physics for scientists and engineers ii, summer semester 2009 physics 2220 physics for scientists...

Physics for Scientists and Engineers II , Summer Semester 2009

Physics 2220

Physics for Scientists and Engineers II


Chapter 23: Electric Fields

• Materials can be electrically charged.• Two types of charges exist: “Positive” and “Negative”.• Objects that are “charged” either have a net “positive” or a net “negative”

charge residing on them.• Two objects with like charges (both positively or both negatively charged)

repel each other.• Two objects with unlike charges (one positively and the other negatively

charged) attract each other.• Electrical charge is quantized (occurs in integer multiples of a fundamental

charge “e”).q = N e (where N is an integer)electrons have a charge q = - e protons have a charge q = + eneutrons have no charge


Material Classification According to Electrical Conductivity

• Electrical conductors: Some electrons (the “free” electrons) can move easily through the material.

• Electrical insulators: All electrons are bound to atoms and cannot move freely through the material.

• Semiconductors: Electrical conductivity can be changed over several orders of magnitude by “doping” the material with small quantities of certain atoms, making them more or less like conductors/insulators.


Shifting Charges in a Conductor by “Induction”

+

++

+

+

-

-

-

-

-

---

Negatively charged rod

uncharged metal sphere

+

++

+

+ --

-

-

-

---

Left side of metal spheremore positively charged

Right side of metal spheremore negatively charged


Coulomb’s Law (Charles Coulomb 1736-1806)

Magnitude of force between two “point charges” q1 and q2 .

2

21

r

qqkF ee

2

212

2

29

102854.8

4

1106987.8

mN

Cxwhere

C

Nmxk

o

oe

r = distance betweenpoint charges

Coulomb constant

Permittivity of free space


Charge

Unit of charge = Coulomb

Smallest unit of free charge: e = 1.602 18 x 10-19 C

Charge of an electron: qelectron = - e = - 1.602 18 x 10-19 C


Vector Form of Coulomb’s Law

Force is a vector quantity (has magnitude and direction).

12221

12 r̂r

qqkF e

unit vector pointing fromcharge q1 to charge q2

Force exerted by charge q1 on charge q2

(force experienced by charge q2 ).


Vector Form of Coulomb’s Law

Force is a vector quantity (has magnitude and direction).

1212221

21221

21 ˆˆ Frr

qqkr

r

qqkF ee

unit vector pointing fromcharge q2 to charge q1

Force exerted by charge q2 on charge q1

(force experienced by charge q1 ).


Directions of forces and unit vectors

+

+

q1

q212r̂

21F

12F

+

-

q1

q221F

12F

21r̂


Calculating the Resultant Forces on Charge q1 in a Configuration of 3 charges

q3

-

+ +

q1 q2a = 1cm

q1 = q2 = +2.0 C

q3 = - 2.0 C

0.5 cm 0.5 cm


Forces acting on q1

q3

-

+ +

q1 q2

21F31F

31211 FFF Total force on q1:


Magnitude of the Various Forces on q1

Nm

C

C

Nm

m

CCkF e

224

212

2

29

221 10596.3100.1

100.41099.8

)010.0(

0.20.2

Nm

C

C

Nm

m

CCkF e

224

212

2

29

231 10192.7105.0

100.41099.8

)25.0010.0(

0.20.2

Note: I am temporarily carrying along extra significant digits in theseintermediate results to avoid rounding errors in the final result.


Adding the Vectors Using a Coordinate System

q3

-

+ +

q1 q2

21F31F

y

x


Adding the Vectors Using a Coordinate System

21F31F

y

x

jiFF ˆ0ˆ2121

jFiFF ˆ2

2ˆ2

2313131

jFiFF

FFF

ˆ2

2ˆ2

2313121

31211


…doing the algebra…

jNiN

jNiNN

jFiFF

FFF

ˆ101.5ˆ105.1

ˆ10086.5ˆ10086.510596.3

ˆ2

2ˆ2

2

22

222

315121

31211

F1 has a magnitude of

NF 21 103.5


Calculating the force on q2 … another example using an even more mathematical approach

q1 = +3.0 C

q2 = - 4.0 C

Charges Location of charges

x1=3.0cm ; y1=2.0cm ; z1=5.0cm

x2=2.0cm ; y2=6.0cm ; z2=2.0cm

In this example, the location of the charges and the distancebetween the charges are harder to visualize Use a more mathematical approach!



12212

21122 r̂

d

qqkFF e

d12=distance between q1 and q2.

.q toq from pointingr unit vecto ˆ 2112 r


Calculating the force on q2 … mathematical approach

We need the distance between the charges. d12 is distance between q1 and q2.

+

x

y

z

1r -

q1

q2

2r

12 rr )()( 121212 rrrrd


cm

cm

cm

cm

cm

cm

cm

cm

cm

rr

0.3

0.4

0.1

0.5

0.2

0.3

0.2

0.6

0.2

12

cmcmcmcm

cm

cm

cm

cm

cm

cm

rrrrd

260.90.160.1

0.3

0.4

0.1

0.3

0.4

0.1

)()(

222

121212

Distance between charges q1 and q2 .




We need the unit vectors between charges. For example, the unitvector pointing from q1 to q2 is easily obtained by normalizing the vector pointing from from q1 to q2.

+

x

y

z

1r -

q1

q2

2r

12 rr

)()( 121212 rrrrd

12

1212ˆ

d

rrr

12r̂



0.3

0.4

0.1

14

1

0.3

0.4

0.1

26

1ˆ

12

1212

cm

cm

cm

cmd

rrr

The needed unit vector:



126

26

26

9

26

16

26

1ˆ ofLength

263

264

261

263

264

261

12

r

You can easily verify that the length of the unit vector is “1”.



NN

N

m

cm

cm

C

C

Nm

cm

CCk

rd

qqkFF

e

e

2

2

2

212

2

29

2

12212

21122

10

2.1

7.1

42.0

44.124

92.165

48.41

0.3

0.4

0.1

48.41

0.3

0.4

0.1

1

10010462.0109876.8

0.3

0.4

0.1

26

1

26

)0.4(0.3

ˆ



N

NNN

N

N

N

N

N

N

FFF

2

222

222

101.2

15485275291721

44.124

92.165

48.41

44.124

92.165

48.41

…and if you want to know just the magnitude of the force on q2 :


.qby divided q charge test aon acting force theas defined is

vector field electric The

:definedFaraday Similarly,

ooe

o

e

FE

q

FE

23.4 The Electric Field

mass.by that divided m mass of particle test aon acting force theas defined is

m. mass theoflocation at the field nalgravitatio The

:2210 Physics fromRemember

g

g

Fg

m

Fg

It is convenient to use positive test charges. Then, the direction ofthe electric force on the test charge is the same as that of the fieldvector. Confusion is avoided.



C

N : field electric of units SI E

again. removed is charge test the

once exists that field actual theis measured field The alone. Q charge source theof field electric the todue

solely is charge test on the force theTherefore, charge. test by the produced field the todue charge testthe

on acting force no is thereHowever, field. electric additionalan produces also charge test thecourse, Of

Q. charge sourceby produced detectsonly charge test The

itself. charge test by the produced NOT is measured The :Note

E

E

+ +

+ ++ +

+ +

+

Source charge

test charge

Q qo E



point.at that placed particle chargedany on force

thecalculatecan you space,in point someat knowyou Once

.directions oppositein are and : q charge negativeFor

. asdirection same in the is : q charge positiveFor

: field electrican in q charge aon Force

E

EqF

EeF

EeF

e


23.4 The Electric Field of a “Point Charge” q

.q toq from pointingr unit vecto theis r̂

and q and qbetween distance theisr where

ˆE:q of place at the qby created field Electric

ˆ:qby q chargeon test exerted Force

0

0

20

0

20

0

rr

qk

q

F

rr

qqkF

ee

ee

q

q0rr̂


23.4 The Electric Field of a Positive “Point Charge” q

+

q0

(Assuming positive test charge q0)

eF

The electric field of a positive point charge points away from it.

+

EP

Force on test charge

Electric field where test chargeused to be (at point P).


23.4 The Electric Field of a Negative “Point Charge” q

-

q0

(Assuming positive test charge q0)

eF

The electric field of a negative point charge points towards it.

-

EP

Force on test charge

Electric field where test chargeused to be (at point P).


23.4 The Electric Field of a Collection of Point Charges

i

i

i

e rr

qk ˆE 2


23.4 The Electric Field of Two Point Charges at Point P

222

212

1

12

ˆˆˆE rr

qkr

r

qkr

r

qk ee

ii

i

e

x

y

P

q1 q2

a b

?E

y



2222

1221 ˆˆE r

yb

qkr

ya

qk ee

x

y

P

q1 q2

r2r1

a b

y 2221 yar

2222 ybr

Pythagoras:



x

y

P

q1 q2

1̂r 2̂r

1C 2C

R

y

a

yay

a

r

a

yrCR

rr

rr p

2211

1

11

11

11

0

011ˆ

pr1pr1

y

b

yby

b

r

b

yrCR

rr

rr p

2211

2

22

22

11

0

011ˆ

y

b

ybyb

qk

y

a

yaya

qk ee 2222

2

22221 11

E

y

0R

0C1

a

0C2

b



y

b

yb

qk

y

a

ya

qk ee

23

22

2

23

22

1E

23

22

2

23

22

1Eyb

bqk

ya

aqk eex

23

22

2

23

22

1Eyb

yqk

ya

yqk eey



23

2223

2223

2223

22

2

23

22

1 2E

ya

qak

ya

qak

ya

qak

yb

bqk

ya

aqk eeeeex

0E

23

2223

2223

22

2

23

22

1

ya

qyk

ya

qyk

yb

yqk

ya

yqk eeeey

Special case: q1= q and q2 = -q AND b = a

q -q+ -

E from + charge

E from - charge

component)-y no (has total E



0E

23

2223

2223

22

2

23

22

1

ya

qak

ya

qak

yb

bqk

ya

aqk eeeex

23

2223

2223

2223

22

2

23

22

1 2E

ya

qyk

ya

qyk

ya

qyk

yb

yqk

ya

yqk eeeeey

Special case: q1= q and q2 = q AND b = a

q q+ +

E from + chargeE from other + charge

component)- xno (has total E


This is called an electric DIPOLE

23

22

2E

ya

qakex

0E y

Special case: q1= q and q2 = -q AND b = a

q -q+ -

E from + charge

E from - charge

component)-y no (has total E

For large distances y (far away from the dipole), y >> a:

32

32

22E

y

qak

y

qak eex E falls off proportional to 1/y3

Fall of faster than field of single charge (only prop. to 1/r2).From a distance the two opposite charges look like they arealmost at the same place and neutralize each other.

physics for scientists and engineers ii, summer semester 2009 physics 2220 physics for scientists...

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