Capacitance and Dielectricssergioandresgarcia.com/pucmm/fis212/4.1.T26.SOL.pdfCapacitance and...

26Capacitance and Dielectrics

CHAPTER OUTLINE

26.1 Defi nition of Capacitance26.2 Calculating Capacitance26.3 Combinations of Capacitors26.4 Energy Stored in a Charged

Capacitor26.5 Capacitors with Dielectrics26.6 Electric Dipole in an Electric Field26.7 An Atomic Description of

Dielectrics

ANSWERS TO QUESTIONS

*Q26.1 (a) False. (b) True. In Q = C∆V, the capacitance is the proportionality constant relating the variables Q and ∆V.

Q26.2 Seventeen combinations:

Individual C C C1 2 3, ,

Parallel C C C C C C C C C1 2 3 1 2 1 3 2 3+ + + + +, , ,

Series-Parallel 1 1

1 2

1

3C CC+

⎛⎝⎜

⎞⎠⎟

+−

, 1 1

1 3

1

2C CC+

⎛⎝⎜

⎞⎠⎟

+−

, 1 1

2 3

1

1C CC+

⎛⎝⎜

⎞⎠⎟

+−

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 Volume is proportional to radius cubed. Increasing the radius by a factor of 31/3 will triple the volume. Capacitance is proportional to radius, so it increases by a factor of 31/3 . Answer (d).

*Q26.4 Let C2 = NC

1 be the capacitance of the large capacitor and C

1 that of the small one.

The equivalent capacitance is

Ceq

= CC NC N NC

N

NCeq =

+=

+=

+1

1 1

1

1 11 1 11/ / ( ) /

This is slightly less than C1, answer (d).

*Q26.5 We fi nd the capacitance, voltage, charge, and energy for each capacitor. (a) C = 20 µF ∆V = 4 V Q = C∆V = 80 µC U = (1/2)Q∆V = 160 µJ (b) C = 30 µF ∆V = Q/C = 3 V Q = 90 µC U = 135 µJ (c) C = Q/∆V = 40 µF ∆V = 2 V Q = 80 µC U = 80 µJ (d) C = 10 µF ∆V = (2U/C)1/2 = 5 V Q = 50 µC U = 125 µJ (e) C = 2U/∆V 2 = 5 µF ∆V = 10 V Q = 50 µC U = 250 µJ (f) C = Q2/2U = 20 µF ∆V = 6 V Q = 120 µC U = 360 µJ Then (i) the ranking by capacitance is c > b > a = f > d > e. (ii) The ranking by voltage ∆V is e > f > d > a > b > c. (iii) The ranking by charge Q is f > b > a = c > d = e. (iv) The ranking by energy U is f > e > a > b > d > c.

75

ISMV2_5104_26.indd 75ISMV2_5104_26.indd 75 6/12/07 3:57:34 PM6/12/07 3:57:34 PM

76 Chapter 26

Q26.6 A capacitor stores energy in the electric fi eld 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 fi nd that very little residual charge remains on each plate. When reassem-bled, the capacitor is suddenly “recharged”—by induction—due to the electric fi eld 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 after they sign a liability waiver.)

*Q26.7 (i) According to Q = C∆V, the answer is (b). (ii) From U = (1/2)C∆V 2, the answer is (a).

*Q26.8 The charge stays constant as C is cut in half, so U = Q2/2C doubles: answer (b).

*Q26.9 (i) Answer (b). (ii) Answer (c). (iii) Answer (c). (iv) Answer (a). (v) Answer (a).

*Q26.10 (i) Answer (b). (ii) Answer (b). (iii) Answer (b). (iv) Answer (c). (v) Answer (b).

Q26.11 The work you do to pull the plates apart becomes additional electric potential energy stored in the capacitor. The charge is constant and the capacitance decreases but the potential difference

increases to drive up the potential energy1

2Q V∆ . The electric fi eld between the plates is

constant in strength but fi lls more volume as you pull the plates apart.

Q26.12 The work done, W Q V= ∆ , is the work done by an external agent, like a battery, to move a charge through a potential difference, ∆V . To determine the energy in a charged capacitor, we must add the 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 ∆V versus 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∆ .

*Q26.13 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, ∆VQ

C

Q

Csdischarge

C

5.00 10 F= = =

×=−10

0 4008 05

.. 00 kV

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

Q26.14 The potential difference must decrease. Since there is no external power supply, the charge on the capacitor, Q, will remain constant—that is, assuming that the resistance of the meter is suffi ciently large. Adding a dielectric increases the capacitance, which must therefore decrease the potential difference between the plates.

*Q26.15 (i) Answer (a). (ii) Because ∆V is constant, Q = C∆V increases, answer (a). (iii) Answer (c). (iv) Answer (c). (v) U = (1/2)C∆V 2 increases, answer (a).

Q26.16 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 emission under high electric fi elds.


Capacitance and Dielectrics 77

Q26.17 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 κ ≈ 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 separation—hence 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 your sheets of insulating dielectric. Making sure that none of the conducting sheets are in contact with their next neighbors, connect every other plate together. Figure Q26.17 illustrates this idea.

ConductorConductor

Dielectric

FIG. Q26.17

This technique is often used when “home-brewing” signal capacitors for radio applications, as they can withstand huge potential differences without fl ashover (without either discharge between plates around the dielectric or dielectric breakdown). One variation on this technique is to sandwich together fl exible materials such as aluminum roof fl ashing 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 fi lled with motor oil (again to prevent fl ashover).

SOLUTIONS TO PROBLEMS

Section 26.1 Defi nition of Capacitance

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

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

P26.2 (a) CQ

V= = × = × =

−−

∆10 0 10

1 00 10 1 006

6.. .

C

10.0 V F Fµ

(b) ∆VQ

C= = ×

×=

−

−

100 10

10100

6

6

C

1.00 F V

Section 26.2 Calculating Capacitance

P26.3 Ek q

re= 2 : q =

×( )( )× ⋅( ) =

4 90 10 0 210

8 99 10

4 2

9

. .

.

N C m

N m C2 200 240. Cµ

(a) σπ

µ= = ×( )

=−q

A

0 240 10

4 0 1201 33

6

2

.

.. C m2

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


P26.4 CA

d= =

( ) ×( ) ×( )−κ ∈0

12 3 21 00 8 85 10 1 00 10. . . C m

N

2

⋅⋅ ( ) =m m

nF2 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

P26.5 (a) ∆V Ed= so E =×

=−

20 0

1011 13

..

V

1.80 m kV m toward the negative plate

(b) E =∈σ

0

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

(c) CA

d= =

× ⋅( )( )−∈0

128 85 10 7 60 1 00. . . C N m cm m 12 2 2 000 cm

m pF

( )×

=−

2

31 80 103 74

..

(d) ∆VQ

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

P26.6 With θ π= , the plates are out of mesh and the overlap area is zero.

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

2. By proportion,

the effective area of a single sheet of charge is π θ−( ) 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 are N plates on each comb, the number of parallel capacitors is 2 1N − and the total capacitance is

C NA N R

= −( ) ∈=

−( ) ∈ −( )2 1

2 10 0effective

distance

π θ 220

22

2

2 1

d

N R

d=

−( ) ∈ −( )π θ.

P26.7 QA

dV= ∈ ( )0 ∆

Q

A

V

d= =

∈ ( )σ 0 ∆

dV

=∈ ( ) =

× ⋅( )( )×

−0

128 85 10 150

30 0 10

∆σ

.

.

C N m V2 2

−−( ) ×( ) =9 41 00 10

4 42 C cm cm m

m2 2 2.

. µ

P26.8 Fy∑ = 0: T mgcosθ − = 0

Fx∑ = 0: T Eqsinθ − = 0

Dividing, tanθ = Eq

mg

so Emg

q= tanθ

and ∆V Edmgd

q= = tanθ

FIG. P26.6

78 Chapter 26



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 kb

ae= ⎛⎝⎜

⎞⎠⎟2 λ ln

λ = = × = ×

=

−−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: ∆VQ

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 ∆VQ

C= = =1 000

100 C

10 F V.

µµ

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 vv f

2 0+

v f =

×( )×

= ×−

−

2 1 00 10

2 101 00 10

4

166.

.J

kgm s

P26.11 (a) Cab

k b ae

=−( ) =

( )( )×( ) −

0 070 0 0 140

8 99 10 0 1409

. .

. . 00 070 015 6

..( ) = pF

(b) CQ

V=

∆ ∆V

Q

C= = ×

×=

−

−

4 00 10

15 6 10256

6

12

.

.

C

F kV

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 Fµ µ µ

(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 Cµ µ

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


80 Chapter 26

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

5 00

1

12 01 2C C Ceq

= + = +. . F Fµ µ

and Ceq = 3 53. Fµ

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

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

So Q Q1 2 31 8= = . Cµ

(b) The potential difference across each is ∆VQ

C11

1

31 86 35= = =..

C

5.00 F V

µµ

and ∆VQ

C22

2

31 82 65= = =..

C

12.0 FV

µµ

P26.14 (a) Capacitors 2 and 3 are in parallel and present equivalent capacitance 6C. This is in

series with capacitor 1, so the battery sees capacitance 1

3

1

62

1

C CC+⎡

⎣⎢⎤⎦⎥

=−

.

(b) If they were initially uncharged, C1 stores the same charge as C2 and C3 together.

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, ∆VQ

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

In parallel with each other, C2 and C3 have equal voltages: ∆ ∆ ∆V V V1 2 3> = .

(d) If C3 is increased, the overall equivalent capacitance increases. More charge moves through the battery and Q increases. As ∆V1 increases, ∆V2 must decrease so Q2 decreases.

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

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 Cp p s1

21 0− + =

CC C C C

C C C Cp 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 Cp p p s1

2 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 Cp p 44

1

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



P26.16 C C Cp = +1 2

and 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 Cp p s1

21 0− + =

and CC C C C

C C C Cp 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 Cp p p s221

2

1

4= − −

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

µµ

µ11

20 05 96

1

..

F F

µµ

⎛⎝⎜

⎞⎠⎟

=−

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

∆VQ

C= = =

− =

89 54 47

15 0 4 47 10 53

..

. . .

C

20.0 F V

µµ

VV

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

89 5 63 2 26 3. . .− = Cµ on 15 0. Fµ and 3 00. Fµ

P26.18 (a) In series , to reduce the effective capacitance:

1

32 0

1

34 8

1

1

2 51 103983

. .

.

F F

F F

µ µ

µµ

= +

=×

=−

C

C

s

s

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

29 8 32 0

2 20

. .

.

F F

F

µ µ

µ

+ =

=

C

C

p

p

FIG. P26.17


82 Chapter 26

P26.19 CQ

V=

∆ so 6 00 10

20 06.

.× =− Q

and Q = 120 Cµ

Q Q1 2120= − Cµ

and ∆VQ

C= : 120 2

1

2

2

− =Q

C

Q

C

or 120

6 00 3 002 2− =Q Q

. .

3 00 120 6 002 2. .( ) −( ) = ( )Q Q

Q2

360

9 0040 0= =

.. Cµ

Q1 120 40 0 80 0= − = C C Cµ µ µ. .

P26.20 For C1 connected by itself, C V1 30 8∆ = . Cµ where ∆V is the battery voltage: ∆VC

= 30 8

1

..

Cµ

For C1 and C2 in series:

1

1 123 1

1 2C CV

+⎛⎝⎜

⎞⎠⎟

=∆ . Cµ

substituting,

30 8 23 1 23 1

1 1 2

. . . C C Cµ µ µC C C

= + C C1 20 333= .

For C1 and C3 in series:

1

1 125 2

1 3C CV

+⎛⎝⎜

⎞⎠⎟

=∆ . Cµ

30 8 25 2 25 2

1 1 3

. . . C C Cµ µ µC C C

= + C C1 30 222= .

For all three:

Q

C C CV

C V

C C C C=

+ +⎛⎝⎜

⎞⎠⎟

=+ +

=1

1 1 1 1

30 8

1 2 3

1

1 2 1 3

∆ ∆ . CC

µ µ1 0 333 0 222

19 8+ +

=. .

.

This is the charge on each one of the three.

P26.21 nCn CC C C

n

=+ + +

=100 1001 1 1 ��

capacitors

nCC

n= 100

so n2 100= and n = 10

FIG. P26.19



P26.22 According to the suggestion, the combination of capacitors shown is equivalent to

Then 1 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 02 2

0

20 0

2

0 02

2 3

2 2 0

2 4 4

+ = +

+ − =

=− ± + 22

402C( )

Only the positive root is physical.

C

C= −( )0

23 1

P26.23 C

C

s

p

= +⎛⎝⎜

⎞⎠⎟ =

= ( ) +

−1

5 00

1

10 03 33

2 3 33 2

1

1

. ..

.

Fµ

.. .

. .

.

00 8 66

2 10 0 20 0

1

8 66

1

2

2

=

= ( ) =

= +

F

F

µ

µC

C

p

eq 00 06 04

1

..⎛

⎝⎜⎞⎠⎟ =

−

Fµ

P26.24 Q C Veq eq= ( ) = ×( )( ) = ×− −∆ 6 04 10 60 0 3 62 106 4. . . F V C

Q Qp eq1 = , so ∆VQ

Cpeq

p1

1

4

6

3 62 10

8 66 1041 8= = ×

×=

−

−

.

..

C

F V

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

P26.25 C

C

s

p

= +⎛⎝⎜

⎞⎠⎟ =

= + +

−1

5 00

1

7 002 92

2 92 4 00

1

. ..

. .

Fµ

66 00 12 9. .= Fµ

FIG. P26.22

FIG. P26.23

FIG. P26.25


84 Chapter 26

Section 26.4 Energy Stored in a Charged Capacitor

P26.26 U C V= 1

22∆

∆V

U

C= =

( )×

= ×−

2 2 300

30 104 47 106

3 J

C V V.

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

2

1

23 00 12 0 216

2 2∆ . . F V Jµ µ

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

2

1

23 00 6 00 54 0

2 2∆ . . . F V Jµ µ

P26.28 U C V= ( )1

22∆

The circuit diagram is shown at the right.

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

U = ×( )( ) =−1

230 0 10 100 0 1506 2

. . J

(b) CC Cs = +

⎛⎝⎜

⎞⎠⎟

= +⎛⎝⎜

⎞⎠⎟

− −1 1 1

25 0

1

5 001 2

1

. . F Fµ µ

11

4 17= . Fµ

U C V

VU

C

= ( )

= =( )

×=−

1

2

2 2 0 150

4 17 10268

2

6

∆

∆.

. V

P26.29 W U Fdx= = ∫

so FdU

dx

d

dx

Q

C

d

dx

Q x

A

Q= =⎛

⎝⎜

⎞

⎠⎟ =

∈⎛

⎝⎜

⎞

⎠⎟ =

∈

2 2

0

2

02 2 2 AA

P26.30 With switch closed, distance ′ =d d0 500. and capacitance

′ = ∈′

= ∈ =CA

d

A

dC0 02

2 .

(a) Q C V C V= ′( ) = ( ) = ×( )( ) =−∆ ∆2 2 2 00 10 100 4006. F V µCC

(b) The force stretching out one spring is

FQ

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 xd=4

, so

kF

x

C V

d d

C V

d= = ( ) ⎛

⎝⎞⎠ = ( )

=×( −

2 4 8 8 2 00 102 2

2

6∆ ∆ . F))( )×( ) =

−

100

8 00 102 50

2

3 2

V

mkN m

..

FIG. P26.28



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

(b) U C V= ( )1

22∆

∆V

U

C= =

×( )×

= ×−

−

2 2 250 10

150 101 83 10

6

123

J

F V.

P26.32 (a) U C V C V C V= ( ) + ( ) = ( )1

2

1

22 2 2∆ ∆ ∆

(b) The altered capacitor has capacitance ′ =CC

2. The total charge is the same as before:

C V C V C VC

V VV∆ ∆ ∆ ∆ ∆ ∆( ) + ( ) = ′( ) + ′( ) ′ =

2

4

3

(c) ′ = ⎛⎝⎜

⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟ =

( )U C

VC

VC

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 U C V= ( )1

22∆ where C R

R

ke

= ∈ =4 0π and ∆Vk Q

R

k Q

Re e= − =0

UR

k

k Q

R

k Q

Re

e e=⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ =1

2 2

2 2

P26.34 (a) The total energy is U U Uq

C

q

C

q

R

Q q= + = + =

∈+

−(1 2

12

1

22

2

12

0 1

11

2

1

2

1

2 4

1

2π))

∈

2

024π R

.

For a minimum we set dU

dq1

0= :

1

2

2

4

1

2

2

41 01

0 1

1

0 2

2 1 1

q

R

Q q

R

R q R Q R

π π∈+ −

∈−( ) =

= −

( )

11 1 11

1 2

q qR Q

R R=

+

4 Then

q Q qR Q

R Rq2 1

2

1 22= − =

+= .

(b) Vk q

R

k R Q

R R R

k Q

R Re e e

11

1

1

1 1 2 1 2

= =+( ) =

+

Vk q

R

k R Q

R R R

k Q

R Re e e

22

2

2

2 1 2 1 2

= =+( ) =

+

and V V1 2 0− =


86 Chapter 26

P26.35 The energy transferred is T Q VET C V J= = ( ) ×( ) = ×1

2

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 m is 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 10 J kg J( ) = ×.

giving m = 9 79. kg

Section 26.5 Capacitors with Dielectrics

P26.36 Q C Vmax max ,= ∆

but ∆V E dmax max=

Also, CA

d=

∈κ 0

Thus, QA

dE d AEmax max max=

∈ ( ) = ∈κ κ0

0

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

and Emax .= ×3 00 106 V m

Therefore,

Q AEmax max . . .= ∈ = ×( ) ×( )− −κ 012 48 85 10 5 00 10 3F m m2 000 10 13 36×( ) =V m nC.

(b) With polystyrene between the plates, κ = 2 56. and Emax .= ×24 0 106 V m.

Q AEmax max . . .= ∈ = ×( ) ×− −κ 012 42 56 8 85 10 5 00 10F m m2(( ) ×( ) =24 0 10 2726. V m nC

P26.37 (a) CA

d= ∈ =

×( ) ×( )− −κ 0

12 42 10 8 85 10 1 75 10

4 00

. . .

.

F m m2

××= × =−

−

108 13 10 81 3

511

m F pF. .

(b) ∆V E dmax max . . .= = ×( ) ×( ) =−60 0 10 4 00 10 2 406 5 V m m kV

P26.38 Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between

them. Suppose the plastic has κ ≈ 3, Emax ~ 107 V m , and thickness 1 mil = 2 54. cm

1 000. Then,

C

A

d= ∈ × ⋅( )( )

×

−

−

κ 0

123 8 85 10 0 4

2 54 10~

. .

.

C N m m2 2 2

556

7 5

10

10 2 54 10 10

mF

V m m

~

~ . ~max max

−

−= ( ) ×( )∆V E d 22 V

P26.39 CA

d= ∈κ 0

or 95 0 103 70 8 85 10 0 070 0

0 025 0 19

12

.. . .

.× =

×( )( )×

−− �

00 3−

� = 1 04. m



P26.40 Originally, CA

d

Q

V i

=∈

=( )

0

∆

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

di=

∈ ( )0 ∆.

Q =× ⋅( ) ×( )( )− −8 85 10 25 0 10 250

1

12 4. . C N m m V2 2 2

..50 10369

2×( ) =− m pC

(b) Finally,

CA

d

Q

Vf

f

= ∈ = ( )κ 0

∆

C f =× ⋅( ) ×( )− −80 0 8 85 10 25 0 10

1 5

12 4. . .

.

C N m m2 2 2

00 10118

2×( ) =− m pF

∆∆ ∆

VQd

A

A V d

Ad

Vf

i i( ) =∈

=∈ ( )

∈=

( )=

κ κ κ0

0

0

250 V

80.00 V= 3 12.

(c) Originally, U C VA V

di ii= ( ) =

∈ ( )1

2 22 0

2

∆∆

Finally, U C VA V

d

A V

df f fi i= ( ) =

∈ ( )=

∈ ( )1

2 2 22 0

2

20

2

∆∆ ∆κ

κ κ

So, ∆∆

U U UA V

df ii= − =

− ∈ ( ) −( )02 1

2

κκ

∆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

P26.41 The given combination of capacitors is equivalent to the circuit diagram shown to the right.

Put charge Q on point A. Then,

Q V V VAB BC CD= ( ) = ( ) = ( )40 0 10 0 40 0. . .F F Fµ µ µ∆ ∆ ∆

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

∆ ∆ ∆V V VAB CD BC= = ( ) =1

43 75. V

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

FIG. P26.41


88 Chapter 26

Section 26.6 Electric Dipole in an Electric Field

P26.42 (a) The displacement from negative to positive charge is

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

The electric dipole moment is

� �p a i j= = ×( ) − +( ) × =− −2 3 50 10 2 60 2 40 109 3q . . ˆ . ˆC m −− +( ) × ⋅−9 10 8 40 10 12. ˆ . î j C m

(b) � � �ττ = × = − +( ) × ⋅⎡

⎣⎤⎦ ×−p E i j9 10 8 40 10 7 8012. ˆ . ˆ .C m ˆ . î j−( ) ×⎡

⎣⎤⎦4 90 103 N C

�ττ = + −( ) × ⋅ = − × ⋅− −44 6 65 5 10 2 09 109 8. ˆ . ˆ . ˆk k kN m N m

(c) U = − ⋅ = − − +( ) × ⋅⎡⎣

⎤⎦ ⋅−� �

p E i j9 10 8 40 10 7 812. ˆ . ˆ .C m 00 4 90 103ˆ . î 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

P26.43 (a) Let x represent the coordinate of the negative charge. Then x a+ 2 cosθ is the coordinate of the positive charge. The force on the negative charge

is �F i− = − ( )qE x ˆ. The force on the positive charge is

�F i i i+ = + +( ) ≈ ( ) + ( )qE x a qE x q

dE

dxa2 2cos ˆ ˆ cos ˆθ θ .

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

dE

dxa p

dE

dx2 cos ˆ cos ˆθ θ .

(b) The balloon creates fi eld along the x-axis of k q

xe2

ˆ.i

Thus,dE

dx

k q

xe=

−( )23

.

At x =16 0. cm, dE

dx=

−( ) ×( ) ×( )( )

= −−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 ˆC m N C m ° .. ˆ3i mN

θ

p F+

F-

E

FIG. P26.43(a)



Section 26.7 An Atomic Description of Dielectrics

P26.44 (a) Consider a gaussian surface in the form of a cylindrical pillbox with ends of area ′ <<A A parallel to the sheet. The side wall of the cylinder passes no fl ux of electric fi eld since this surface is everywhere parallel to the fi eld. Gauss’s law becomes

EA EAQ

AA′+ ′ =

∈′, so E

Q

A=

∈2 directed away from the positive sheet.

(b) In the space between the sheets, each creates fi eld Q

A2∈ away from the positive and

toward the negative sheet. Together, they create a fi eld of

EQ

A=

∈

(c) Assume that the fi eld is in the positive x-direction. Then, the potential of the positive plate relative to the negative plate is

∆V dQ

Adx= − ⋅ = −

∈⋅ −( )

−

+

−∫

� �E s i i

plate

plate

plat

ˆ ˆee

plate+

∫ = +∈Qd

A

(d) Capacitance is defi ned by: CQ

V

Q

Qd A

A

d

A

d= =

∈= ∈ =

∈∆

κ 0 .

P26.45 Emax occurs at the inner conductor’s surface.

Ek

ae

max = 2 λfrom an equation derived about this situation in Chapter 24.

∆V kb

ae= ⎛⎝

⎞⎠2 λ ln from Example 26.1.

EV

a b a

V E ab

a

max

max max

ln

ln .

= ( )

= ⎛⎝

⎞⎠ = ×

∆

∆ 18 0 106 V mm m kV( ) ×( ) ⎛⎝

⎞⎠ =−0 800 10

3 00

0 80019 03. ln

.

..

Additional Problems

P26.46 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 CA

d10=

∈ and the

lower CA

d20

2=

∈. Charge fl ows from ground onto each of the outside

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

Then Q

C

Q

C

Q d

A

Q d

A1

1

2

2

1

0

2

0

2= =∈

=∈

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

(a) QQ Q

2 3 3= −. .On the lower plate the charge is

QQ Q

1

2

3

2= −. On the upper plate the charge is33

.

(b) ∆VQ

C

Qd

A= =

∈1

1 0

2

3

FIG. P26.46

d

2d


90 Chapter 26

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

P1

P2P2P3

Each capacitor by itself has capacitance

CA

d= ∈ =

×( ) ×⋅ ×

− −κ 0

12 41 8 85 10 7 5 10

1 19 1

. .

.

C m

N m

2 2

2 005 58

3− =m

pF.

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 P4 carries charge: Q C V= = × ( ) =−∆ 5 58 10 66 912. .F 12 V pC

P26.48 From the Example about a cylindrical capacitor,

V V kb

a

V

b a e

b

− = −

− = − ×( )

2

345 2 8 99 10 19

λ ln

.kV Nm C2 2 .. ln

. .

40 1012

2 8 99 1 4 10

6

3

×( )= − ( ) ×

− C mm

0.024 m

J CC V

V

( ) = − ×

= × − ×

ln .

. .

500 1 564 3 10

3 45 10 1 56 10

5

5 5Vb VV V= ×1 89 105.

P26.49 (a) We use the equation U = Q2/2C to fi nd the potential energy of the capacitor. As we will see, the potential difference ∆V changes as the dielectric is withdrawn. The initial and

fi nal energies are UQ

Cii

=⎛⎝⎜

⎞⎠⎟

1

2

2

and UQ

Cff

=⎛

⎝⎜⎞

⎠⎟1

2

2

.

But the initial capacitance (with the dielectric) is C Ci f= κ . Therefore, UQ

Cfi

=⎛⎝⎜

⎞⎠⎟

1

2

2

κ .

Since the work done by the external force in removing the dielectric equals the change in

potential energy, we have W U UQ

C

Q

C

Q

Cf ii i i

= − =⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

=⎛⎝

1

2

1

2

1

2

2 2 2

κ ⎜⎜⎞⎠⎟

−( )κ 1 .

To express this relation in terms of potential difference ∆Vi , we substitute Q C Vi i= ( )∆ , and

evaluate: W C Vi i= ( ) −( ) = ×( )( )−1

21

1

22 00 10 100 5 0

2 9 2∆ κ . .F V 00 1 00 4 00 10 5−( ) = × −. . .J

The positive result confi rms that the fi nal energy of the capacitor is greater than the initial energy. The extra energy comes from the work done on the system by the external force that pulled out the dielectric.

(b) The fi nal potential difference across the capacitor is ∆VQ

Cff

= .

Substituting CC

fi=

κ and Q C Vi i= ( )∆ gives ∆ ∆V Vf i= = ( ) =κ 5 00 100 500. .V V

Even though the capacitor is isolated and its charge remains constant, the potential difference across the plates does increase in this case.



*P26.50 (a)

a

C1

b

50 mF 20 mF

30 mF

FIG. P26.50a

(b ) 1

1 30 1 2020

1/ / ( )+ +=

CµF gives

1

30+

+=1

20

1

201C so

1

20

1

601+=

C

and C1 = 40 0. Fµ

(c)

FIG. P26.50c

50 mF

30 mF 60 mF

6 V

For the 50 µF, ∆V = 6 00. V and Q = C∆V = 300 Cµ .

For the 30 µF and 60 µF, the equivalent capacitance is 20 µF and Q = C∆V =

20 µF 6 V = 120 Cµ .

For 30 µF, ∆V = Q/C = 120 µC/30 µF = 4 00. .V

For 20 µF and 40 µF, ∆V = 120 µC/60 µF = 2 00. V .

For 20 µF, Q = C∆V = 20 µF 2 V = 40 0. Cµ .

For 40 µF, Q = C∆V = 40 µF 2 V = 80 0. Cµ .

P26.51 κ = 3 00. , EV

dmaxmax.= × =2 00 108 V m

∆

For CA

d=

∈= × −κ 0 60 250 10. F

ACd C V

E= = =

×( )( )−

κ κ∈ ∈0 0

60 250 10 4 000

3 0

∆ max

max

.

. 00 8 85 10 2 00 100 18812 8. .

.×( ) ×( ) =− m2


92 Chapter 26

*P26.52 (a) The partially fi lled capacitor constitutes two capacitors in series, with separate capacitances

1

1

6 50 0∈ ∈A

d f

A

fd( )

.

−and and equivalent capacitance

11

6 5

6 5

6 5 6 5

0 0

0 0

d fA

fdA

A

d df fd

A( )

.

.

. .− +=

− +=

∈ ∈

∈ ∈dd f

f6 5

6 5 5 525 0 1 0 846 1.

. .. ( . )

−= − −Fµ

(b) For f = 0, the capacitor is empty so we can expect capacitance 25 0. Fµ , and the

expression in part (a) agrees with this.

(c) For f = 1 we expect 6.5(25.0 µF) = 162 µF. The expression in (a) becomes

25.0 µF(1 – 0.846)–1 = 162 F, in agreementµ .

(d) The charge on the lower plate creates an electric fi eld in the liquid given by

E

Q

A A= =

κµ

∈ ∈0 0

300 C

6.5

The charge on the upper plate creates an electric fi eld in the vacuum according to

E

Q

A A= =

∈ ∈0 0

300 Cµ

The change in strength of the fi eld at the upper surface of the liquid is described by

300 300

0 0 0

C

6.5

Cµ µA

Q

A Ainduced

∈ ∈ ∈+ =

which gives Qinduced

= 254 C, independent of .µ f The induced charge

will be opposite in sign to the charge on the top capacitor plate and

the same in sign as the charge on the lower plate .

P26.53 (a) Put charge Q on the sphere of radius a and –Q on the other sphere. Relative to V = 0 at infi nity,

the potential at the surface of a is Vk Q

a

k Q

dae e= −

and the potential of b is Vk Q

b

k Q

dbe e= − + .

The difference in potential is V Vk Q

a

k Q

b

k Q

d

k Q

da be e e e− = + − −

and CQ

V V a b da b

=−

=∈

( ) + ( ) − ( )⎛⎝⎜

⎞⎠⎟

4

1 1 20π

(b) As d → ∞, 1

d becomes negligible compared to 1

a. Then,

Ca b

=∈

+4

1 10π and

1 1

4

1

40 0C a b=

∈+

∈π π

as for two spheres in series.



P26.54 The initial charge on the larger capacitor is

Q C V= = ( ) =∆ 10 150F 15 V Cµ µ

An additional charge q is pushed through the 50-V battery, giving the smaller capacitor charge q and the larger charge 150 Cµ + q .

Then 505

150

10V

F

C

F= + +q q

µµµ

500 2 150

117

C C

C

µ µµ= + +

=q q

q

So across the 5- Fµ capacitor ∆Vq

C= = =117

23 3C

5 FV

µµ

.

Across the 10- Fµ capacitor ∆V = + =150 11726 7

C C

10 FV

µ µµ

.

P26.55 Gasoline: 126 000 1 0541 00

3Btu gal J Btugal

3.786 10( )( )

× −

.

m

m

kgJ kg3

3⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

= ×1 00

6705 24 107..

Battery: 12 0 100 3 600

16 02 70 105.

..

J C C s s

kgJ 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.

*P26.56 (a) The portion of the device containing the dielectric has plate area � x and

capacitance Cx

d10=

∈κ �. The unfi lled part has area � � −( )x and

capacitance Cx

d20=

∈ −( )� �. The total capacitance is C C

dx1 2

0 2 1+ =∈

+ −( )⎡⎣ ⎤⎦� � κ .

(b) The stored energy is UQ

C

Q d

x= =

∈ + −( )( )1

2 2 1

2 2

02� � κ

.

(c) � �

� �F i i= − ⎛

⎝⎞⎠ = −( )

∈ + −( )( )dU

dx

Q d

xˆ ˆ .

2

02 2

1

2 1

κκ

When x = 0, the original value of the force

is Q d2

03

1

2

κ −( )∈ �

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

The force decreases to its fi nal valueQ d2

03 2

1

2

κκ−( )

∈ �i .

continued on next page


94 Chapter 26

(d) At x = �2

,�

�F i=

−( )∈ +( )2 1

1

2

03 2

Q d κκ

ˆ.

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

Q C VV

d= =

∈0

02

∆∆�

Then the force is � �F i=

∈ ( ) −( )+( )

2 1

10

2

2

∆V

d

κκ

ˆ

�F =

×( ) ( ) −( )−2 8 85 10 0 05 2 000 4 5 112 2. . .C m Nm2

NNm m CN2 0 002 4 5 1

2052 2. .ˆ ˆ

( ) +( ) =i iµ

*P26.57 The portion of the capacitor nearly fi lled by metal has

capacitance κ ∈ ( ) → ∞0 �x

d

and stored energy Q

C

2

20→

The unfi lled portion has

capacitance ∈ −( )0 � � x

d

The charge on this portion is Qx Q

=−( )�

�0

(a) The stored energy is

UQ

C

x Q

x d

Q x d= =

−( )⎡⎣ ⎤⎦∈ −( )

=−( )

∈

20

2

0

02

02 2 2

� �

� ��

��3

(b) FdU

dx

d

dx

Q x d Q d= − = −−( )

∈

⎛

⎝⎜⎜

⎞

⎠⎟⎟ = +

∈02

03

02

02 2

�� 3

�

�F =

∈Q d0

2

032

to the right (into the capacitor)

(c) Stress = =∈

F

d

Q

� �02

042

(d) u EQ Q

= ∈ = ∈∈

⎛⎝⎜

⎞⎠⎟

= ∈∈

⎛⎝⎜

⎞⎠⎟

=1

2

1

2

1

202

00

2

00

02

2σ

�002

042∈ �

.

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

precisely the same.

*P26.58 One capacitor cannot be used by itself—it 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

µ µµ

( ) + ( ) =− − . When 90 V is

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

FIG. P26.58



P26.59 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

µµ( )( )

−( ) =

*P26.60 Consider a strip of width dx and length W at position x from the front left corner. The

capacitance of the lower portion of this strip isκ1 0∈ W dx

t x L.

/ The capacitance of the upper portion

is κ 2 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

tκ κ

κ κκ

∈+ −

∈

=∈

( ) xx tL tx+ −κ κ1 1

The whole capacitance is a combination of elements in parallel:

CWL dx

tx tL t

L=

∈−( ) +

=−( )∫

κ κκ κ κ κ κ

κ κ1 2 0

2 1 1 2 10

11 22 0 2 1

2 1 10

1 2 0

∈ −( )−( ) +

=∈

∫WL tdx

tx tL

WL

L κ κκ κ κ

κ κκ 22 1

2 1 1 0

1 2 0

2−( ) −( ) +⎡⎣ ⎤⎦ =∈−κ

κ κ κ κ κκ κt

tx tLWLL

ln11

2 1 1

1

1 2 0

0( )−( ) +

+⎡

⎣⎢

⎤

⎦⎥

=∈

t

tL tL

tL

WL

lnκ κ κ

κ

κ κκ 22 1

2

1

1 2 0

2 11−( )⎡

⎣⎢

⎤

⎦⎥ =

∈− −( )κ

κκ

κ κκ κt

WL

tln

( )ln

κκκ

κ κκ κ

κκ

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 answer should be the same with κ

1 and κ

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

the solution to part (a).

(c) Let κ1 = κ

2 (1 + x). Then C =

+ ∈+[ ]κ κ

κ2 2 0

2

11

( )ln

x WL

xtx .

As x approaches zero we have CWL

xtx

WL

t=

+ ∈=

∈κ κ( )1 0 0 0 as was to be shown.

P26.61 (a) CA

d

Q

V00 0

0

=∈

=∆

When the dielectric is inserted at constant voltage,

C CQ

V

UC V

UC V C V

= =

=( )

=( )

=( )

κ

κ

00

00 0

2

0

2

0 02

2

2 2

∆

∆

∆ ∆

and U

U0

= κ

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 0κ

so the charge increases according to Q

Q0

= κ .


96 Chapter 26

P26.62 Assume a potential difference across a and b, and notice that the potential difference across the 8 00. Fµ capacitor must be zero by symmetry. Then the equivalent capacitance can be deter-mined from the following circuit:

FIG. P26.62

Cab = 3 00. Fµ

P26.63 Initially (capacitors charged in parallel),

q C V1 1 6 00 250 1 500= ( ) = ( )( ) =∆ . F V Cµ µ

q C V2 2 2 00 250 500= ( ) = ( )( ) =∆ . F V Cµ µ

After reconnection (positive plate to negative plate),

′ = − =q q qtotal C1 2 1 000 µ and ∆ ′ = ′ = =Vq

Ctotal

total

C

8.00 FV

1 000125

µµ

Therefore,

′ = ′( ) = ( )( ) =q C V1 1 6 00 125 750∆ . F V Cµ µ

′ = ′( ) = ( )( ) =q C V2 2 2 00 125 250∆ . F V Cµ µ

*P26.64 Let charge λ per length be on one wire and –λ be on the other. The electric fi eld due to the charge on the positive wire is perpendicular to the wire, radial, and of magnitude

Er+ =

∈λ

π2 0

The potential difference between the surfaces of the wires due to the presence of this charge is

∆V ddr

rD r

r

10 02 2

= − ⋅ = −∈

=∈−

+

−∫ ∫

� �E r

wire

wireλ

πλ

πlln

D r

r

−⎛⎝

⎞⎠

The presence of the linear charge density −λ on the negative wire makes an identical contribution to the potential difference between the wires. Therefore, the total potential difference is

∆ ∆V VD r

r= ( ) =

∈−⎛

⎝⎞⎠2 1

0

λπ

ln

With D much larger than r we have nearly ∆VD

r=

∈⎛⎝

⎞⎠

λπ 0

ln

and the capacitance of this system of two wires, each of length �, is

CQ

V V D r D r= = =

∈( ) [ ] =∈

[ ]∆ ∆λ λ

λ ππ� � �

0

0

ln ln

The capacitance per unit length is C

D r�=

∈[ ]π 0

ln /



P26.65 By symmetry, the potential difference across 3C is zero, so the circuit reduces to

FIG. P26.65

CC C

C Ceq = +⎛⎝

⎞⎠ = =

−1

2

1

4

8

6

4

3

1

P26.66 The condition that we are testing is that the capacitance increases by less than 10%, or,

′ <C

C1 10.

Substituting the expressions for C and ′C from Example 26.1, we have

′ =

( )

( )=

( )C

C

k b a

k b a

b a

be

e

�

�2 1 10

21

ln / .

ln /

ln /

ln / ...

101 10

a( ) <

This becomes

ln . ln.

. ln .b

a

b

a

b

a⎛⎝

⎞⎠ < ⎛

⎝⎞⎠ = ⎛

⎝⎞⎠ +1 10

1 101 10 1 10 lln

.. ln . ln .

1

1 101 10 1 10 1 10⎛

⎝⎞⎠ = ⎛

⎝⎞⎠ − ( )b

a

We can rewrite this as,

− ⎛⎝

⎞⎠ < − ( )

⎛⎝

⎞⎠ >

0 10 1 10 1 10

11 0 1

. ln . ln .

ln . ln

b

a

b

a.. ln . .10 1 10 11 0( ) = ( )

where we have reversed the direction of the inequality because we multiplied the whole expres-sion by –1 to remove the negative signs. Comparing the arguments of the logarithms on both sides of the inequality, we see that

b

a> ( ) =1 10 2 8511 0. ..

Thus, if b a> 2 85. , the increase in capacitance is less than 10% and it is more effective to increase � .


98 Chapter 26

ANSWERS TO EVEN PROBLEMS

P26.2 (a) 1 00. Fµ (b) 100 V

P26.4 11.1 nF; 26.6 C

P26.6 2 1 02N R

d

−( ) ∈ −( )π θ

P26.8 mgd

q

tanθ

P26.10 Yes; its total energy is suffi cient to make the trip; 1.00 × 106 m/s.

P26.12 (a) 17 0. Fµ (b) 9 00. V (c) 45 0. Cµ and 108 Cµ

P26.14 (a) 2C (b) Q1 > Q

3 > Q

2 (c) ∆V

1 > ∆V

2 = ∆V

3 (d) Q

1 and Q

3 increase, Q

2 decreases

P26.16 C C

C Cp pp s2 4

2

+ − and

C CC Cp p

p s2 4

2

− −

P26.18 (a) 398 Fµ in series (b) 2 20. Fµ in parallel

P26.20 19 8. Cµ

P26.22 3 12

0−( ) C

P26.24 83.6 Cµ

P26.26 4 47. kV

P26.28 (a) See the solution. Stored energy = 0.150 J (b) See the solution. Potential difference = 268 V

P26.30 (a) 400 Cµ (b) 2 50. kN m

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

P26.34 (a) qR Q

R R11

1 2

=+

and qR Q

R R22

1 2

=+

(b) See the solution.

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

P26.38 ~10 6− F and ~102 V for two 40 cm by 100 cm sheets of aluminum foil sandwiching a thin sheet of plastic.

P26.40 (a) 369 pC (b) 118 pF, 3.12 V (c) –45.5 nJ

P26.42 (a) − +( ) ⋅9 10 8 40. ˆ . ˆi j pC m (b) − ⋅20 9. ˆnN mk (c) 112 nJ (d) 228 nJ

P26.44 See the solution.

P26.46 (a) −2Q/3 on upper plate, −Q/3 on lower plate (b) 2Qd/3∈0A

P26.48 189 kV



P26.50 (a) See the solution. (b) 40.0 µF (c) 6.00 V across 50 µF with charge 300 µF; 4.00 V across 30 µF with charge 120 µF; 2.00 V across 20 µF with charge 40 µF; 2.00 V across 40 µF with charge 80 µF

P26.52 (a) 25.0 µF (1 − 0.846 f )−1 (b) 25.0 µF; the general expression agrees. (c) 162 µF; the general expression agrees. (d) It has the same sign as the lower capacitor plate and its magnitude is 254 µC, independent of f.

P26.54 23 3. V; 26 7. V

P26.56 (a) ∈ + −( )⎡⎣ ⎤⎦0

2 1� �x

d

κ (b)

Q d

x

2

022 1∈ � �+ −( )⎡⎣ ⎤⎦κ

(c) Q d

x

2

02 2

1

2 1

�

� �

κκ−( )

+ −( )⎡⎣ ⎤⎦∈to the right

(d) 205 µN right

P26.58 One capacitor cannot be used by itself—it 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 (a) κ κκ κ

κκ

1 2 0

1 2

1

2

∈−( )

WL

tln (b) The capacitance is the same if k

1 and k

2 are interchanged, as it should be.

P26.62 3 00. Fµ




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