence of an electric field is a property of its source

bullPresence of test charge is not necessary for the field to exist

0q

FE

Electric chargeCondense charge E = (kQr2)r

bullInstead we choose to represent the electric field with lines whose direction indicates the direction of the field bullNotice that as we

move away from the charge the density of lines decreases

bullThese are called

bullElectric Field Lines

Electric field lines

Lines of force

Quiz The field directionA charge +q is placed at (01)A charge ndashq is placed at (0-1)What is the direction of the field at (10)

A) i + jB) i - jC) -jD) -i

Superposition amp Electric FieldSuperposition amp Electric Field

1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark

Fish to detect object

Electric Field of a single charge

bull+

r

E

bull+Q0

bull+Q0

bull+Q0

bullNote the Electric Field is defined everywhere even if there is no test charge is not there

bull+Q0

bullElectric field from

test particles

bullElectric Field from isolated charges

(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q

bullUsing the Field to determine the force

bullCh16 bull49

bullExample 3A A +100 C point charge is separated from a

bull-50 C charge by a distance of 050 m as shown below (A) First calculate the electric field at midway between the two charges (B) Find the force on an electron that is placed at this point and then calculate the acceleration when it is released

bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50

bullExample 2 Particles of charge Q1 = +500 C Q 2 = -600 C and Q3 = +800 C are placed on the corners of a square of side 0400 m as shown below Calculate the force on Q2

(Magnitude and direction)bull+

bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F

bullNote that the charges and distances are the same as in Example 1 so we do not need to use Coulombs Law again

NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E

In part A we found that E = 21x107 NC and is directed to the right

q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )

Electric field of a dipole along the y-axishellip

r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field

bullElectric ldquofield linesrdquo tell a positive

bullcharge which way to move

bullFor example a positive charge itself

bullhas field lines pointing away from it

bullbecause this is how a positively-charged

bullldquotest-particlerdquo would respond if placed

bullin the vicinity (repulsive force)

bullRun Away

bull+

Electric line

Two charges are placed on the x axis The first with a charge of +Q is at the origin

bullhe second with a charge of -2Q is at x = 100 m Where on the x axis is the electric field equal to zero

bullThe answer to go with is x = 241 m This corresponds to 241 m to the left of the +Q charge The other point is between the charges It corresponds to the point where the fields from the two charges have the same magnitude but they both point in the same direction there

so they dont cancel out

bullexampleexample

bullFind electric field at point P in the figure

bullsolution solution

CNr

kqE 18

4

108109 99

2

8730621

61tan 1

Solution (details given in class)

A vector may be decomposed into its x- and y-components as shown

2 2 2

cos

sinx

y

x y

A A

A A

A A A

bullEY =8Nc +3456N C

bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample

bullFind electric field at point P3 in the figure

bullsolutionsolution

Example Problem

Solution (details given in class)0tot E

bull30degbullE1bullE2

bullE3

bullThree identical charges (q = ndash50 mC) lie along a circle of radius 20 m at angles of 30deg 150deg and 270deg as shown What is the resultant electric field at the center of the circle

Example

bullFind the electric field at point p in

figure due to the charges shownbullSolution

bullEx = E1 - E2 = -36104NCEy = E3 = 288104NC

Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1


bullExample Electric Fields Around Example Electric Fields Around ChargesCharges

Electric Field lines

bullRank the electric field strength in order from smallest to largest

bullA E1 lt E2 lt E3 = E4

bullB E3 = E4 lt E2 lt E1

bullC E2 = E3 lt E4 lt E1

bullD E1 lt E4 lt E2 = E3

1 A test charge of +3 microC is at a point P where an external electric field is directed to the right and has a magnitude of 4times106 NC If the test charge is replaced with another test charge of ndash3 microC what happens to the external electric field at P

A It is unaffected

B It reverses direction

C It changes in a way that cannot be determined

bullElectric Field due to a Point Charge Electric Field due to a Point Charge Q

rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00

Direction is radial outward for +|Q|

bull inward for -|Q|

Magnitude constant on any spherical shell

Flux through any shell enclosing Q is the same EAAA = EBAB

bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0

bullElectric Field due to a group of Electric Field due to a group of individual chargeindividual charge

bullExample Electric Field of a DipoleExample Electric Field of a Dipole

y

ar

kqE

2)( yar

kqE

2)(bullThe total field at P is

yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4

bullExample bullFind the electric field due to electric dipole shown in figure 310

along x-axis at point p which is a distance r from the origin then assume rgtgta

bullSolution

bullWhen xgtgta then

bullExample

bullIn figure shown locate the point at which the electric field is zero Assume a = 50cm

bullE1 = E2

bulld = 30cm

Electric Field LinesThe lines must begin on a positive

charge and terminate on a negative charge In the case of an excess of one type of charge some lines will begin or end infinitely far away

The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge

No two field lines can cross

Electric Field

3 Rank the magnitudes E of the electric field at points A B and C shown in the figure

A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC

bullSeptember 18 2007

bullC

bullA

bullB

Properties of Electric fieldsProperties of Electric fieldsConcentric spherical shells

E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q

bullSpherical Symmetry

bull

bullSpherical Symmetry charge distribution

bullThe electric field produced by the sphere must be the same as that

bullOf the point charge This is true everywhere outside the sphere

rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2

The field look like two parallel planes so it is the same in both cases and that

bullExcerpt n of the electric field above is Imagine that instead of a point particle the charge q gt 0 were to be distributed uniformly on the surface of a spherical shell By symmetry the field outside the shell must be the same as that of a point charge as shown below The field is zero everywhere inside the shell Therefore the magnitude of the field due the charged spherical shell has a discrete jump at the surface of the shell (see below) Since the electric field obeys the superposition principle you can use the above discussion about field due to a charged spherical shell to describe a charged solid sphere To do so represent the solid sphere as a sequence of concentric spherical shell s To find the total field at any one point simply add the values of the field due to each of the concentric shells

Example

Three particles of unknown charge are placed at rest in a

uniform electric field as shown The acceleration of

each particle after being placed in the field is indicated

What is the sign of the electric charge of each particle

a)+A B neutral - C

b) ndashA -B C neutral

C)A neutral - B +C

D)Non of the above

Spherical SymmetryNow consider a Gaussian surface with radius r2 gt RAgain by spherical symmetry we know that the electric

field will be radial and perpendicular to the Gaussian surface

Gaussrsquo Law gives us

Solving for E we find

3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge

bullsame as a point charge

bullField Due to a Thin Spherical bullShellbullUse spheres as the Gaussian surfaces

bull1048708 When r gt a the charge inside the surface is Q and

bullE = keQ r2

bull1048708 When r lt a the charge inside the surface is 0 and E = 0

Ch 23-8 Applying Gaussrsquo Law Planar Symmetry

Two conducting plates with charge density 1

All charges on the two faces of the plates For two oppositely charged plates placed near each

other E field outer side of the plates is zero while inner side the E-field= 2 10

1

0

2iE

bull85

bullPhys 133

Find the electric field inside capacitor

E inside

0E outside 0

Ch Applying Gaussrsquo Law Planar Symmetry

Two non-conducting plates with charge density +

All charges on the one face of the plates

For two oppositely charged plates placed near each other E field outer side of the plates is EL or ER with EL= ER= E(+)- E(+) and between the plates the field EB = E(+)+ E(-)

bull87

bullConsider a conducting slab in an external field E

bullIf the field inside the conductor were not zero free electrons in the conductor would experience an electrical force

bullThese electrons would accelerate

bullThese electrons would not be in equilibrium

bullTherefore there cannot be a field inside the conductor

bullPhysics 133

bullFor two oppositely charged plates placed near each other E field outer side of the plates is zero while inner side the E-field

Calculate E1 E2 and ETOTAL at point ldquoCrdquo

q = 12 nC

See Fig 2123 Electric field at ldquoCrdquo set up by charges q1 and q1

(an electric dipole)

At ldquoCrdquo E1= 64 (10)3 NC E2 = 64 (10)3 NC

EC = 49 (10)3 NC

in the +x-directionA

C

Lab 2

Need TABLE of ALL vector component

VALUES

bull92

Potential Difference in a Uniform Field

bull The V in a uniform field varies with the displacement from a reference point

bull V = E dbull The displacement is moved in the

direction of the fieldbull Any displacement perpendicular to

the field does not change the electrical potential energy

The electric potential energy Start Then So

The electric potential

Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0

Potential difference depends only on the source charge distribution (Consider points i and f without the presence of the test charge

The difference in potential energy exists only if a test charge is moved between the points

Joules JVolt V

Coulomb C

bullBecause of this potential difference is often referred to as ldquovoltagerdquo

b a baba b a

U U WV V V

q q

bullSo what is an electron Volt (eV)

bullIn addition 1 NC = 1 Vm - we can interpret the electric field as a measure of the rate of change with position of the electric potential

Potential Due to a Point ChargePotential Due to a Point Charge

Start with (set Vf=0 at and Vi=V at R)

We have

Then

So

A positively charged particle produces a positive electric potential

A negatively charged particle produces a negative electric potential

204

1

r

qE

204

1

r

qE

f

i R

f

iif EdrdsEsdEVVV )0cos(

r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40

Potential due to a group of point charges

Use superposition

For point charges

The sum is an algebraic sum not a vector sumE may be zero where V does not equal to zeroV may be zero where E does not equal to zero

n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq

Just as with potential energy only differences in electric potential are meaningful

Relative reference choose arbitrary zero reference level for ΔU or ΔV

Absolute reference start with all charge infinitely far away and set Ui = 0 then we have and at any point in an electric field

where W is the work done by the electric field on a charged particle as that particle moves in from infinity to point f

SI Unit of electric potential Volt (V) 1 volt = 1 joule per coulomb 1 J = 1 VC and 1 J = 1 N m Electric field 1 NC = (1 NC)(1 VCJ)(1 JNm) = 1 Vm Electric energy 1 eV = e(1 V) = (160times10-19 C)(1 JC) = 160times10-19 J

Electric Potential

WU qWV

bull99

Electric potential and capacitance

bullCalculate the electric potential V at 10 cm from a -60C charge

bullEquation

bullV = kq

bull rbull

bullAnswer bullV = -54x

106V

bullCalculate the electric potential V at the midpoint between a 250C charge and a -450 C separated by a distance of 60 cm

bullEquation bullV = kq

bull r

bullAnswerbullV = -60x106V

bullOctober 3 2007

Electric Potential Energy of a System of Point Charges

Start with (set Ui=0 at and Uf=U at r)

We have

If the system consists of more than two charged particles calculate U for each pair of

charges and sum the terms algebraically

r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3

Electric Potential DifferencebullThe electric potential energy depends on the charge present

bullWe can define and electric potential V which does not depend on charge by using a ldquotestrdquo charge

EdQU 0

bullChange in potential is change in potential energy for a test charge divided by the unit charge

EdQ

UV

0

bullRemember that for uniform field

0Q

UV

bull101

E Potential amp Potential Energy vsElectric Field amp Coulomb Force

bullIf we know the potential field this allows us to calculate changes in potential energy for any charge introduced

bullCoulomb Force is thus Electric Field multiplied by charge

bullElectric Field is Coulomb Force divided by test charge

bullD Potential is D Energy divided by test charge

bullD Energy is D Potential multiplied by test charge

0Q

UV

0Q

FE

VQU EF Q

bull102

Potential Difference Near a Point Charge

bull An electric potential exists at some point in an electric field regardless of whether there is a charge at that point

bull The electric potential at a point depends on only two quantities the charge responsible for the electric potential and the distance (r)from this charge to the point in question

bull V = ke qr

bull103

Electric Potential units

bullSI Units of Electric Potential

0Q

UV

EdV

bullUnits are JC

bullAlternatively called Volts (V)

bullWe have seen

dVE bullThus E also has units of Vmbull104

Electric Potential

bull Since the electrical potential energy can change depending on the amount of charge you are moving it is helpful to describe the electrical potential energy per unit charge

bull Electric potential the electrical potential energy associated with a charged particle divided by the charge of the particle

bull A larger charge would involve a larger amount of PEe but the ratio of that energy to the charge is the same as it would be if a smaller charge was in that same place

bull105

bull106

The Voltbull The commonly encountered unit joulescoulomb is

called the volt abbreviated V after the Italian physicist Alessandro Volta (1745 - 1827)

bull With this definition of the volt we can express the units of the electric field as

bull For the remainder of our studies we will use the unit Vm for the electric field

1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E

Potential Difference The Change in Energy per Unit Charge

bull V = PEeqbull The unit for Potential Difference (voltage) is the

voltbull 1 volt is equivalent to one joule per coulombbull As a 1 C charge moves through a potential

difference of 1 V the charge gainsloses 1 J of energy

bull107






bull108

Potential Difference in a Uniform field

E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109

Electric Potential of a single charge

bull+

r bullif V = 0 at rA=

r

QV

1

4 0

bullEbullB

bullAbullIt can be shown that

20

1

4 r

QE

bullRemember that

rEV bullso

bullThis looks a bit like the formulae for the potential in a Uniform Field

||EdVAC bullPotential energy Arbitrary shape

bullPotential difference

bullArbitrary shapebull110

Electric Potential is a scalar field

bullElectric Potential is a scalar field

bullit is defined everywhere

bullbut it does not have any direction

bullit doesnrsquot depend on a charge being there

bull111

bullGeneral Points for either positive or negative charges

bullThe Potential increases if you move in the direction opposite to the electric field

bullandbullThe Potential decreases if you move in the same direction as the electric field

Electric Potential

bull112

Electron-Volts

bull Another unit of energy that is commonly used in atomic and nuclear physics is the electron-volt

bull One electron-volt is defined as the energy a charge-field system gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 voltndash 1 eV = 160 x 10-19 J

bull113

bull114

bull If a charged particle moves perpendicular to electric field lines no work is done

bull If the work done by the electric field is zero then the electric potential must be constant

bull Thus equipotential surfaces and lines must always be perpendicular to the electric field lines

bullif d E

General Considerations

V We

q0 V is constant

Electrical Potential Energy of Two Charges

bull V1 is the electric potential due to q1 at some point P

bull The work required to bring q2 from infinity to P without acceleration is q2V1

bull This work is equal to the potential energy of the two particle system r

qqkVqPE 21

e12

bull115

Electric Potential of Multiple Point Charges

bull Superposition principle appliesbull The total electric potential at some point P due to

several point charges is the algebraic sum of the electric potentials due to the individual chargesndash The algebraic sum is used because potentials are

scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r

bull In the drawing on the right q = 20 μC and d = 096 m Find the total potential at the location P assuming that the potential is zero at infinity

V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample

bullAt locations A and B find the total electric potential

bull119

bull Examplebull The work done by the electric force as the

bull test charge (+20x10-6C) moves from A to

bull B is +50x10-5J

(a)Find the difference in EPE between these

bull points

(b)Determine the potential difference between

bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE

J1005EPEEPE 5 ABAB W

V 25C1020

J10056-

5

o

ABAB q

WVV

bull120

192 The Electric Potential Difference

BAABW EPEEPE bull(a)


bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121

bullWhat is the potential difference between points A and B

bullΔVAB = VB - VA

bulla) ΔVAB gt 0 b) ΔVAB = 0 c) ΔVAB lt 0

bullE

bullA

bullBbullC

bullExample 4bullPoints A B and C lie in a uniform electric field

bullSince points A and B are in the same relative horizontal location in the electric field there is on potential difference between them

bullThe electric field E points in the direction of decreasing potential

bull122

bullE

bullA

bullBbullC

bullPoint C is at a higher potential than point A bullTrue

False


bullAs stated previously the electric field points in the direction of decreasing potential

bullSince point C is further to the right in the electric field and the electric field is pointing to the right point C is at a lower potential

bullThe statement is therefore false

bull123

bullIf a negative charge is moved from point A to point B its electric potential energy

bulla) Increases b) decreases c) doesnrsquot

change

bullE

bullA

bullBbullC


bullThe potential energy of a charge at a location in an electric field is given by the product of the charge and the potential at the location

bullAs shown in Example 4 the potential at points A and B are the same

bullTherefore the electric potential energy also doesnrsquot change

bull124

Units for EnergybullThere is an additional unit that is used for energy in addition to that of joules

bullA particle having the charge of e (16 x 10-19 C) that is moved through a potential difference of 1 Volt has an increase in energy that is given by

eV

joulesqVW

1

1061 19

bull125

Problem

bullShow that the amount of work required to assemble four identical point charges of magnitude Q at the corners of a square of side s is 541keQ2s

1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126

194 Equipotential Surfaces and Their Relation to the Electric Field

bullExample

bullThe plates of the capacitor are separated by

bulla distance of 0032 m and the potential difference

bullbetween them is VB-VA=-64V Between the

bulltwo equipotential surfaces shown in color there

bullis a potential difference of -30V Find the spacing

bullbetween the two colored surfaces

mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127

Electrical Potential Energy in a Uniform Electric Field

bull If a charge is released in a uniform electric field at a constant velocity there is a change in the electrical potential energy associated with the chargersquos new position in the field

bull PEe = -qE d The unit Joules

bull The negative sign indicates that the electrical potential energy will increase if the charge is negative and decrease if the charge is positive bull128

bull129

Potential Energy in 3 charges

r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU

bullEnergy when we bring in Q2

bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU

bullSo finally we find

bullCalculate the work done bringing a 250 mC charge and a -450 mC from infinity to a distance of 60 cm apart

Equation

PE = kq1q2

bull rbullAnswer = 9x109(250x10-6)(-

450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm

Equation bullPE = kq1q2

r

Answer bullPE = -216J

bullPE = 18 J

bullPE = -18 J

bullPE = -216J

bullCalculate the potential difference across the 25 cm path

bullEquation

bullV =Ed bullAnswer bull625 V

bullWhat is the a-particlersquos speed after 25 cm of travel

bullEquation bullqEd = frac12 mv2bullAnswerbull77x104ms

POTENTIAL ENERGY IN A UNIFORM FIELDbullThe Electric Field points in the

direction of a positive test charge

+ Charge - Charge

Along E Loses PEe Gains PEe

Opposite E Gains PEe Loses PEe

bull133

Clicker Question bull In the figure a proton

moves from point i to point f in a uniform electric field directed as shown Does the electric field do positive negative or no work on the proton

A positive B negative C no work is done on the

proton

bull134

ExampleExample Finding the Electric Potential at Point P (apply Finding the Electric Potential at Point P (apply VV==kkeeqqrr))

bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V

bullSuperposition Vp=V1+V2

bullVp=112104 V+(-360103 V)=76103 V

bull135bullT Norah Ali Almoneef

bull136

PROBLEMSbull As a particle moves 10 m along an electric field of

strength 75 NC its electrical potential energy decreases by 48 x10 ndash 16 J What is the particlersquos charge

bull What is the potential difference between the initial and final locations of the particle in the problem above

bull An electron moves 45 m in the direction of an electric field of strength 325 NC Determine the change in electrical potential energy

1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m

q = Must be a + q since PEe was lostPEe = - qEd q = (- 48 x10 -16)(-(75)(10)) =Q = +64 E -19 C2) V = PEeq (-48 x10 -16 J)(64 E -19 C)

V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC

PEe = PEe = - qEd -(-16 x10 -19 C)(325 NC)(45 m) = 23 x10 -16 J

bullElectric potential

bullExample An electron accelerates from rest through an electric potential of 2 V What is the final speed of the electron

bullAnswer The electron acquires 2 eV kinetic energy We have

bulland since the mass of the electron is me = 91 times 10ndash31 kg the speed is

J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary

Electric Potential Energy a point charge moves from i to f in an electric field the change in electric potential energy is

Electric Potential Difference between two points i and f in an electric field

Equipotential surface the points on it all have the same electric potential No work is done while moving charge on it The electric field is always directed perpendicularly to corresponding equipotential surfaces

Finding V from E Potential due to point charges Potential due to a collection of point charges Potential due to a continuous charge distribution Potential of a charged conductor is constant

everywhere inside the conductor and equal to its value to its value at the surface

Calculating E from V Electric potential energy of system of point charges

WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141

bullT Norah Ali Almoneef bull142142

bullThe Potential due to a Point Charge

bull144

bullThe three charges in Fig with q1 = 8 nC q2 = 2 nC and q3 = - 4 nC are separated by distances r2 = 3 cm and r3 = 4 cm How much work is required to move q1 to infinity

bull145

CapacitanceA water tower holds water A capacitor

holds charge(and hence the amount) of the water The

voltage across a capacitor depends on the am The pressure at the base of the water tower depends on the height out of charge held by the capacitor

CapacitanceThe ability of a capacitor to store charge is called

capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)

Cameras use capacitors to supply quick bursts of energy to flash bulbs

CapacitanceWe define capacitance as the amount of

charge stored per volt C = Qstored V

UNITS Farad = Coulomb VoltJust as the capacity of a water tower

depends on the size and shape so the capacitance of a capacitor depends on its size and shape Just as a big water tower can contain more water per foot (or per unit pressure) so a big capacitor can store more charge per volt

Capacitance

Capacitance is measured in farads (F) A one-farad capacitor can store one coulomb of

charge when the voltage across its plates is one volt

One farad is a large amount of capacitance so the microfarad (μF) is frequently used in place of the farad

Parallel Plate Capacitor

How a capacitor works insideThe amount of charge a capacitor can store

depends on several factors

1 The voltage applied to the capacitor2 The insulating ability of the material

between the positive and negative plates3 The area of the two plates (larger areas

can hold more charge)4 The separation distance between the

plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV

bullThe constant of proportionality is called ldquocapacitancerdquo

bullFor a parallel plate capacitor the capacitance is0A

Cd

bullNote this is independent of the

charge and the potential difference

Factors affecting capacitanceSize of the capacitor (A d)Geometric arrangement

PlatesCylinders

Material between conductorsAirPaper

Wax

Coulomb CFarad F

Volt V

ba

QC

V

bullA Farad is a lot of capacitance Typical capacitors are

bullldquomicro nano pico-Farad

CapacitanceWhat happens when a water tower is over-

filled It can break due to the pressure of the water pushing on the walls

What happens when an electric capacitor is ldquoover-filledrdquo or equivalently a higher voltage is placed across the capacitor than the listed maximum voltage It will ldquobreakrdquo by having the charge ldquoescaperdquo This escaping charge is like lightning - a spark that usually destroys the capacitor

Parallel Plate CapacitorFor a parallel plate capacitor we can pull charge from one plate (leaving a Q on

that plate) and deposit it on the other plate (leaving a +Q on that plate) Because of

the charge separation we have a voltage difference between the plates V The

harder we pull (the more voltage across the two plates) the more charge we pull C

= Q V Note that C is NOT CHANGED by either Q or V C relates Q and V

V or V When we deal with height h we usually

refer to the change in height h between the base and the top Sometimes we do

refer to the height as measured from some reference point It is usually clear from

the context whether h refers to an actual h or a h

With voltage the same thing applies We often just use V to really mean V You

should be able to determine whether we really mean V or V when we say V

bullCalculate the capacitance

bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation

bullHow much charge flows from a 12 V battery when connected to a

bull 20 F capacitorbullEquation bullq = CVbullAnswer

bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors

Storing a charge between the plates

bull Electrons on the left plate are attracted toward the positive terminal of the voltage source

bull This leaves an excess of positively charged holes

bull The electrons are pushed toward the right plate

bull Excess electrons leave a negative charge

bull+ bull-

bull+ bull_bull+ bull_

Capacitance of parallel plates

bull+Q bull-Q

bullIntutivelybullThe bigger the plates the more surface area over which the capacitor can store charge C A

bullE

bullMoving plates togeth`er Initially E is constant (no charges moving) thus V=Ed decreases charges flows from battery to

increase V C 1d bull Never Readybull

+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd

ExampleHow strong is the electric field between the

plates of a 080 F air gap capacitor if they are 20 mm apart and each has a charge of 72

C

Example Parallel Plate Capacitor

Consider a parallel plate capacitor made from two plates each 5 cm x 5 cm separated by 2

mm with vacuum in between What is the capacitance of this capacitor

Further if a power supply puts 20 volts across this capacitor what is the amount of charged

stored by this capacitor

Capacitance

bullThe constant of proportionality C is the capacitance which is a property of the conductor

VQ VCQ V

QC

bullExperiments show that the charge in a capacitor is proportional to the electric potential difference (voltage)

between the plates

bull166d

A

V

QC

dA

QEdV

0

0

CapacitanceNote that if we doubled the voltage we

would not do anything to the capacitance Instead we would double the charge stored on the capacitor

However if we try to overfill the capacitor by placing too much voltage across it the positive and negative plates will attract each other so strongly that they will spark across the gap and destroy the capacitor Thus capacitors have a maximum voltage

bull168

Example Thundercloudbull Suppose a thundercloud with horizontal dimensions of 20 km by

30 km hovers over a flat area at an altitude of 500 m and carries a charge of 160 C

bull Question 1ndash What is the potential difference between

the cloud and the groundbull Question 2

ndash Knowing that lightning strikes requireelectric field strengths of approximately25 MVm are these conditions sufficientfor a lightning strike

bull169

Examplebull Question 1bull We can approximate the cloud-ground system as a parallel plate

capacitor whose capacitance is

bull The charge carried by the cloud is 160 C which means that the ldquoplate surfacerdquo facing the earth has a charge of 80 C

bull 720 million volts

C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170

bull Question 2bull We know the potential difference between the cloud and ground

so we can calculate the electric field

bull E is lower than 25 MVm so no lightning cloud to groundndash May have lightning to radio tower or treehellip

E V

d

72 108 V

500 m15 MVm

bull171

A parallel-plate capacitor has an area of 500 cm2 and the plates are separated by a distance of 250 mm


bullEquationbullAnswer C = 18 pFd

AC O

bullT Norah Ali Almoneef

bull172

The two plates of a capacitor hold +5000C and -5000C respectively when the potential difference is 200V What is the capacitance

bullEquation bullAnswe

r

bull25 mF


bull 20 mF capacitorbullEquation q = CV bullAnswer 240 C

V

qC

Electrostaticrdquo ForceElectrostaticrdquo Force



1 22E c

q qF k

r










bull bull7

bullNotesNotes










bullr

bullndash bull+



bulldirections



bullforce doubles









bull11


X

Y

bull12









form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



1 22E c

q qF k

r










bull bull7

bullNotesNotes










bullr

bullndash bull+



bulldirections



bullforce doubles









bull11


X

Y

bull12









form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC






bull bull7

bullNotesNotes










bullr

bullndash bull+



bulldirections



bullforce doubles









bull11


X

Y

bull12









form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullNotesNotes










bullr

bullndash bull+



bulldirections



bullforce doubles









bull11


X

Y

bull12









form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bullr

bullndash bull+



bulldirections



bullforce doubles









bull11


X

Y

bull12









form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bullforce doubles









bull11


X

Y

bull12









form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull11


X

Y

bull12









form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull12









form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC






form

rr

qqkF ˆ

221

2

rrr

rr

r

rrr

12

12

ˆ

bull14

bullx

bully

bullq1

bullq2

bullr2

bullr1

bullr

rr

qqkF ˆ

221

1



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



znF

zF

zF

zF

ynF

yF

yF

yF

xnF

xF

xF

xF

nnet

113121

113121

113121

131211

FFFF




bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bull17

bullExampleExample



bullie huge


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


axis


q2 = 035 C x2 = 040 m



x1 x2



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



x1 x2

bullx3 lt x1

bullx1 lt x3 lt x2

bullx3 gt x2


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


22

21

202

21

)40(

)40(

x

q

x

q

x

qkq

x

qkq o

21 FF

= 012m or =

072m X

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullexample



bullcharges



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bullr = 0193 m)

510 510

bull9

145

1054

103102109145

1

2

559

221

r

rN

r

qkqF

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullExampleExample




bull-8

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

example




bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bullexampleexample



])1015(

1032

)1025(

1097[1085109

22

6

22

669

cF

LeftthetoNF 261


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bull27


bullAnswer 145 N










stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC








stringbull28

mgTFd

kqTF

y

x

cos

sin

lefton Ball

2

2

bullx

bully


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


tan

cos

sin

2

2

22

dkq

mg

mgdkq

TT

bull29





bulld=2 l sin



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bull30



bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bull31




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC




bull35

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull+Q0

Electric Field

rr

E ˆ||4

12

0

Q

0Q

FE



rr

F

E














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC














bullElectric field



0q

FE







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC







Lines of force


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


A) i + jB) i - jC) -jD) -i


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


1E

bullQ1

1r1rbullQ2

2r

2E

i

ii

iQr

rE ˆ

||4

12

0

02202

20012

01

10

00 ˆ

||ˆ

||4

1r

rr

rF

QQQQ

02202

2012

01

1

0

ˆ||

ˆ||4

1r

rr

rE

QQ

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Shark



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bull+

r

E

bull+Q0

bull+Q0

bull+Q0


bull+Q0


test particles


(interactiv

e)

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

EEF Q

EF Q

bull+Q

bull-Q


bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullCh16 bull49



bullE1

bullE2

bull+bullQ1

bull_bullQ2

21

1 r

QkE

22

2 r

QkE

C

Nm

C

C

mN 72

6

2

29 1041

250

101001009

C

Nm

C

C

mN 62

6

2

29 1027

250

10501009

21 EEE CN

CN

CN 767 101210271041

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullCh16 bull50



bullQ1

bull_bullQ2

bull+ bullQ3

21F

23F


NF 7121

NF 7223

223

221 FFF NNN 23)72()71( 22

F

23

21tanF

F

23

211tanF

F o

N

N32

72

71tan 1

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullCh16 bull51

bull+

bullQ1

bull_bullQ2E


q

FE

EeEqF

CNCF 719 10121061 N121043

amF

bull( to left )

m

Fa 2

1831

12

10731019

1043s

mkg

N

bull( to left )


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


r E

r E

r E total

r E net E i

i

r E

r E

r E K

q

r2

r E K

q

r2

r E net E i

i

r E

r E

r E net

x

r E

x

r E

x

Kq

r2 cos Kq

r2 cos

2Kq

r2 cos

2Kq

r2

d

r

2

Kqd

r3

2Kqd

d2 y 2 3

2

r r r d2 y 2

r E

r E K

q

r2

r E total

y

r E

y

r E

y0

r

r

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

The Electric Field

bull53

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullElectric Field








bullRun Away

bull+

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Electric line





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





bullexampleexample



CNr

kqE 18

4

108109 99

2

8730621

61tan 1



2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


2 2 2

cos

sinx

y

x y

A A

A A

A A A


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bull EX =228 NC

bullEnet = 117NC

= 1712

bull

bullexampleexample



Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Example Problem



bullE3


Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Example




Ep = [(36104)2+(288104)2 ] = 461NC

= 141o

2

1










A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC









A It is unaffected




rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


rr

QqF ˆ

4

12

0

0

rr

Q

q

FE ˆ

4

12

00





bullQr

bullqo

bullBbullA

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

nFFFF 002010

n

n

EEE

q

F

q

F

q

F

q

FE

21

0

0

0

02

0

01

0

0

i

i

i

i rr

qE ˆ

4

12

0



y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


y

ar

kqE

2)( yar

kqE


yarar

kqEE

)(

1

)(

1

22

y

ar

karqE ]

)(

4[

222

yr

kqaE

3

4



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bullSolution


bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullExample


bullE1 = E2

bulld = 30cm





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





Electric Field


A) ECgtEBgtEA

B) EBgtECgtEA

C) EAgtECgtEB

D) EBgtEAgtEC

E) EAgtEBgtEC


bullC

bullA

bullB


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


E = 4k

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull74

bull+Q

bull-Q

bull+Q

bull-Q

bull+Q

bull-Q


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bull




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC




rr 2

KQE

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

xA

QE

xA

QE

Ar

rA

2

4

4

4

2

2



Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


Example





a)+A B neutral - C


C)A neutral - B +C

D)Non of the above





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





3342

200 4 RQrEdAE

22r

kQE


bullarea

bulloutside

bullfull charge




bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bullE = keQ r2






1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





1

0

2iE

bull85

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullPhys 133


E inside

0E outside 0





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





bull87






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC






bullPhysics 133



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



q = 12 nC




EC = 49 (10)3 NC


C

Lab 2


VALUES

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull92








Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



Electric Potential

q

UV

q

U

q

U

q

UVVV if

if

sdFdW

sdEqdW

0

sdEqWf

i

0

f

iif sdEqWUUU

0

f

isdE

q

UV

0



Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Joules JVolt V

Coulomb C


b a baba b a

U U WV V V

q q





We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



We have

Then

So



204

1

r

qE

204

1

r

qE

f

i R

f


r

qrV

04

1)(

R

q

r

qdr

r

qV

RR

002

0 4

11

4

1

40


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


Use superposition

For point charges


n

ii

n

i

r

i

rVsdEsdEV

11

n

i i

in

ii r

qVV

101 4

1

bullq bullq






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC






Electric Potential

WU qWV

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull99



bullEquation

bullV = kq

bull rbull


106V



bull r


bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullOctober 3 2007



We have



r

qV 1

04

1

r

qqVqU 21

02 4

1

)(4

1

23

32

13

31

12

21

0231312 r

qq

r

qq

r

qqUUUU

WWapp appif WUUU

WUUU if rEqrFW

bullq1

bullq2

bullq3



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



EdQU 0


EdQ

UV

0


0Q

UV

bull101







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC







0Q

UV

0Q

FE

VQU EF Q

bull102




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC




bull V = ke qr

bull103



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



0Q

UV

EdV

bullUnits are JC


bullWe have seen


Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Electric Potential




bull105

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull106





1 V = 1 J

1 C

m

V

C

Jm

C

N

][

][][

qF

E





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





bull107






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC






bull108


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


E

bull+Q bull+Q

bull+Q

bullA bullB

bullC0|| dFWBC

|||| QEddFWAB

BCABAC WWW

||QEd

||QEdU AC bulld||

||EdVAC

bull109


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bull+


r

QV

1

4 0

bullEbullB


20

1

4 r

QE

bullRemember that

rEV bullso










bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC






bull111




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC




Electric Potential

bull112

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Electron-Volts



bull113

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull114




bullif d E


V We

q0 V is constant





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





qqkVqPE 21

e12

bull115




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC




scalar quantities

bull116

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

0 1 2 3V V V V

bull+Q3

bull+Q2

bull+Q110r

20r

30r

bull0

31 20

10 20 30

kQkQ kQV

r r r

Ni

0i 1 i0

kQV

r


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


V 49

2960

29

2

2

2

10

10103

69

V

d

qKV

d

q

d

q

d

q

d

qKV

d

qK

d

qK

d

qK

d

qKV

bull118

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

V 240

m 600

C1008CmN10998

m 200

C1008CmN10998 82298229

AV

V 0

m 400

C1008CmN10998

m 400

C1008CmN10998 82298229

BV

bullExample


bull119



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bull B is +50x10-5J


bull points


bull these points

BAABW EPEEPE

o

AB

o

A

o

BAB q

W

qqVV

EPEEPE


V 25C1020

J10056-

5

o

ABAB q

WVV

bull120




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC




bull(b) V 25C1020

J10056-

5

o

ABAB q

WVV

bull121


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bullΔVAB = VB - VA


bullE

bullA

bullBbullC




bull122

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullE

bullA

bullBbullC


False





bull123



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



change

bullE

bullA

bullBbullC





bull124



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



eV

joulesqVW

1

1061 19

bull125

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Problem


1 2 3 4

12 13 23 14 24 34

2 2 2

2 2

0

1 10 1 1 1

2 2

24 541

2

e e e

e e

U U U U U

U U U U U U U

k Q k Q k QU

s s s

k Q k QU

s s

bull126


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bullExample







mV1002m 0032

V 64 3

s

VE

m1051mV1002

V 03 33

E

Vs

bull127





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





bull129


r

QV

04

1

bullQ2bullQ1

bullQ3

12

1

02212 4

1

r

QQVQU

12

21

012 4

1

r

QQU

3312 VQUU

231312 UUUU

23

32

13

31

12

21

04

1

r

QQ

r

QQ

r

QQU


bullNow bring in Q3

23

2

13

1

0312 4

1

r

Q

r

QQU



Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


Equation

PE = kq1q2


450x10-6)

bull 60x10-2

bullPE = -17x103J

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull20C bull30 C

-C

bull50 cm

bull30 cm

bull40 cm


r


bullPE = 18 J

bullPE = -18 J

bullPE = -216J


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bullEquation






+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



+ Charge - Charge



bull133




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC




proton

bull134


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bull50 C bull-20 C

V10603)m04()m03(

)C1002()CNm10998(

V10121m04

C1005)CNm10998(

3

22

6229

2

46

2291

V

V


bullVp=112104 V+(-360103 V)=76103 V


bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull136





1) E = 75 NC PEe = - 48 x10 -16 J d = 10 m


V = - 750 V3) q = - 16 x10 -19 C d = 45 m E = 325 NC






J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





J 10 32eV

J 10 16 eV] 2[

2

1 19192

vme

ms 10 48kg 10 91

J 10 322 531ndash

19

v

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullOctober 3 2007

Summary







WUUU if

q

U

q

U

q

UVVV if

if

r

qrV

04

1)(

n

i i

in

ii r

qVV

101 4

1

r

dqdVV

04

1

s

VEs

z

VEz

x

VEx

y

VEy

r

qqVqU 21

02 4

1

f

isdE

q

UV

0

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull141



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



bull144


bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull145





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





capacitance (C)

Charge

(C)

Capacitance

(coulombsvolt)

q = C V

Voltage (volts)






Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC





Capacitance










plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC







plates

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull+Q

bull-Qbulla

bullb

bulldo

AE A

o

E

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull+Q

bull-Qbulla

bullb

bulld

bao o

QdV d

A

baQ V

baQ CV



Cd




PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


PlatesCylinders


Wax

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

Coulomb CFarad F

Volt V

ba

QC

V


















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC
















bullEquation

pFC

C

d

AC

811052

10510858

3

412

0

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bullEquation



bull240 C

FCVV

qC

25200

105000 6

bullq =20F x12V =

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull161

Capacitors






bull+ bull-



bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC


bull+Q bull-Q


bullE



+V

bull162

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull163

bullExample







mV1002m 0032

V 64 3

d

VE

m1051mV1002

V 03 33

E

Vd



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC



C






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC






Capacitance


VQ VCQ V

QC


between the plates

bull166d

A

V

QC

dA

QEdV

0

0




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC




bull168






bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull169





C 0A

d

(885middot10-12 Fm)(2000 m)(3000 m)

500 m011 F

V q

C

80 C

011 F72 108 V

bull++++++++++++bull++++++++++++

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull170




E V

d

72 108 V

500 m15 MVm

bull171




AC O


bull172



r

bull25 mF



V

qC

bull171




AC O


bull172



r

bull25 mF



V

qC


bull172



r

bull25 mF



V

qC