Solutions to Chapter26

8/10/2019 Solutions to Chapter26

1/24

26Capacitance and Dielectrics

CHAPTER OUTLINE

26.1 Definition of Capacitance

26.2 Calculating Capacitance

26.3 Combinations of Capacitors

26.4 Energy Stored in a Charged

Capacitor

26.5 Capacitors with Dielectrics

26.6 Electric Dipole in an Electric Field

26.7 An Atomic Description of

Dielectrics

ANSWERS TO QUESTIONS

*Q26.1 Volume is proportional to radius cubed. Increasing the

radius by a factor of 31/3will triple the volume. Capaci-

tance is proportional to radius, so it increases by a factor

of 31/3. Answer (d).

Q26.2 Seventeen combinations:

Individual C C C1 2 3, ,

Parallel C C C C C C C C C 1 2 3 1 2 1 3 2 3+ + + + +, , ,

Series-Parallel1 1

1 2

1

3C C

C+

+

,1 1

1 3

1

2C C

C+

+

,1 1

2 3

1

1C C

C+

+

1 1

1 2 3

1

C C C+ +

,1 1

1 3 2

1

C C C+ +

,1 1

2 3 1

1

C C C+ +

Series

1 1 1

1 2 3

1

C C C+ +

,

1 1

1 2

1

C C+

,

1 1

2 3

1

C C+

,

1 1

1 3

1

C C+

*Q26.3 We find the capacitance, voltage, charge, and energy for each capacitor.

(a) C=20 F V= 4 V Q = CV= 80 C U= (1/2)QV= 160 J

(b) C=30 F V= Q/C= 3 V Q = 90 C U= 135 J

(c) C=Q/V=40 F V= 2 V Q = 80 C U= 80 J

(d) C=10 F V= (2U/C)1/2= 5 V Q = 50 C U= 125 J

(e) C=2U/V2= 5 F V= 10 V Q = 50 C U= 250 J

(f) C=Q2/2U= 20 F V= 6 V Q = 120 C U= 360 J

Then (i) the ranking by capacitance is c > b > a = f > d > e.

(ii) The ranking by voltage Vis e > f > d > a > b > c.

(iii) The ranking by charge Qis f > b > a = c > d = e.

(iv) The ranking by energy Uis f > e > a > b > d > c.

Q26.4 A capacitor stores energy in the electric field between the plates. This is most easily seen when

using a dissectible capacitor. If the capacitor is charged, carefully pull it apart into its compo-

nent pieces. One will find that very little residual charge remains on each plate. When reassem-

bled, the capacitor is suddenly rechargedby inductiondue to the electric field set up and

stored in the dielectric. This proves to be an instructive classroom demonstration, especially

when you ask a student to reconstruct the capacitor without supplying him/her with any rubber

gloves or other insulating material. (Of course, this is afterthey sign a liability waiver.)

71


2/24

72 Chapter 26

*Q26.5 (i) Answer (b). (ii) Answer (c). (iii) Answer (c). (iv) Answer (a). (v) Answer (a).

*Q26.6 The charge stays constant as Cis cut in half, so U = Q2/2Cdoubles: answer (b).

*Q26.7 (i) According to Q = CV, the answer is (b).

(ii) From U =(1/2)CV2, the answer is (a).

Q26.8 The work done, W Q V= , is the work done by an external agent, like a battery, to move a chargethrough a potential difference, V. To determine the energy in a charged capacitor, we must addthe work done to move bits of charge from one plate to the other. Initially, there is no potential

difference between the plates of an uncharged capacitor. As more charge is transferred from one

plate to the other, the potential difference increases as shown in the textbook graph

of Vversus Q, meaning that more work is needed to transfer each additional bit of charge.

The total work is the area under the curve on this graph, and thus W Q V=1

2 .

*Q26.9 Let C=the capacitance of an individual capacitor, and Cs represent the equivalent capacitance of

the group in series. While being charged in parallel, each capacitor receives charge

Q C V= = ( )( )= charge F V C5 00 10 800 0 4004. .

While being discharged in series, V Q

C

Q

Csdischarge

C

5.00 10 F= = =

=

10

0 4008 0

5

.. 00 kV

(or 10 times the original voltage). Answer (b).

Q26.10 The potential difference must decrease. Since there is no external power supply, the charge on

the capacitor, Q, will remain constantthat is, assuming that the resistance of the meter is

sufficiently large. Adding a dielectric increasesthe capacitance, which must therefore decrease

the potential difference between the plates.

*Q26.11 (i) Answer (a). (ii) Because Vis constant, Q =CVincreases, answer (a).

(iii) Answer (c). (iv) Answer (c). (v) U= (1/2)CV2increases, answer (a).

Q26.12 Put a material with higher dielectric strength between the plates, or evacuate the space between

the plates. At very high voltages, you may want to cool off the plates or choose to make them of a

different chemically stable material, because atoms in the plates themselves can ionize, showing

thermionic emissionunder high electric fields.

Q26.13 The primary choice would be the dielectric. You would want to choose a dielectric that has a large

dielectric constant and dielectric strength, such as strontium titanate, where 233 (Table 26.1).A convenient choice could be thick plastic or Mylar. Secondly, geometry would be a factor. To maxi-

mize capacitance, one would want the individual plates as close as possible, since the capacitance is

proportional to the inverse of the plate separationhence the need for a dielectric with a high dielectric

strength. Also, one would want to build, instead of a single parallel plate capacitor, several capacitors

in parallel. This could be achieved through stacking the plates of the capacitor. For example, you can

alternately lay down sheets of a conducting material, such as aluminum foil, sandwiched between yoursheets of insulating dielectric. Making sure that none of the conducting sheets are in contact with their

next neighbors, connect every other plate together. Figure Q26.13 illustrates this idea.

continued on next page


3/24

Capacitance and Dielectrics 73

ConductorConductor

Dielectric

FIG. Q26.13

This technique is often used when home-brewing signal capacitors for radio applications, as

they can withstand huge potential differences without flashover (without either discharge between

plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich

together flexible materials such as aluminum roof flashing and thick plastic, so the whole product

can be rolled up into a capacitor burrito and placed in an insulating tube, such as a PVC pipe, and

then filled with motor oil (again to prevent flashover).

SOLUTIONS TO PROBLEMS

Section 26.1 Definition of Capacitance

P26.1 (a) Q C V= = ( )( )= = 4 00 10 12 0 4 80 10 486 5. . . .F V C 00 C

(b) Q C V= = ( )( )= = 4 00 10 1 50 6 00 10 6 06 6. . . .F V C 00 C

P26.2 (a) C Q

V= =

= =

10 0 10

1 00 10 1 006

6. . .C

10.0 VF F

(b) V Q

C= =

=

100 10

10100

6

6

C

1.00 FV

Section 26.2 Calculating Capacitance

P26.3 (a) V Ed= so E=

=20 0

1011 1

3

..

V

1.80 m kV m toward the negative plate

(b) E=

0

so = ( ) ( )=1 11 10 8 85 10 98 34 12. . .N C C N m nC m2 2 22

(c) C A

d= =

( )( )0128 85 10 7 60 1 00. . .C N m cm m 12 2 2 000 cm

mpF

( )

=

2

31 80 103 74

..

(d) V Q

C= so Q=( ) ( )=20 0 3 74 10 74 712. . .V F pC


4/24

P26.4 With = , the plates are out of mesh and the overlap area is zero.

With = 0, the overlap area is that of a semi-circle,R2

2. By proportion,

the effective area of a single sheet of charge is ( )R2

2.

When there are two plates in each comb, the number of adjoining

sheets of positive and negative charge is 3, as shown in the sketch. When

there areNplates on each comb, the number of parallel capacitors

is 2 1N and the total capacitance is

C N A N R

= ( )

= ( ) ( )

2 12 10 0effective

distance

22 022

2

2 1

d

N R

d=

( ) ( ) .

P26.5 E k q

r

e=2

: q= ( )( )

( ) =

4 90 10 0 210

8 99 10

4 2

9

. .

.

N C m

N m C2 200 240. C

(a) = =

( ) =

q

A

0 240 10

4 0 120 1 33

6

2

.

. . C m2

(b) C r= = ( )( )=4 4 8 85 10 0 120 13 3012 . . . pF

P26.6 C A

d= =

( ) ( ) ( ) 012 3 21 00 8 85 10 1 00 10. . .C m

N

2

( ) =

m mnF

2 80011 1.

The potential between ground and cloud is

V Ed

Q C V

= = ( )( )=

= (

3 00 10 800 2 40 106 9. .N C m V

))= ( ) ( )=11 1 10 2 40 10 26 69 9. . .C V V C

P26.7 Q A

dV=

( )0

Q

A

V

d= =

( ) 0

d V

= ( )

= ( )( )

0

128 85 10 150

30 0 10

.

.

C N m V2 2

( ) ( )=

9 41 00 104 42

C cm cm m m

2 2 2..

P26.8 Fy = 0: T mgcos = 0 Fx = 0: T Eqsin = 0

Dividing, tan=Eq

mg

so E mg

q= tan

and V Ed mgd

q= =

tan

FIG. P26.4

74 Chapter 26


5/24


P26.9 (a) Ck b ae

=( )

=( ) ( )

=

2

50 0

2 8 99 10 7 27 2 589ln /

.

. ln . / .22 68. nF

(b) Method 1: V k b

ae

=

2 ln

= =

=

=

q

V

8 10 10

1 62 10

2 8 99

67.

.

.

C

50.0 m C m

( ) ( )

=10 1 62 10

7 27

2 583 029 7. ln

.

.. kV

Method 2: V Q

C= =

=

8 10 10

2 68 103 02

6

9

.

.. kV

*P26.10 The original kinetic energy of the particle is

K m= = ( ) ( ) = 1

2

1

22 10 2 10 4 00 102 16 6

2 4v kg m s J.

The potential difference across the capacitor is V Q

C= = =

1000100

C

10 FV.

For the particle to reach the negative plate, the particle-capacitor system would need energy

U q V= = ( ) ( )= 3 10 100 3 00 106 4C V J.

Since its original kinetic energy is greater than this, the particle will reach the negative plate .

As the particle moves, the system keeps constant total energy

K U K U +( ) = +( )+ at plate at plate:

4 00 10 3 10 1001

22 104 6 16. + ( ) +( )= ( ) J C V vvf

2 0+

vf

= ( )

=

2 1 00 10

2 101 00 10

4

16

6.

.J

kgm s

P26.11 (a) C ab

k b ae= ( )= ( )( )

( ) 0 070 0 0 140

8 99 10 0 1409. .

. . 00 070 0 15 6. .( )= pF

(b) C Q

V=

V

Q

C= =

=

4 00 10

15 6 10256

6

12

.

.

C

FkV

Section 26.3 Combinations of Capacitors

P26.12 (a) Capacitors in parallel add. Thus, the equivalent capacitor has a value of

C C Ceq

= + = + =1 2

5 00 12 0 17 0. . .F F F

(b) The potential difference across each branch is the same and equal to the voltage of the battery.

V= 9 00. V

(c) Q C V5 5 00 9 00 45 0= =( )( )= . . .F V C

and Q C V12 12 0 9 00 108= =( )( )= . .F V C


6/24

76 Chapter 26

P26.13 (a) In series capacitors add as1 1 1 1

5 00

1

12 01 2C C Ceq= + = +

. .F F

and Ceq= 3 53. F

(c) The charge on the equivalent capacitor is Q C Veq eq= =( )( )= 3 53 9 00 31 8. . .F V C

Each of the series capacitors has this same charge on it.

So Q Q1 2 31 8= = . C

(b) The potential difference across each is V Q

C1

1

1

31 86 35= = =

..

C

5.00 FV

and V Q

C2

2

2

31 82 65= = =

..

C

12.0 FV

P26.14 (a) In series , to reduce the effective capacitance:

1

32 0

1

34 8

1

1

2 51 10398

3

. .

.

F F

F F

= +

=

=

C

C

s

s

(b) In parallel , to increase the total capacitance:

29 8 32 0

2 20

. .

.

F F

F

+ =

=

C

C

p

p

P26.15 C C Cp= +1 2 1 1 1

1 2C C Cs= +

Substitute C C Cp2 1= 1 1 1

1 1

1 1

1 1C C C C

C C C

C C Cs p

p

p

= +

= +

( )

Simplifying, C C C C C p p s12

1 0 + =

CC C C C

C C C C p p p s

p p p s1

2

24

2

1

2

1

4=

=

We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus

sign, we would get the same two answers with their names interchanged.)

C C C C C p p p s12 21

2

1

4

1

29 00

1

49 00= + = ( ) + ( ). .pF pF ( )( )=

= =

9 00 2 00 6 00

1

2

12 1

. . .pF pF pF

C C C C p p44

1

29 00 1 50 3 002C C Cp p s = ( ) =. . .pF pF pF


7/24


P26.16 C C Cp= +1 2

and1 1 1

1 2C C Cs= + .

Substitute C C Cp2 1= :1 1 1

1 1

1 1

1 1C C C C

C C C

C C Cs p

p

p

= +

= +

( ) Simplifying, C C C C C p p s1

2

1 0 + =

and CC C C C

C C C C p p p s

p p p s1

2

24

2

1

2

1

4=

= +

where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values

for the capacitances, with the names reversed).

Then, from C C Cp2 1=

C C C C C p p p s2 212

14=

P26.17 (a)1 1

15 0

1

3 00Cs= +

. .

C

C

C

s

p

eq

== + =

= +

2 50

2 50 6 00 8 50

1

8 50

.

. . .

.

F

F

F

11

20 05 96

1

..

F F

=

(b) Q C V= =( )( )= 5 96 15 0 89 5. . .F V C on 20 0. F

V Q

C= = =

=

89 54 47

15 0 4 47 10 53

..

. . .

C

20.0 FV

VV

F V C on 6.00Q C V= =( )( )= 6 00 10 53 63 2. . . F

89 5 63 2 26 3. . . = Con 15 0. F and 3 00. F

FIG. P26.17


8/24

78 Chapter 26

P26.18 (a) Capacitors 2 and 3 are in parallel and present equivalent capacitance 6C. This is in

series with capacitor 1, so the battery sees capacitance1

3

1

62

1

C CC+

=

.

(b) If they were initially uncharged, C1stores the same charge as C2 and C3together.

With greater capacitance, C3

stores more charge than C2

. Then Q Q Q1 3 2

> > .

(c) The C C2 3||( )equivalent capacitor stores the same charge as C1. Since it has greater

capacitance, V Q

C= implies that it has smaller potential difference across it than C1.

In parallel with each other, C2and C3have equal voltages: V V V1 2 3> = .

(d) If C3 is increased, the overall equivalent capacitance increases. More charge moves

through the battery and Qincreases. As V1increases, V2must decrease so Q2decreases.

Then Q3must increase even more: Q Q Q3 1 2and increase; decreases .

P26.19 C Q

V=

so 6 00 10

20 0

6..

= Q

and Q= 120 C

Q Q1 2120= C

and V Q

C= : 120 2

1

2

2

=

Q

C

Q

C

or120

6 00 3 00

2 2 =Q Q

. .

3 00 120 6 002 2. .( ) ( )=( )Q Q

Q2 3609 00

40 0= =.

. C

Q1 120 40 0 80 0= =C C C . .

P26.20 According to the suggestion, the combination

of capacitors shown is equivalent to

Then1 1 1 1

0 0 0

0 0 0

0 0

C C C C C

C C C C C

C C C

= ++

+

= + + + +

+( )

C C C C C C

C C C C

CC C

0 0

2 2

0

2

0 0

2

0 0

2

2 3

2 2 0

2 4 4

+ = +

+ =

= + 22

4

0

2C( )

Only the positive root is physical.

C C= ( )0

23 1

FIG. P26.19

FIG. P26.20


9/24


P26.21 C

C

s

p

= +

=

= ( ) +

1

5 00

1

10 03 33

2 3 33 2

1

1

. ..

.

F

.. .

. .

.

00 8 66

2 10 0 20 0

1

8 66

1

2

2

=

= ( )=

= +

F

F

C

C

p

eq00 0 6 04

1

. .

=

F

P26.22 Q C Veq eq= ( )= ( )( )= 6 04 10 60 0 3 62 106 4. . .F V C

Q Qp eq1= , so VQ

Cp

eq

p

1

1

4

6

3 62 10

8 66 1041 8= =

=

.

..

C

FV

Q C Vp3 3 162 00 10 41 8= ( )= ( )( )= . .F V 83.6 C

P26.23 nCn C

C C Cn

=+ + +

=100 100

1 1 1

capacitors

nC C

n=

100 so n2 100= and n= 10

Section 26.4 Energy Stored in a Charged Capacitor

P26.24 U C V= ( )1

2

2

The circuit diagram is shown at the right.

(a) C C Cp= + = + =1 2 25 0 5 00 30 0. . .F F F

U= ( )( ) =1

230 0 10 100 0 1506

2. . J

(b) CC C

s= +

= +

1 1 1

25 0

1

5 001 2

1

. .F F

11

4 17= . F

U C V

V U

C

= ( )

= = ( )

=

1

2

2 2 0 150

4 17 10268

2

6

.

.V

FIG. P26.21

FIG. P26.24


10/24

80 Chapter 26

P26.25 W U Fdx = =

so F dU

dx

d

dx

Q

C

d

dx

Q x

A

Q= =

=

=

2 2

0

2

02 2 2 AA

P26.26 With switch closed, distance =d d0 500. and capacitance

= = =C A

dA

dC0 02 2 .

(a) Q C V C V = ( )= ( )= ( )( )= 2 2 2 00 10 100 4006. F V CC

(b) The force stretching out one spring is

F Q

A

C V

A

C V

A d d

C V=

=

( )

=

( )

( ) =

(2

0

2 2

0

2 2

02

4

2

2 2 ))2

d

One spring stretches by distance x d=

4, so

k Fx

C Vd d

C Vd= = ( ) = ( ) = (

2 4 88 2 00 102 2

2

6

. F

))( )

( ) =100

8 00 102 50

2

3 2

V

mkN m

..

P26.27 (a) U C V= ( ) = ( )( ) =1

2

1

23 00 12 0 216

2 2 . .F V J

(b) U C V= ( ) = ( )( ) =1

2

1

23 00 6 00 54 0

2 2 . . .F V J

P26.28 U C V=1

2

2

V UC

= = ( )

= 2 2 30030 10

4 47 106

3J C V

V.

P26.29 U C V= ( )1

2

2 where C R R

ke= =4 0 and V

k Q

R

k Q

R

e e= =0

U R

k

k Q

R

k Q

Re

e e=

=1

2 2

2 2


11/24


P26.30 (a) The total energy is U U U q

C

q

C

q

R

Q q= + = + =

+

(1 2

1

2

1

2

2

2

1

2

0 1

11

2

1

2

1

2 4

1

2

))

2

0

24 R.

For a minimum we setdU

dq10= :

1

2

2

4

1

2

2

4 1 01

0 1

1

0 2

2 1 1

q

R

Q q

R

R q R Q R

+

( )=

=

( )

11 1 11

1 2

q q R Q

R R=

+

4 Then

q Q q R Q

R Rq2 1

2

1 2

2= =

+ = .

(b) V k q

R

k R Q

R R R

k Q

R R

e e e1

1

1

1

1 1 2 1 2

= =+( )

=+

V k q

R

k R Q

R R R

k Q

R R

e e e2

2

2

2

2 1 2 1 2

= =

+( )

=

+ and V V1 2 0 =

P26.31 (a) Q C V= = ( ) ( )= 150 10 10 10 1 50 1012 3 6F V C.

(b) U C V= ( )1

2

2

V

U

C= =

( )

=

2 2 250 10

150 101 83 10

6

12

3J

FV.

P26.32 (a) U C V C V C V = ( ) + ( ) = ( )

1

2

1

2

2 2 2

(b) The altered capacitor has capacitance =C C2

.The total charge is the same as before:

C V C V C V C

V V V

( ) + ( )= ( ) + ( ) =2

4

3

(c) =

+

= ( )

U C V

C V

C V1

2

4

3

1

2

1

2

4

34

3

2 2 2

(d) The extra energy comes from work put into the system by the agent pulling the capacitor

plates apart.

P26.33 The energy transferred is T Q VET C V J= = ( ) ( )= 12

1

250 0 1 00 10 2 50 108 9 . . .

and 1% of this (or Eint J= 2 50 107. ) is absorbed by the tree. If mis the amount of water boiled

away,

then E m mint J kg C C C= ( ) ( ) + 4 186 100 30 0 2 26 10 . .66 72 50 10J kg J( )= .

giving m= 9 79. kg


12/24

82 Chapter 26

Section 26.5 Capacitors with Dielectrics

P26.34 Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between

them. Suppose the plastic has 3, Emax ~ 107 V m,and thickness 1 mil =

2 54. cm

1 000. Then,

C

A

d=

( )( )

0123 8 85 10 0 4

2 54 10~

. .

.

C N m m2 2 2

55

6

7 5

10

10 2 54 10 10

m F

V m m

~

~ . ~max max

= ( ) ( )V E d 22 V

P26.35 (a) C A

d=

=

( ) ( ) 012 42 10 8 85 10 1 75 10

4 00

. . .

.

F m m2

= =

108 13 10 81 3

5

11

mF pF. .

(b) V E dmax max . . .= = ( ) ( )=

60 0 10 4 00 10 2 406 5 V m m kV

P26.36 Q C Vmax max ,=

but V E dmax max=

Also, C A

d=

0

Thus, Q A

dE d AEmax max max=

( )=

0 0

(a) With air between the plates, = 1 00.

and Emax .= 3 00 106 V m

Therefore,

Q AEmax max . . .= = ( ) ( ) 0

12 48 85 10 5 00 10 3F m m2 000 10 13 36( )=V m nC.

(b) With polystyrene between the plates, = 2 56. andEmax .= 24 0 106

V m. Q AEmax max . . .= = ( )

012 42 56 8 85 10 5 00 10F m m2(( ) ( )=24 0 10 2726. V m nC

P26.37 C A

d=

0

or 95 0 103 70 8 85 10 0 070 0

0 025 0 1

9

12

.. . .

. =

( )( )

00 3

= 1 04. m


13/24


P26.38 Originally, C A

d

Q

Vi

=

=( )

0

(a) The charge is the same before and after immersion, with value QA V

d

i= ( )0 .

Q

=

( ) ( )( ) 8 85 10 25 0 10 250

1

12 4. .C N m m V2 2 2

..50 10369

2( ) = mpC

(b) Finally,

C A

d

Q

Vf

f

=

=( )

0

Cf= ( ) ( ) 80 0 8 85 10 25 0 10

1 5

12 4. . .

.

C N m m2 2 2

00 10118

2( ) =

m

pF

V Qd

A

A V d

Ad

Vf

i i( ) =

= ( )

=

( )=

0

0

0

250 V

80.00V= 3 12.

(c) Originally, U C V A V

di i

i= ( ) = ( )1

2 2

2 0

2

Finally, U C V A Vd

A Vd

f f fi i= ( ) = ( ) = ( )12 2 2

2 0

2

20

2

So,

U U U A V

df i

i= = ( ) ( )0

21

2

U= ( ) ( )( ) 8.85 10 C N m 25.0 10 m 250 V12 2 2 4 2 2 779.0

2 1.50 10 m 80.045.5 nJ

( )

( )( ) = 2

P26.39 The given combination of capacitors is equivalent to the circuit

diagram shown to the right.

Put charge Qon pointA. Then,

Q V V V AB BC CD=( ) =( ) =( )40 0 10 0 40 0. . .F F F

So, V V VBC AB CD= =4 4 , and the center capacitor will break down first, at VBC= 15 0. V.When this occurs,

V V VAB CD BC= = ( )=1

43 75. V

and V V V V AD AB BC CD= + + = + + =3 75 15 0 3 75 22 5. . . .V V V V .

FIG. P26.39


14/24

84 Chapter 26

Section 26.6 Electric Dipole in an Electric Field

P26.40 (a) The displacement from negative to positive charge is

2 1 20 1 10 1 40 1 30a= +( ) ( ) = . . . . i j i jmm mm 22 60 2 40 10 3. . i j+( ) m

The electric dipole moment is

p a i j= = ( ) +( ) = 2 3 50 10 2 60 2 40 109 3q . . . C m +( ) 9 10 8 40 10 12. . i j C m

(b) = = +( )

p E i j9 10 8 40 10 7 8012. . .C m . i j( ) 4 90 10

3 N C

= + ( ) = 44 6 65 5 10 2 09 109 8. . . k k kN m N m

(c) U= = +( )

p E i j9 10 8 40 10 7 812. . .C m 00 4 90 103 . i j( ) N C

U= +( ) =71 0 41 2 10 1129. . J nJ

(d)p= ( ) +( ) = 9 10 8 40 10 12 4 102 2 12 12. . .C m C m

E

p

= ( ) +( ) =

=

7 80 4 90 10 9 21 102 2 3 3. . .

max

N C N C

U EE= =

=

114 114

228

nJ, nJU

U U

min

max min nJ

P26.41 (a) Letxrepresent the coordinate of the negative

charge. Then x a+ 2 cos is the coordinate of thepositive charge. The force on the negative charge

isF i= ( )qE x . The force on the positive charge is

F i i i+= + +( ) ( ) + ( )qE x a qE x qdE

dx a2 2cos

cos

.

The force on the dipole is altogether F F F i i= + = ( ) = + q

dE

dxa p

dE

dx2 cos cos .

(b) The balloon creates field along thex-axis ofk q

x

e

2.i

Thus,dE

dx

k q

x

e= ( )2

3.

Atx=16 0. cm, dEdx

= ( ) ( ) ( )

( )=

2 8 99 10 2 00 10

0 1608 7

9 6

3

. .

.. 88 MN C m

F i= ( ) ( ) = 6 30 10 8 78 10 0 559 6. . cos C m N C m .. 3i mN

p F+

F-

E

FIG. P26.41(a)


15/24


Section 26.7 An Atomic Description of Dielectrics

P26.42 (a) Consider a gaussian surface in the form of a cylindrical pillbox with ends of area


16/24


17/24


P26.47 (a) Each face of P2 carries charge, so the three-plate system is equivalent to

P1P2

P2

P3

Each capacitor by itself has capacitance

C A

d= = ( )

012 41 8 85 10 7 5 10

1 19 1

. .

.

C m

N m

2 2

2 005 58

3 =

mpF.

Then equivalent capacitance = + =5 58 5 58 11 2. . . pF .

(b) Q C V C V = + = ( )= 11 2 10 12 13412. F V pC

(c) Now P3 has charge on two surfaces and in effect three capacitors are in parallel:

C= ( )=3 5 58 16 7. .pF pF

(d) Only one face of P4carries charge:Q C V= = ( )= 5 58 10 66 912. .F 12 V pC

P26.48 Imagine the center plate is split along its midplane and pulled apart.

We have two capacitors in parallel, supporting the same V and

carrying total charge Q. The upper has capacitance C A

d1

0=

and the

lower C A

d2

0

2=

. Charge flows from ground onto each of the outside

plates so that Q Q Q1 2+ = V V V1 2= =

ThenQ

C

Q

C

Q d

A

Q d

A

1

1

2

2

1

0

2

0

2= =

=

Q Q1 22= 2 2 2Q Q Q+ =

(a) Q Q Q

2

3 3

= . .On the lower plate the charge is

Q Q Q

1

2

3

2= . On the upper plate the charge is

33.

(b) V Q

C

Qd

A= =

1

1 0

2

3

P26.49 Gasoline: 126 000 1 0541 00

3Btu gal J Btu

gal

3.786 10( )( )

.

m

m

kgJ kg

3

3

=

1 00

6705 24 107

..

Battery:

12 0 100 3 600

16 0 2 70 105.

. .

J C C s s

kg J kg

( )( )( )

=

Capacitor:12

20 100 12 0

0 10072 0

. .

..

F V

kgJ kg

( )( )=

Gasoline has 194 times the specific energy content of the battery and 727 000 times thhat of the

capacitor.

FIG. P26.48

d

2d


18/24


19/24


(c)

F i i=

=

( )

+ ( )( )dU

dx

Q d

x

.2

0

2 2

1

2 1

Whenx= 0, the original value of the force

isQ d

2

0

3

1

2

( )

i. As the dielectric slides in, the charges on the plates redistribute themselves.

The force decreases to its final valueQ d2

0

3 2

1

2

( )

i .

(d) Atx=

2,

F i=

( ) +( )

2 1

1

2

0

3 2

Q d

.

For the constant charge on the capacitor and the initial voltage we have the relationship

Q C V V

d= =

0

0

2

Then the force is F i=

( ) ( )+

2 1

1

0

2

2

V

d

F= ( ) ( ) ( )

2 8 85 10 0 05 2 000 4 5 112 2

. . .C m Nm2

NNm m CN

2 0 002 4 5 1205

2 2. .

( ) +( ) =i i

*P26.53 The portion of the capacitor nearly filled by metal has

capacitance ( ) 0

x

d

and stored energyQ

C

2

20

The unfilled portion has

capacitance ( )0 x

d

The charge on this portion is Q x Q

= ( )

0

(a) The stored energy is

U Q

C

x Q

x d

Q x d= =

( )[ ] ( )

= ( )

20

2

0

0

2

0

32 2 2

(b) F dU

dx

d

dx

Q x d Q d= = ( )

= +

0

2

0

3

0

2

02 2

3

F=

Q d02

032 to the right (into the capacitor)

(c) Stress = =

F

d

Q

0

2

0

42

(d) u E Q Q

= =

=

=1

2

1

2

1

20

2

0

0

2

00

0

2

2

00

2

0

42 .

The answers to parts (c) and (d) are

precisely the same.


20/24

90 Chapter 26

*P26.54 Consider a strip of width dxand length Wat positionxfrom the front left corner. The

capacitance of the lower portion of this strip is1 0 W dx

t x L.

/The capacitance of the upper portion

is2 0

( )W dx

t x L1.

/The series combination of the two elements has capacitance

1

1 0 2 0

1 2 0

2txWL dx

t L xWL dx

WL dx

t

+

=

( ) xx tL tx+ 1 1

The whole capacitance is a combination of elements in parallel:

C WL dx

tx tL t

L

=

( ) + =

( )

1 2 0

2 1 1 2 10

11 22 0 2 1

2 1 10

1 2 0

( )( ) +

=

WL tdx

tx tL

WL

L

22 12 1 1 0

1 2 0

2( ) ( ) + =

ttx tL

WLLln

11

2 1 1

1

1 2 0

0( )

( ) ++

=

t

tL tL

tL

WL

ln

22 1

2

1

1 2 0

2 11( )

=

( )

t

WL

tln

( )ln

2

1

1

1 2 0

1 2

1

=

( )

WL

tln

22

(b) The capacitor physically has the same capacitance if it is turned upside down, so the answershould be the same with

1and

2interchanged. We have proved that it has this property in

the solution to part (a).

(c) Let 1=

2(1 +x). Then C =

+ +[ ]

2 2 0

2

11

( )ln

x WL

xtx .

Asxapproaches zero we have C WL

xtx

WL

t=

+ =

( )1 0 0 0 as was to be shown.

P26.55 (a) C A

d

Q

V0

0 0

0

=

=

When the dielectric is inserted at constant voltage,

C C Q

V

UC V

UC V C V

= =

= ( )

= ( )

= ( )

0

0

0

0 0

2

0

2

0 0

2

2

2 2

andU

U0=

The extra energy comes from (part of the) electrical work done by the battery in separating

the extra charge.

(b) Q C V0 0 0=

and Q C V C V = = 0 0 0

so the charge increases according toQ

Q0= .


21/24


P26.56 Assume a potential difference acrossaandb, and notice that the potential difference across the

8 00. F capacitor must be zero by symmetry. Then the equivalent capacitance can be deter-mined from the following circuit:

FIG. P26.56

Cab= 3 00. F

P26.57 Call the unknown capacitance Cu

Q C V C C V

CC V

V V

u i u f

u

f

i f

= ( )= +( )( )

= ( )

( ) ( )

=

100 0 30 0

100 30 04 29

. .

..

F V

V VF

( )( )( )

=

*P26.58 One capacitor cannot be used by itselfit would burn out. She can

use two capacitors in parallel, connected in series to another two

capacitors in parallel. One capacitor will be left over. The equivalent

capacitance is1

200 2001001 1

F FF

( ) +( ) = . When 90 V is

connected across the combination, only 45 V appears across each individual capacitor.

P26.59 By symmetry, the potential difference across 3Cis zero, so the circuit reduces to

FIG. P26.59

CC C

C Ceq= +

= =

1

2

1

4

8

6

4

3

1

FIG. P26.58


22/24


23/24


P26.22 83.6 C

P26.24 (a) See the solution. Stored energy = 0.150 J (b) See the solution. Potential

difference = 268 V

P26.26 (a) 400 C (b) 2 50. kN m

P26.28 4 47. kV

P26.30 (a) q R Q

R R1

1

1 2

=+

and q R Q

R R2

2

1 2

=+

(b) See the solution.

P26.32 (a) C(V)2 (b) 4V/3 (c) 4C(V)2/3 (d) Positive work is done by the agent pulling the

plates apart.

P26.34 ~10 6

Fand ~102 V for two 40 cm by 100 cm sheets of aluminum foil sandwiching a thin sheet

of plastic.

P26.36 (a) 13 3. nC (b) 272 nC

P26.38 (a) 369 pC (b) 118 pF, 3.12 V (c) 45.5 nJ

P26.40 (a) +( ) 9 10 8 40. . i j pC m (b) 20 9. nN mk (c) 112 nJ (d) 228 nJ

P26.42 See the solution.

P26.44 189 kV

P26.46 (a) 25.0 F (1 0.846f )1 (b) 25.0 F; the general expression agrees. (c) 162 F; the

general expression agrees. (d) It has the same sign as the lower capacitor plate and its

magnitude is 254 C, independent off.

P26.48 (a) 2Q/3 on upper plate, Q/3 on lower plate (b) 2Qd/30A

P26.50 23 3. V; 26 7. V

P26.52 (a) + ( ) 0

2 1 x

d

(b)

Q d

x

2

0

22 1 + ( ) (c)

Q d

x

2

0

2 2

1

2 1

( )

+ ( ) to the right

(d) 205 N right

P26.54 (a)

1 2 0

1 2

1

2

( )

WL

tln (b) The capacitance is the same if k1and k2are interchanged, as it should be.

P26.56 3 00. F

P26.58 One capacitor cannot be used by itselfit would burn out. She can use two capacitors in parallel,

connected in series to another two capacitors in parallel. One capacitor will be left over. Each of

the four capacitors will be exposed to a maximum voltage of 45 V.

P26.60 See the solution.


24/24

Documents

Solutions to Chapter26