chapter 23 part1.ppt

7/25/2019 chapter 23 part1.ppt

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1Norah Ali Al-moneefking Saud unversity

23.1 Properties of Electric harges23.2 harging !"#ects $y %nduction23.3 oulom"&s 'a(23.) *he Electric +ield

23., Electric +ield 'ines23. otion of harged Particles in a /niformElectric +ield

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Electric harge

*ypes Positive lass ru""ed (ith silk issing electrons

Negative

4u""er0Plastic ru""ed (ith fur E5tra electrons

Ar"itrary choice convention attri"uted to 6

/nits amount of charge is

measured in 7oulom"s8 Empirical !"servations

'ike charges repel /nlike charges attract


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harge in the Atom

Protons 9:; Electrons 9-;

%ons Polar olecules


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23.1 Properties of Electric harges onservationelectricity is the implication that electric charge is al(ays conserved.

< *hat is= (hen one o"#ect is ru""ed against another= charge is notcreated in the process. *he electri>ed state is due to a transfer ofcharge from one o"#ect to the other.

< !ne o"#ect gains some amount of negative charge (hile the othergains an e?ual amount of positive charge.

@uantiation*he smallest unit of charge is that on an electron or proton. 9e B

1., 5 1-1C;

%t is impossi"le to have less charge than this

%t is possi"le to have integer multiples of this charge

Q Ne= )Norah Ali Al-moneef

king Saud unversity10102,


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onductors and %nsulators

onductor transfers charge on contact %nsulator does not transfer charge on contact Semiconductor might transfer charge on

contact

DNorah Ali Al-moneefking Saud unversity

23.2 harging !"#ects $y %nd

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harge *ransfer Processes

onduction Polariation %nduction

,Norah Ali Al-moneefking Saud unversity

10102,


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23-3 oulom"&s 'a( Empirical !"servations

+ormal Statement

1 2F q q2

1Fr

irection of the force is along the line #oining the t(o charges

1 2

12 21221

kq q

F rr=

r

Norah Ali Al-moneefking Saud unversity

10102,


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F

Consider two electric charges: q1and q2 The electric force F between these two

charges separated by a distance r is given by

Coulombs Law

The constant is called Coulombs constantand is given by

221

r

qkqF=

229/CNm109=k

The coulomb constant is also written as

!is the "electric permittivity of vacuum#

$ fundamental constant of nature

2

212

00 Nm

C1085.8where

4

1 ==

k

2

21

0

4

1

r

qqF

=

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C

ou"le one of the charges

force dou"les hange sign of one of the charges

force changes direction

hange sign of bothcharges force stays the same ou"le the distance "et(een

charges force four times (eaker

ou"le "oth charges force four times stronger

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1

%&ample:

'hat is the force between two charges of 1 C separated by 1 meter(

Ans(er F.CC 51CN=

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oulom"&s 'a( E5ample

Ghat is the magnitude of the electricforce of attraction "et(een an ironnucleus 9?B:2,e; and its innermostelectron if the distance "et(een them is

1.D 5 1-12m

*he magnitude of the oulom" force is + B k@1@20r2 B 9C. 5 1CN H m202;92,;91., 5 1I1C;

91., 5 1I1C;091.D51I12m;2 B 2. 5 1I3N.


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12

E5ample - *he JeliumNucleus

Part 1 *he nucleus of a helium atom has t(o protons and

t(o neutrons. Ghat is the magnitude of the electric force"et(een the t(o protons in the helium nucleus6

Ans(er DF N

Part 2: What if the distance is doubled; how will the forcechange?Ans(er 1).D N

)nverse square law: )f the distance is doubled then the force isreduced by a factor of *+

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13

onsider t(o charges located on the 5 a5is

*he charges are descri"ed "y ?1B .1D 5 B . m ?2B .3D 5 B .) m

Ghere do (e need to put a third

charge for that charge to "e at an e?uili"rium point6At the e?uili"rium point= the forces from the t(o charges (ill

cancel.

x1

x2

-Position


10102,

mx

x

qq

kx

qq

k

16.0

)4.0()( 232

2

31

=

=

Jere the forces from ?1and ?2can "alance.

q3

xx

x x4.0


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Zero Resultant Force, Example

The magnitudes of the individual forces will beequal

,irections will be opposite

'ill result in a quadratic Choose the root that gives the forces inopposite directions

Two fi&ed charges- 1mC and ./mC are separated by 1!cm as shown in

the figure 0a where may a third charge be located so that no force acts on

it(

cmx

xx

x

qqk

x

qqk

7.13

)10(

103

)(

101

)10()(

2

6

2

6

2

32

2

31

=

+

=

+

=

10102, 1)Norah Ali Al-moneefking Saud unversity


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10102, Norah Ali Al-moneefking Saud unversity

1D

two charges are located on the positive x-axis of a coordinate system, as shown inthe figure. harge q!"#n is #cm from the origin, and charge q#"-$n is %cm

from the origin. &hat is the total force exerted 'y these two charges on a chargeq$"(n located at the origin)

*he total force on q$is the vector sum of the forces due to q!and q#individually.

*he total force is directed to the left, with magnitude !.%!x!+-%.

Example:


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1,

E5ample - harged Pendulums

onsider t(o identical charged "allshanging from the ceiling "y stringsof e?ual length 1.D m 9ine?uili"rium;. Each "all has a chargeof 2D . *he "alls hang at an angleB 2D(ith respect to the vertical.

Ghat is the mass of the "alls6

Step 1*hree forces act on each"all oulom" force= gravity and the

tension of the string.

mgTF

d

kqTF

y

x

=

=

cos

si

!"e#$o%&""

2

2

5

y

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1

-Pendulums 92;Step 2:The balls are in equilibrium

positions. That means the sum of all

forces acting on the ball is zero

$&

/

cos

si

2

2

22

d

kqmg

mg

dkq

T

T

=

=

Ans(er m B ., kg

A similar analysis applies to the "all on the right.

d=2 l sin

10102,

mgT

d

kqT

=

=

cos

si2

2


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1F

+orce

oulom"&s 'a( that descri"es the electric force and

Ne(ton&s gravitational la( have a similar functionalform

$oth forces vary as the inverse s?uare of the distance "et(een the o"#ects. ravitation is al(ays attractive. k and give the strength of the force.

Felectric = kq

1q2

r2

Felectric = kq

1q2

r2 Fgravity = G

m1m

2

r2

Fgravity = Gm

1m

2

r2

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1C

Example:

n electron is released a'ove the surface of the Earth. second electrondirectly 'elow it exerts an electrostatic forceon the first electron ust greatenough to cancel out the gravitational forceon it. /ow far 'elow the firstelectron is the second)

e

e

mg

Fe

r " )

==

==

=

)8.9)(1011.9(

)106.1()109(

31

2199

21

2

21

x

x

r

mg

qqkrmgr

qqk

mgFE

(.! m

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( )( )

N102.8

m103.5

C1060.1

C

N.m109.

4

1

8

211

219

2

29

2

2

0

=

==

r

eFe

ompare the electrostaticand gravitationalthe forces

( ) ( )

( )N106.3m103.5

k1067.1k1011.9

k

N.m107.6

47

211

2731

2

211

2'

=

==

gg

gr

mmGF

pe

Fe/Fg" # x !+$0*he force of gravity is much wea1er than the electrostatic force

*he electron and proton of a hydrogen atom are separated 2on the average3 'y a

distance of approximately (.$ x!+-!! m. Find the magnitudes of the electric force

and the gravitational force 'etween the two particles.

10102, 2Norah Ali Al-moneefking Saud unversity


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2110102,


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Electric +orces and KectorsElectric +ields and +orces are A'' vectors= thus

all rules applying to vectors must "e follo(ed.onsider three point charges, 4!" 5.++ x!+-0 2located at the origin3,4$"(.++x!+-0, and 4#" -#.++x!+-0, located at the corners of a R67/* triangle. 4#is located at y" $ m while 4$is located %m to the right of 4#. Find the resultantforce on 4$.

4!

4# 4$

$m

%m

(m

4$

&hich way does 4#push 4$)&hich way does 4!push 4$)

Fon $ due to #

Fon $ due to !

" $8

" tan-!2$9%3



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=

=

23

2

999

234

)102)(100.5()1099.8(

F

xxxF

=

=

13

2

99

913

5)105)(106()1099.8(

F

xxxF

===

=

+=

==

=

=

)($&

)()(F

1062.6)37si(

1018.3

)37cos(

1

22

res"$&$

9

13

9

2313

x

y

res

yx

y

x

x

F

FDirection

F

FF

NxFF

NxF

FFF

4!

4# 4$

$m

%m

(m

4$

Fon $ due to #

Fon $ due to !

" $8" tan-!2$9%3

5.6 x10-9N

1.1x10-8N

F$,!cos$8

F$,!sin$8

7.34x10-9N

2*+/ !above the 3&



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oulom"&s 'a( E5ample 4 5 2+! mC

L 5 !+1! m 'hat is the magnitude and

direction of the net force on one ofthe charges(

F1

+

F2

F3

Q

x

y

L

++

+

L

QQ

Q

2)Norah Ali Al-moneefking Saud unversity

&e find the magnitudes of the individual forces on the charge at the upperright corner:

F!" F#" kQQ9L#" kQ#9L#" 20 x!+0 ; m#9#325 x!+$3#92+.!++ m3# " $.#% x!+8 .F$" kQQ92L


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2D

along the diagonal, or away from the center of the s4uare.

From the symmetry, each of the other forces will have the samemagnitude and a direction away from the center: *he net force oneach charge is" 5.#+ =!+8 away from the center of the s4uare.

.

( ) ( )

45$&

102.6107385.4107385.42 2

10385.4101.1451024.3

10385.4101.1451024.3

101.145707.0101.62si

101.145707.0101.62cos

45so1

*

*$&

0

01

7

777

777

77

33+

77

33,

0

F

F

45so1*

*$&

==

=+=

=+=

=+=

====

====

===

===

F

F

F

F

F

F

x

y

y

x

NF

N

N

N

N

10102,

E ample +our


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2,

E5ample - +ourhargesonsider four charges placed at the

corners of a s?uare (ith sides of

length 1.2D m as sho(n on theright. Ghat is the magnitude of theelectric force on ?)resulting from

the electric force from the

remaining three charges6

Ans(er

+ 9on ?); B .C1, N

L and the direction6

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#$-8> *hree point charges are located at the corners of an e4uilateraltriangle. alculate the net electric force on the 8.++ u charge.

HOMEWORK :

#$-?: *wo small 'eads having positive charges $q and q are fixed at theopposite ends of a hori@ontal insulating rod extending from the origin to thepointx =d.a third small charged 'ead is free to slide on the rod. t whatposition is the third 'ead in e4uili'rium) an it 'e in sta'le e4uili'rium)



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#$-!#> n o'ect having a net charge of #%.+ is placed in a uniform electricfield of 5!+ 9 that is directed vertically. &hat is the mass of this o'ect if itAfloatsB in the field)

$-!?> *wo #.++u point charges are located on thex axis. Cne is atx =!.++m, and the other is atx =-!.++ m. 2a3 Determine the electric field on the yaxis at y =+.(++ m. 2'3 alculate the electric force on a - $.++u chargeplaced on the y axis at y =+.(++ m.

#$-%!> n electron and a proton are each placed at rest in an electric field of(#+ 9. alculate the speed of each particle %?.+ ns after 'eing released.

#$-%%> *he electrons in a particle 'eam each have a 1inetic energy of !.5+ x !+-!8. &hat are the magnitude and direction of the electric field that stops theseelectrons in a distance of !+.+ cm)

10102, 2FNorah Ali Al-moneefking Saud unversity

chapter 23 part1.ppt

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