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26

CHAPTER OUTLINE26.1 Definition 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

Capacitance and Dielectrics

ANSWERS TO QUESTIONS

Q26.1 Nothing happens to the charge if the wires are disconnected. Ifthe wires are connected to each other, charges in the singleconductor which now exists move between the wires and theplates until the entire conductor is at a single potential and thecapacitor is discharged.

Q26.2 336 km. The plate area would need to be 1

0∈ m2 .

Q26.3 The parallel-connected capacitors store more energy, since theyhave higher equivalent capacitance.

Q26.4 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-Parallel1 1

1 2

1

3C CC+

FHG

IKJ +−

, 1 1

1 3

1

2C CC+

FHG

IKJ +−

, 1 1

2 3

1

1C CC+

FHG

IKJ +−

1 1

1 2 3

1

C C C++

FHG

IKJ−

, 1 1

1 3 2

1

C C C++

FHG

IKJ−

, 1 1

2 3 1

1

C C C++

FHG

IKJ−

Series1 1 1

1 2 3

1

C C C+ +

FHG

IKJ−

, 1 1

1 2

1

C C+

FHG

IKJ−

, 1 1

2 3

1

C C+

FHG

IKJ−

, 1 1

1 3

1

C C+

FHG

IKJ−

Q26.5 This arrangement would decrease the potential difference between the plates of any individualcapacitor by a factor of 2, thus decreasing the possibility of dielectric breakdown. Depending on theapplication, this could be the difference between the life or death of some other (most likely moreexpensive) electrical component connected to the capacitors.

Q26.6 No—not just using rules about capacitors in series or in parallel. See Problem 72 for an example. Ifconnections can be made to a combination of capacitors at more than two points, the combinationmay be irreducible.

77

78 Capacitance and Dielectrics

Q26.7 A capacitor stores energy in the electric field between the plates. This is most easily seen when usinga “dissectable” capacitor. If the capacitor is charged, carefully pull it apart into its component pieces.One will find that very little residual charge remains on each plate. When reassembled, the capacitoris suddenly “recharged”—by induction—due to the electric field set up and “stored” in thedielectric. This proves to be an instructive classroom demonstration, especially when you ask astudent to reconstruct the capacitor without supplying him/her with any rubber gloves or otherinsulating material. (Of course, this is after they sign a liability waiver).

Q26.8 The work you do to pull the plates apart becomes additional electric potential energy stored in thecapacitor. The charge is constant and the capacitance decreases but the potential difference increases

to drive up the potential energy 12

Q V∆ . The electric field between the plates is constant in strength

but fills more volume as you pull the plates apart.

Q26.9 A capacitor stores energy in the electric field inside the dielectric. Once the external voltage source isremoved—provided that there is no external resistance through which the capacitor candischarge—the capacitor can hold onto this energy for a very long time. To make the capacitor safeto handle, you can discharge the capacitor through a conductor, such as a screwdriver, provided thatyou only touch the insulating handle. If the capacitor is a large one, it is best to use an externalresistor to discharge the capacitor more slowly to prevent damage to the dielectric, or welding of thescrewdriver to the terminals of the capacitor.

Q26.10 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 potentialdifference between the plates of an uncharged capacitor. As more charge is transferred from oneplate to the other, the potential difference increases as shown in Figure 26.12, meaning that morework is needed to transfer each additional bit of charge. The total work is the area under the curve of

Figure 26.12, and thus W Q V=12

∆ .

Q26.11 Energy is proportional to voltage squared. It gets four times larger.

Q26.12 Let C = the capacitance of an individual capacitor, and Cs represent the equivalent capacitance of thegroup in series. While being charged in parallel, each capacitor receives charge

Q C V= = × =−∆ charge F V C500 10 800 0 4004e ja f . .

While being discharged in series, ∆VQC

QCs

discharge C

5.00 10 F kV= = =

×=−10

0 4008 005

..

(or 10 times the original voltage).

Q26.13 Put a material with higher dielectric strength between the plates, or evacuate the space between theplates. At very high voltages, you may want to cool off the plates or choose to make them of adifferent chemically stable material, because atoms in the plates themselves can ionize, showingthermionic emission under high electric fields.

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

Q26.15 Each polar molecule acts like an electric “compass” needle, aligning itself with the external electricfield set up by the charged plates. The contribution of these electric dipoles pointing in the samedirection reduces the net electric field. As each dipole falls into a configuration of lower potentialenergy it can contribute to increasing the internal energy of the material.

Chapter 26 79

Q26.16 The material of the dielectric may be able to support a larger electric field than air, without breakingdown to pass a spark between the capacitor plates.

Q26.17 The dielectric strength is a measure of the potential difference per unit length that a dielectric canwithstand without having individual molecules ionized, leaving in its wake a conducting path fromplate to plate. For example, dry air has a dielectric strength of about 3 MV/m. The dielectric constantin effect describes the contribution of the electric dipoles of the polar molecules in the dielectric tothe electric field once aligned.

Q26.18 In water, the oxygen atom and one hydrogen atom considered alone have an electric dipole momentthat points from the hydrogen to the oxygen. The other O-H pair has its own dipole moment thatpoints again toward the oxygen. Due to the geometry of the molecule, these dipole moments add tohave a non-zero component along the axis of symmetry and pointing toward the oxygen.

A non-polarized molecule could either have no intrinsic dipole moments, or have dipolemoments that add to zero. An example of the latter case is CO2. The molecule is structured so thateach CO pair has a dipole moment, but since both dipole moments have the same magnitude andopposite direction—due to the linear geometry of the molecule—the entire molecule has no dipolemoment.

Q26.19 Heating a dielectric will decrease its dielectric constant, decreasing the capacitance of a capacitor.When you heat a material, the average kinetic energy per molecule increases. If you refer back to theanswer to Question 26.15, each polar molecule will no longer be nicely aligned with the appliedelectric field, but will begin to “dither”—rock back and forth—effectively decreasing its contributionto the overall field.

Q26.20 The primary choice would be the dielectric. You would want to chose a dielectric that has a largedielectric constant and dielectric strength, such as strontium titanate, where κ ≈ 233 (Table 26.1). Aconvenient choice could be thick plastic or mylar. Secondly, geometry would be a factor. Tomaximize capacitance, one would want the individual plates as close as possible, since thecapacitance is proportional to the inverse of the plate separation—hence the need for a dielectricwith a high dielectric strength. Also, one would want to build, instead of a single parallel platecapacitor, several capacitors in parallel. This could be achieved through “stacking” the plates of thecapacitor. For example, you can alternately lay down sheets of a conducting material, such asaluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none ofthe conducting sheets are in contact with their next neighbors, connect every other plate together.Figure Q26.20 illustrates this idea.

ConductorConductor

Dielectric

FIG. Q26.20

This technique is often used when “home-brewing” signal capacitors for radio applications, asthey can withstand huge potential differences without flashover (without either discharge betweenplates around the dielectric or dielectric breakdown). One variation on this technique is to sandwichtogether flexible materials such as aluminum roof flashing and thick plastic, so the whole productcan be rolled up into a “capacitor burrito” and placed in an insulating tube, such as a PVC pipe, andthen 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 48 06 5. . . . F V C Ce ja f µ

(b) Q C V= = × = × =− −∆ 4 00 10 1 50 6 00 10 6 006 6. . . . F V C Ce ja f µ

P26.2 (a) CQV

= =×

= × =−

−

∆10 0 10

1 00 10 1 006

6.. .

C10.0 V

F Fµ

(b) ∆VQC

= =××

=−

−100 10

10100

6

6 C

1.00 F V

Section 26.2 Calculating Capacitance

P26.3 Ek qr

e= 2 : q =×

× ⋅=

4 90 10 0 210

8 99 100 240

4 2

9

. .

..

N C m

N m C C

2 2

e ja fe j

µ

(a) σπ

µ= =×

=−q

A0 240 10

4 0 1201 33

6

2.

..a f C m2

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

P26.4 (a) C R= ∈4 0π

RC

k Ce=∈

= = × ⋅ × =−

48 99 10 1 00 10 8 99

0

9 12

π. . . N m C F mm2 2e je j

(b) C R= ∈ =× ×

⋅=

− −

44 8 85 10 2 00 10

0 2220

12 3

ππ . .

. C m

N m pF

2

2

e je j

(c) Q CV= = × = ×− −2 22 10 100 2 22 1013 11. . F V Ce ja f

P26.5 (a)QQ

RR

1

2

1

2=

Q QRR

Q Q1 21

22 21 3 50 7 00+ = +

FHG

IKJ = =. . Cµ

Q Q2 12 00 5 00= =. . C Cµ µ

(b) V VQC

QC1 2

1

1

2

21

45 00

0 5008 99 10 89 9= = = =

×= × =−

.

.. .

C

8.99 10 m F m V kV

9

µ

e j a f

Chapter 26 81

P26.6 CA

d=

∈=

× ×

⋅=

−κ 0

12 3 21 00 8 85 10 1 00 10

80011 1

. . ..

a fe je ja f C m

N m m nF

2

2

The potential between ground and cloud is

∆

∆

V Ed

Q C V

= = × = ×

= = × × =−

3 00 10 800 2 40 10

11 1 10 2 40 10 26 6

6 9

9 9

. .

. . .

N C m V

C V V C

e ja fa f e je j

P26.7 (a) ∆V Ed=

E =×

=−20 0

1011 13

..

V1.80 m

kV m

(b) E =∈σ

0

σ = × × ⋅ =−1 11 10 8 85 10 98 34 12. . . N C C N m nC m2 2 2e je j

(c) CA

d=∈

=× ⋅

×=

−

−0

12 2

3

8 85 10 7 60 1 00

1 80 103 74

. . .

..

C N m cm m 100 cm

m pF

2 2 2e je jb g

(d) ∆VQC

=

Q = × =−20 0 3 74 10 74 712. . . V F pCa fe j

P26.8 CA

d=

∈= × −κ 0 1560 0 10. F

dA

Cd

=∈

=× ×

×= × =

− −

−

−

κ 012 12

15

9

1 8 85 10 21 0 10

60 0 103 10 10 3 10

a fe je j. .

.. . m nm

P26.9 QA

dV=

∈0 ∆a f QA

Vd

= =∈

σ 0 ∆a f

dV

=∈

=× ⋅

× ×=

−

−0

12

9 4

8 85 10 150

30 0 10 1 00 104 42

∆a f e ja fe je jσ

µ.

. ..

C N m V

C cm cm m m

2 2

2 2 2


P26.10 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−a fR2

2When there are two plates in each comb, the number of adjoining

sheets of positive and negative charge is 3, as shown in the sketch. Whenthere are N plates on each comb, the number of parallel capacitors is 2 1N −and the total capacitance is

FIG. P26.10

C NA N R

dN R

d= −

∈=

− ∈ −=

− ∈ −2 1

2 1 22

2 10 02

02

a f a f a f a f a feffective

distanceπ θ π θ

.

P26.11 (a) Cke

ba

= =×

=2

50 0

2 8 99 102 68

9 7 272 58

ln.

. ln.

.

.c h e j c h nF

(b) Method 1: ∆V kbae= FHGIKJ2 λ ln

λ = =×

= ×

= × × FHGIKJ =

−−

−

q

V

8 10 101 62 10

2 8 99 10 1 62 107 272 58

3 02

67

9 7

..

. . ln..

.

C50.0 m

C m

kV∆ e je j

Method 2: ∆VQC

= =××

=−

−8 10 102 68 10

3 026

9..

. kV

P26.12 Let the radii be b and a with b a= 2 . Put charge Q on the inner conductor and –Q on the outer.

Electric field exists only in the volume between them. The potential of the inner sphere is Vk Q

aae= ;

that of the outer is Vk Q

bbe= . Then

V Vk Q

ak Q

bQ b a

aba be e− = − =

∈−FHGIKJ4 0π

and CQ

V Vab

b aa b=

−=

∈−

4 0π.

Here Ca

aa=

∈= ∈

4 280

2

0π

π aC

=∈8 0π

.

The intervening volume is Volume = − = FHGIKJ =FHGIKJ ∈

=∈

43

43

743

743 8

7384

3 3 33

3 303

3

203π π π π

π πb a a

C C

Volume =× ⋅

× ⋅= ×

−

−

7 20 0 10

384 8 85 102 13 10

6 3

2 12 316

.

..

C N m

C N m m

2

2 2

3e je jπ

.

The outer sphere is 360 km in diameter.

P26.13 (a) Cab

k b ae=

−=

× −=a f

b ga fe jb g

0 070 0 0 140

8 99 10 0 140 0 070 015 6

9

. .

. . .. pF

(b) CQV

=∆

∆VQC

= =××

=−

−4 00 1015 6 10

2566

12..

C F

kV

Chapter 26 83

P26.14 Fy∑ = 0 : T mgcosθ − = 0

Fx∑ = 0 : T Eqsinθ − = 0

Dividing, tanθ =Eqmg

so Emgq

= tanθ

and ∆V Edmgd

q= =

tanθ.

P26.15 C R= ∈ = × ⋅ × = ×− −4 4 8 85 10 6 37 10 7 08 10012 6 4π π . . . C N m m F2e je j

Section 26.3 Combinations of Capacitors

P26.16 (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 thebattery.

∆V = 9 00. V

(c) Q C V5 5 00 9 00 45 0= = =∆ . . . F V Cµ µb ga f

and Q C V12 12 0 9 00 108= = =∆ . . F V Cµ µb ga f

P26.17 (a) In series capacitors add as

1 1 1 15 00

112 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 µb ga f .

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 ∆VQC1

1

1

31 86 35= = =

..

C5.00 F

Vµµ

and ∆VQC2

2

2

31 82 65= = =

..

C12.0 F

Vµµ

.


P26.18 The circuit reduces first according to the rule for capacitors inseries, as shown in the figure, then according to the rule forcapacitors in parallel, shown below.

C C C Ceq = + +FHG

IKJ = =1

12

13

116

1 83.

⇒

FIG. P26.18

P26.19 C C Cp = +1 21 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

= +−

=− +

−e j.

Simplifying, C C C C Cp p s12

1 0− + = .

CC C C C

C C C Cp p p s

p p p s1

22

4

212

14

=± −

= ± −

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

C C C C C C C

p p p s

p p p p s

12 2

2 12

12

14

12

9 0014

9 00 9 00 2 00 6 00

12

14

12

9 00 1 50 3 00

= + − = + − =

= − = − − = − =

. . . . .

. . .

pF pF pF pF pF

pF pF pF

b g b g b gb g

b g

P26.20 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

= +−

=− +

−e j.

Simplifying, C C C C Cp p s12

1 0− + =

and CC C C C

C C C Cp p p s

p p p s1

22

4

212

14

=± −

= + −

where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values forthe capacitances, with the names reversed).

Then, from C C Cp2 1= −

C C C C Cp p p s221

214

= − − .

Chapter 26 85

P26.21 (a)1 1

15 01

3 00Cs= +

. .

CC

C

s

p

eq

== + =

= +FHG

IKJ =−

2 502 50 6 00 8 50

18 50

120 0

5 961

.. . .

. ..

F F

F F F

µµ

µ µµ

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

∆

∆

VQC

Q C V

= = =

− =

= = =

89 54 47

15 0 4 47 10 53

6 00 10 53 63 2

..

. . .

. . .

C20.0 F

V

V

F V C on 6.00 F

µµ

µ µ µb ga f89 5 63 2 26 3. . .− = Cµ on 15 0. Fµ and 3 00. Fµ

FIG. P26.21

*P26.22 (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

31

62

1

C CC+LNM

OQP =−

.

(b) If they were initially unchanged, 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||b g equivalent capacitor stores the same charge as C1 . Since it has greater

capacitance, ∆VQC

= 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 throughthe 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 .

P26.23 CQV

=∆

so 6 00 1020 0

6..

× =− Q

and Q = 120 Cµ

Q Q1 2120= − Cµ

and ∆VQC

= :120 2

1

2

2

−=

QC

QC

or120

6 00 3 002 2−=

Q Q. .

3 00 120 6 002 2. .a fb g a f− =Q Q

Q23609 00

40 0= =.

. Cµ Q1 120 40 0 80 0= − = C C Cµ µ µ. .

FIG. P26.23


P26.24 (a) In series , to reduce the effective capacitance:

132 0

134 8

1

12 51 10

3983

. .

.

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

P26.25 nCn CC C C

n

=+ + +

=100 100

1 1 1

capacitors

nCC

n=

100 so n2 100= and n = 10

*P26.26 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:

11 1

23 11 2C C

V+

FHG

IKJ =∆ . 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:

11 1

25 21 3C C

V+

FHG

IKJ =∆ . Cµ

30 8 25 2 25 2

1 1 3

. . . C C Cµ µ µC C C

= + C C1 30 222= . .

For all three:

QC C C

VC V

C C C C=

+ +

FHG

IKJ =

+ +=

+ +=

11 1 1 1

30 81 0 333 0 222

19 81 2 3

1

1 2 1 3∆

∆ .. .

. C

Cµ

µ .

This is the charge on each one of the three.

P26.27 C

C

C

C

s

p

p

eq

= +FHG

IKJ =

= + =

= =

= +FHG

IKJ =

−

−

15 00

110 0

3 33

2 3 33 2 00 8 66

2 10 0 20 0

18 66

120 0

6 04

1

1

2

1

. ..

. . .

. .

. ..

F

F

F

F

µ

µ

µ

µ

a fa f

FIG. P26.27

Chapter 26 87

P26.28 Q C Veq eq= = × = ×− −∆a f e ja f6 04 10 60 0 3 62 106 4. . . F V C

Q Qp eq1 = , so ∆VQ

Cpeq

p1

1

4

63 62 108 66 10

41 8= =××

=−

−..

. C F

V

Q C Vp3 3 162 00 10 41 8= = × =−∆e j e ja f. . F V 83.6 Cµ

P26.29 C

C

s

p

= +FHG

IKJ =

= + + =

−15 00

17 00

2 92

2 92 4 00 6 00 12 9

1

. ..

. . . .

F

F

µ

µ

FIG. P26.29

*P26.30 According to the suggestion, the combination ofcapacitors shown is equivalent to

Then1 1 1 1

0 0 0

0 0 0

0 0C C C C CC C C C C

C C C= +

++ =

+ + + ++b g

C C C C C C

C C C C

CC C C

0 02 2

02

0 02

0 02

02

2 3

2 2 0

2 4 4 2

4

+ = +

+ − =

=− ± + e j

Only the positive root is physical

CC

= −0

23 1e j

FIG. P26.30

Section 26.4 Energy Stored in a Charged Capacitor

P26.31 (a) U C V= = =12

12

3 00 12 0 2162 2∆a f b ga f. . F V Jµ µ

(b) U C V= = =12

12

3 00 6 00 54 02 2∆a f b ga f. . . F V Jµ µ

P26.32 U C V=12

2∆

∆VUC

= =×

= ×−2 2 300

30 104 47 106

3 J

C V V

a f.


P26.33 U C V=12

2∆a fThe 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 = × =−12

30 0 10 100 0 1506 2. .e ja f J

(b) CC Cs = +FHG

IKJ = +FHG

IKJ =

− −1 1 1

25 01

5 004 17

1 2

1 1

. ..

F F F

µ µµ

U C V

VUC

=

= =×

=−

12

2 2 0 150

4 17 10268

2

6

∆

∆

a fa f.

. V

FIG. P26.33

P26.34 Use UQC

=12

2

and CA

d=∈0 .

If d d2 12= , C C2 112

= . Therefore, the stored energy doubles .

*P26.35 (a) Q C V= = × × = ×− −∆ 150 10 10 10 1 50 1012 3 6 F V Ce je j .

(b) U C V=12

2∆a f

∆VUC

= =×

×= ×

−

−2 2 250 10

150 101 83 10

6

123

J

F V

e j.

P26.36 uUV

E= = ∈12 0

2

1 00 10 12

8 85 10 3 000

2 51 10 2 51 101 000

2 51

712 2

3 3

..

. . .

×= ×

= × = × FHG

IKJ =

−−

− −

V

V

e jb g

e j m m L

m L3 3

3

P26.37 W U Fdx= = zso F

dUdx

ddx

QC

ddx

Q xA

QA

= =FHGIKJ = ∈

FHG

IKJ = ∈

2 2

0

2

02 2 2

Chapter 26 89

P26.38 With switch closed, distance ′ =d d0 500. and capacitance ′ =∈

′=

∈=C

Ad

Ad

C0 022 .

(a) Q C V C V= ′ = = × =−∆ ∆a f a f e ja f2 2 2 00 10 100 4006. F V Cµ

(b) The force stretching out one spring is

FQ

AC V

AC V

A d dC V

d=

∈=

∈=

∈=

2

0

2 2

0

2 2

0

2

24

22 2∆ ∆ ∆a f a fb g

a f.

One spring stretches by distance xd

=4

, so

kFx

C Vd d

C V

d= = F

HGIKJ = =

×

×=

−

−

2 4 8 8 2 00 10 100

8 00 102 50

2 2

2

6 2

3 2

∆ ∆a f a f e ja fe j

.

..

F V

m kN m .

P26.39 The energy transferred is H Q VET C V J= = × = ×12

12

50 0 1 00 10 2 50 108 9∆ . . .a fe jand 1% of this (or ∆Eint J= ×2 50 107. ) is absorbed by the tree. If m is the amount of water boiledaway,

then ∆E m mint J kg C C C J kg J= ⋅° ° − ° + × = ×4 186 100 30 0 2 26 10 2 50 106 7b ga f e j. . .

giving m = 9 79. kg .

*P26.40 (a) U C V C V C V= + =12

12

2 2 2∆ ∆ ∆a f a f a f

(b) The altered capacitor has capacitance ′ =CC2

. The total charge is the same as before:

C V C V C VC

V∆ ∆ ∆ ∆a f a f a f a f+ = ′ + ′2

∆∆

′ =VV4

3.

(c) ′ = FHGIKJ + F

HGIKJ =U C

VC

VC

V12

43

12

12

43

43

2 2 2∆ ∆ ∆a f

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

P26.41 U C V=12

2∆a f where C RRke

= ∈ =4 0π and ∆Vk QR

k QR

e e= − =0

URk

k QR

k QRe

e e=FHGIKJFHGIKJ =

12 2

2 2


*P26.42 (a) The total energy is U U UqC

qC

qR

Q q

R= + = + =

∈+

−

∈1 212

1

22

2

12

0 1

12

02

12

12

12 4

12 4π πb g

.

For a minimum we set dUdq1

0= :

12

24

12

2

41 01

0 1

1

0 2

2 1 1 1 1 11

1 2

qR

Q q

R

R q R Q R q qR Q

R R

π π∈+

−

∈− =

= − =+

b g a f

Then q Q qR Q

R Rq2 1

2

1 22= − =

+= .

(b) Vk qR

k R QR R R

k QR R

e e e1

1

1

1

1 1 2 1 2= =

+=

+b gV

k qR

k R QR R R

k QR R

e e e2

2

2

2

2 1 2 1 2= =

+=

+b gand V V1 2 0− = .

Section 26.5 Capacitors with Dielectrics

P26.43 (a) CA

d=

∈=

× ×

×= × =

− −

−−κ 0

12 4

511

2 10 8 85 10 1 75 10

4 00 108 13 10 81 3

. . .

.. .

F m m

m F pF

2e je j

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

P26.44 Q C Vmax max= ∆ ,

but ∆V E dmax max= .

Also, CA

d=

∈κ 0 .

Thus, QA

dE d AEmax max max=

∈= ∈

κκ0

0b g .

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

and Emax .= ×3 00 106 V m .

Therefore,

Q AEmax max . . . .= ∈ = × × × =− −κ 012 4 68 85 10 5 00 10 3 00 10 13 3 F m m V m nC2e je je j .

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

Q AEmax max . . . .= ∈ = × × × =− −κ 012 4 62 56 8 85 10 5 00 10 24 0 10 272 F m m V m nC2e je je j

Chapter 26 91

P26.45 CA

d=

∈κ 0

or 95 0 103 70 8 85 10 0 070 0

0 025 0 109

12

3.. . .

.× =

×

×−

−

−

e jb g

= 1 04. m

P26.46 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. cm1 000

. Then,

CA

d

V E d

=∈ × ⋅

×

= ×

−

−−

−

κ 012

56

7 5 2

3 8 85 10 0 4

2 54 1010

10 2 54 10 10

~. .

.~

~ . ~max max

C N m m

m F

V m m V

2 2 2e je j

e je j∆

P26.47 Originally, CA

dQV i

=∈

=0

∆a f .

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

di=

∈0 ∆a f.

Q =× ⋅ ×

×=

− −

−

8 85 10 25 0 10 250

1 50 10369

12 4

2

. .

.

C N m m V

m pC

2 2 2e je ja fe j

(b) Finally,

CA

dQVf

f

=∈

=κ 0

∆a f C f =× ⋅ ×

×=

− −

−

80 0 8 85 10 25 0 10

1 50 10118

12 4

2

. . .

.

C N m m

m pF

2 2 2e je je j

∆∆ ∆

VQd

A

A V d

Ad

Vf

i ia f a f a f=

∈=∈

∈= = =

κ κ κ0

0

0

2503 12

V80.0

V. .

(c) Originally, U C VA V

di ii= =

∈12 2

2 02

∆∆a f a f

.

Finally, U C VA V

d

A V

df f fi i= =

∈=∈1

2 2 22 0

2

20

2

∆∆ ∆a f a f a fκ

κ κ.

So, ∆∆

U U UA V

df ii= − =

− ∈ −02 1

2

a f a fκκ

∆U = −× ⋅ ×

×= −

− −

−

8 85 10 25 0 10 250 79 0

2 1 50 10 80 045 5

12 4 2

2

. . .

. ..

C N m m V

m nJ

2 2 2e je ja f a fe ja f

.


P26.48 (a) C CA

d= =

∈=

× ×

×=

− −

−κκ

00

12 4

3

173 8 85 10 1 00 10

0 100 101 53

a fe je j. .

..

m

m nF

2

(b) The battery delivers the free charge

Q C V= = × =−∆a f e ja f1 53 10 12 0 18 49. . . F V nC .

(c) The surface density of free charge is

σ = =××

= ×−

−−Q

A18 4 10

101 84 10

9

44.

. C

1.00 m C m2

2 .

The surface density of polarization charge is

σ σκ

σp = −FHGIKJ = −FHG

IKJ = × −1

11

1173

1 83 10 4. C m2 .

(d) We have EE

= 0

κ and E

Vd0 =∆

; hence,

EVd

= =×

=−

∆κ

12 0

1 00 10694

4

.

.

V

173 m V ma fe j

.

P26.49 The given combination of capacitors is equivalent to the circuit diagramshown 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 µb g b g b g∆ ∆ ∆ . FIG. P26.49

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

∆ ∆ ∆V V VAB CD BC= = =14

3 75b g . V

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

Section 26.6 Electric Dipole in an Electric Field

P26.50 (a) The displacement from negative to positive charge is

2 1 20 1 10 1 40 1 30 2 60 2 40 10 3a = − + − − = − + × −. . . . . .i j i j i je j e j e j mm mm m.

The electric dipole moment is

p a i j i j= = × − + × = − + × ⋅− − −2 3 50 10 2 60 2 40 10 9 10 8 40 109 3 12q . . . . . C m C me je j e j .

(b) ττττ = × = − + × ⋅ × − ×−p E i j i j9 10 8 40 10 7 80 4 90 1012 3. . . .e j e j C m N C

ττττ = + − × ⋅ = − × ⋅− −44 6 65 5 10 2 09 109 8. . .k k ke j N m N m

continued on next page

Chapter 26 93

(c) U = − ⋅ = − − + × ⋅ ⋅ − ×−p E i j i j9 10 8 40 10 7 80 4 90 1012 3. . . .e j e j C m N C

U = + × =−71 0 41 2 10 1129. .a f J nJ

(d) p = + × ⋅ = × ⋅− −9 10 8 40 10 12 4 102 2 12 12. . .a f a f C m C m

E

p E

= + × = ×

= = = −

− =

7 80 4 90 10 9 21 10

114 114

228

2 2 3 3. . .

max min

max min

a f a f N C N C

nJ, nJ

nJ

U U

U U

P26.51 (a) Let x represent the coordinate of the negative charge.Then x a+ 2 cosθ is the coordinate of the positivecharge. The force on the negative charge isF i− = −qE xa f . The force on the positive charge is

F i i i+ = + + ≈ +qE x a qE x qdEdx

a2 2cos cosθ θa f a f a f .E

θ

p F+

F-

FIG. P26.51(a)

The force on the dipole is altogether F F F i i= + = =− + qdEdx

a pdEdx

2 cos cosθ θa f .

(b) The balloon creates field along the x-axis of k qx

e2 i .

Thus, dEdx

k q

xe=

−23

a f.

At x = 16 0. cm , dEdx

=− × ×

= − ⋅−2 8 99 10 2 00 10

0 1608 78

9 6

3

a fe je ja f

. .

.. MN C m

F i i= × ⋅ − × ⋅ ° = −−6 30 10 8 78 10 0 55 39 6. . cos . C m N C m mNe je j

Section 26.7 An Atomic Description of Dielectrics

P26.52 20

π r Eq

=∈

in

so Er

=∈

λπ2 0

∆

∆

∆

V dr

drrr

E r

V

V

r

r

r

r

= − ⋅ =∈

=∈FHGIKJ

∈=

= × × FHGIKJ

=

z z

−

E r1

2

1

2

2 2

2

1 20 10 0 100 1025 0

0 200

579

0 0

1

2

0

6 3

λπ

λπ

λπ

ln

. . ln.

.

maxmax

max

inner

V m m

V

e je j

FIG. P26.52


P26.53 (a) Consider a gaussian surface in the form of a cylindrical pillbox with ends of area ′ <<A Aparallel to the sheet. The side wall of the cylinder passes no flux of electric field since thissurface is everywhere parallel to the field. Gauss’s law becomes

EA EAQA

A′ + ′ =∈

′ , so EQ

A=

∈2 directed away from the positive sheet.

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

A2∈ away from the positive and

toward the negative sheet. Together, they create a field of

EQA

=∈

.

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

∆V dQA

dxQd

A= − ⋅ = −

∈⋅ − = +

∈−

+

−

+

z zE s i iplate

plate

plate

plate

e j .

(d) Capacitance is defined by: CQV

QQd A

Ad

Ad

= =∈

=∈

=∈

∆κ 0 .

Additional Problems

P26.54 (a) C = +LNM

OQP + +LNM

OQP =

− −13 00

16 00

12 00

14 00

3 331 1

. . . .. Fµ

(c) Q C Vac ac ac= = =∆b g b ga f2 00 90 0 180. . F V Cµ µ

Therefore, Q Q3 6 180= = Cµ

Q C Vdf df df= = =∆d i b ga f1 33 90 0 120. . F V Cµ µ FIG. P26.54

(b) ∆VQC3

3

3

18060 0= = =

C3.00 F

Vµµ

.

∆

∆

∆

VQC

VQC

VQC

66

6

22

2

44

4

18030 0

12060 0

12030 0

= = =

= = =

= = =

C6.00 F

V

C2.00 F

V

C4.00 F

V

µµµµµµ

.

.

.

(d) U C VT eq= = × =−12

12

3 33 10 90 0 13 42 6 2∆a f e ja f. . . V mJ

Chapter 26 95

*P26.55 (a) Each face of P2 carries charge, so the three-plate system is equivalent to

P1P2

P2P3

Each capacitor by itself has capacitance

CA

d=

∈=

× ×

⋅ ×=

− −

−κ 0

12 4

3

1 8 85 10 7 5 10

1 19 105 58

. .

..

C m

N m m pF

2 2

2

e j.

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

(b) Q C V C V= + = × =−∆ ∆ 11 2 10 12 13412. F V pCa f

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

C = =3 5 58 16 7. . pF pFb g .

(d) Only one face of P4 carries charge:

Q C V= = × =−∆ 5 58 10 66 912. . F 12 V pCa f .

*P26.56 From the example about a cylindrical capacitor,

V V kba

V

V

b a e

b

b

− = −

− = − × ×

= − ×

= − ×

= × − × = ×

−

2

345 2 8 99 10 1 40 1012

2 8 99 1 4 10 500

1 564 3 10

3 45 10 1 56 10 1 89 10

9 6

3

5

5 5 5

λ ln

. . ln

. . ln

.

. . .

kV Nm C C m m

0.024 m

J C

V

V V V

2 2e je ja fe j

*P26.57 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 flows from ground onto each of the outside

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

ThenQC

QC

Q dA

Q dA

1

1

2

2

1

0

2

0

2= =

∈=∈

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

d

2d

FIG. P26.57

(a) QQ Q

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

QQ Q

123

23

= −. .On the upper plate the charge is

(b) ∆VQC

QdA

= =∈

1

1 0

23


P26.58 (a) We use Equation 26.11 to find the potential energy of the capacitor. As we will see, thepotential difference ∆V changes as the dielectric is withdrawn. The initial and final

energies are UQCi

i=FHGIKJ

12

2

and UQCf

f=FHGIKJ

12

2

.

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

i=FHGIKJ

12

2

κ .

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

potential energy, we have W U UQC

QC

QCf i

i i i= − =

FHGIKJ −FHGIKJ =FHGIKJ −

12

12

12

12 2 2

κ κa f .To express this relation in terms of potential difference ∆Vi , we substitute Q C Vi i= ∆b g , and

evaluate: W C Vi i= − = × − = ×− −12

112

2 00 10 100 5 00 1 00 4 00 102 9 2 5∆b g a f e ja f a fκ . . . . F V J .

The positive result confirms that the final energy of the capacitor is greater than the initialenergy. The extra energy comes from the work done on the system by the external force thatpulled out the dielectric.

(b) The final potential difference across the capacitor is ∆VQCf

f= .

Substituting CC

fi=

κ and Q C Vi i= ∆b g gives ∆ ∆V Vf i= = =κ 5 00 100 500. V Va f .

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

P26.59 κ = 3 00. , EVdmaxmax.= × =2 00 108 V m

∆

For CA

d=

∈= × −κ 0 60 250 10. F

ACd C V

E=

∈=

∈=

×

× ×=

−

−κ κ0 0

6

12 8

0 250 10 4 000

3 00 8 85 10 2 00 100 188

∆ max

max

.

. . ..

e jb ge je j

m2

*P26.60 The original kinetic energy of the particle is

K mv= = × × = ×− −12

12

2 10 2 10 4 00 102 16 6 2 4 kg m s Je je j . .

The potential difference across the capacitor is ∆VQC

= = =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 4 C V Je ja f . .

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+ = ++ −a f a fat plate at plate : 4 00 10 3 10 10012

2 10 04 6 16 2. × + − × + = × +− − − J C Ve ja f e jv f

v f =×

×= ×

−

−

2 1 00 10

2 101 00 10

4

166

..

J

kg m s

e j.

Chapter 26 97

P26.61 (a) CA

d11 0 2

=∈κ

; CA

d22 0 2

2=

∈κ; C

Ad3

3 0 22

=∈κ

1 1

1 12

2 3

12 3

2 3

0 2 3

2 3

12 3

10 1 2 3

2 3

C CC C

C CA

d

C CC C

Ad

+FHG

IKJ =

+=∈

+FHG

IKJ

= + +FHG

IKJ =

∈+

+FHG

IKJ

−

−

κ κκ κ

κ κ κκ κ FIG. P26.61

(b) Using the given values we find: Ctotal F pF= × =−1 76 10 1 7612. . .

*P26.62 The initial charge on the larger capacitor is

Q C V= = =∆ 10 150 F 15 V Cµ µa f .

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

Then 505

15010

V F

C F

= ++q q

µµµ

500 2 150117 C C

Cµ µ

µ= + +

=q q

q

So across the 5- Fµ capacitor ∆VqC

= = =117

23 3 C

5 F V

µµ

. .

Across the 10- Fµ capacitor ∆V =+

=150 117

26 7 C C

10 F V

µ µµ

. .

P26.63 (a) Put charge Q on the sphere of radius a and –Q on the other sphere. Relative to V = 0 atinfinity,

the potential at the surface of a is Vk Q

ak Q

dae e= −

and the potential of b is Vk Qb

k Qdb

e e=−

+ .

The difference in potential is V Vk Q

ak Q

bk Q

dk Q

da be e e e− = + − −

and CQ

V V a b da b=

−=

∈+ −

FHG

IKJ

41 1 2

0πb g b g b g .

(b) As d →∞ , 1d

becomes negligible compared to 1a

. Then,

Ca b

=∈+

41 1

0π and

1 14

140 0C a b

=∈

+∈π π

as for two spheres in series.


P26.64 (a) Cd

x xd

x=∈

− + =∈

+ −0 0 2 1a f a fκ κ

(b) U C VV

dx= =

∈FHG

IKJ

+ −12

12

12 02

2∆∆a f a f a fκ

(c) F i= −FHGIKJ =

∈−

dUdx

Vd

02

21

∆a f a fκ to the left (out of the capacitor)

(d) F =× −

×= ×

−

−−

2 000 8 85 10 0 050 0 4 50 1

2 2 00 101 55 10

2 12

33

b g e jb ga fe j

. . .

.. N

P26.65 The portion of the capacitor nearly filled by metal has

capacitanceκ ∈

→∞0 xda f

and stored energyQ

C

2

20→ .

The unfilled portion has

capacitance∈ −0 x

da f

.

The charge on this portion is Qx Q

=−a f 0 .

(a) The stored energy is

UQ

C

x Q

x dQ x d

= =−

∈ −=

−

∈

20

2

0

02

032 2 2

a fa f

a f.

(b) FdUdx

ddx

Q x d Q d= − = −

−

∈

FHG

IKJ = + ∈

02

03

02

032 2

a f

F =∈Q d0

2

032

to the right (into the capacitor)

(c) Stress = =∈

Fd

Q02

042

(d) u EQ Q

= ∈ = ∈∈FHGIKJ = ∈

∈

FHGIKJ =

∈12

12

12 20

20

0

2

00

02

202

04

σ

Chapter 26 99

P26.66 Gasoline: 126 000 1 0541 00 1 00

6705 24 103

7 Btu gal J Btu gal

3.786 10 m m kg

J kg3

3

b gb g . ..

×FHG

IKJFHG

IKJ = ×−

Battery:12 0 100 3 600

16 02 70 105.

..

J C C s s

kg J kg

b gb gb g= ×

Capacitor:12

20 100 12 00 100

72 0. .

..

F V kg

J kga fa f

=

Gasoline has 194 times the specific energy content of the battery and 727 000 times that of thecapacitor.

P26.67 Call the unknown capacitance Cu

Q C V C C V

CC V

V V

u i u f

uf

i f

= = +

=−

=−

=

∆ ∆

∆

∆ ∆

b g b gd id i

b g d ib ga fa f10 0 30 0100 30 0

4 29. .

..

F V V V

Fµ

µ

*P26.68 She can clip together a series combination of parallel combinations of two100- Fµ capacitors. The equivalent capacitance is

1

200 2001001 1 F F

Fµ µ

µb g b g− −+= . When 90 V is connected across the

combination, only 45 V appears across each individual capacitor. FIG. P26.68

P26.69 (a) CA

dQV0

0 0

0=∈

=∆

When the dielectric is inserted at constant voltage,

C CQV

UC V

UC V C V

= =

=

= =

κ

κ

00

00 0

2

02

0 02

2

2 2

∆

∆

∆ ∆

;

b g

b g e j

andUU0

=κ .

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

(b) Q C V0 0 0= ∆

and Q C V C V= =∆ ∆0 0 0κ

soQQ0

=κ .


P26.70 The vertical orientation sets up two capacitors in parallel, with equivalent capacitance

CA

d

A

dA

dp =∈

+∈

=+FHGIKJ∈0 0 02 2 1

2b g b gκ κ

where A is the area of either plate and d is the separation of the plates. The horizontal orientationproduces two capacitors in series. If f is the fraction of the horizontal capacitor filled with dielectric,the equivalent capacitance is

1 1 1

0 0 0Cfd

A

f d

A

f f dAs

=∈

+−

∈=

+ −LNMM

OQPP ∈κ

κκ

b g b g, or C

f fA

ds = + −

LNMM

OQPP∈κ

κ 10

b g .

Requiring that C Cp s= gives κ κ

κ+

=+ −

12 1f fb g , or κ κ κ+ + − =1 1 2a f b gf f .

For κ = 2 00. , this yields 3 00 2 00 1 00 4 00. . . .− =a f f , with the solution f =23

.

P26.71 Initially (capacitors charged in parallel),

q C V1 1 6 00 250 1 500= = =∆a f b ga f. F V Cµ µ

q C V2 2 2 00 250 500= = =∆a f b ga f. F V Cµ µ .

After reconnection (positive plate to negative plate),

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

= =VqC

total

total

C8.00 F

V1 000

125µµ

.

Therefore,

′ = ′ = =q C V1 1 6 00 125 750∆a f b ga f. F V Cµ µ

′ = ′ = =q C V2 2 2 00 125 250∆a f b ga f. F V Cµ µ .

P26.72 Assume a potential difference across a and b, and notice that the potential difference across the8 00. Fµ capacitor must be zero by symmetry. Then the equivalent capacitance can be determinedfrom the following circuit:

⇒ ⇒

FIG. P26.72

Cab = 3 00. Fµ .

Chapter 26 101

P26.73 Emax occurs at the inner conductor’s surface.

Ekae

max =2 λ

from Equation 24.7.

∆V kbae= FHGIKJ2 λ ln from Example 26.2

EV

a b a

V E aba

max

max max

ln

ln . . ln.

.. .

=

= FHGIKJ = × × F

HGIKJ =

−

∆

∆

b ge je j18 0 10 0 800 10

3 000 800

19 06 3 V m m kV

P26.74 Ea

=2κλ

; ∆Vba

= FHGIKJ2κλ ln

∆V E aba

dVda

Eba

ab a

ba

max max

maxmax

ln

ln

= FHGIKJ

= FHGIKJ +FHGIKJ −FHGIKJ

LNMM

OQPP =

102

lnbaFHGIKJ = 1 or

ba

e= 1 so abe

=

P26.75 By symmetry, the potential difference across 3C is zero, so the circuit reduces to

CC C

C Ceq = +FHGIKJ = =−1

21

486

43

1

.

FIG. P26.75


P26.76 The electric field due to the charge on the positive wire is perpendicular to the wire, radial, and ofmagnitude

Er+ = ∈

λπ2 0

.

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

∆V ddrr

D ddD d

d

10 02 2

= − ⋅ = −∈

=∈

−FHGIKJ

−

+

−z zE r

wire

wire λπ

λπ

ln .

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

∆ ∆V VD d

d= =

∈−FHGIKJ2 1

0b g λ

πln

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

CQV V D d d D d d

= = =∈ −

=∈−∆ ∆

λ λλ π

π

0

0

b g a f a fln ln.

The capacitance per unit length is: C

D d d=

∈−

π 0

ln a f .

*P26.77 The condition that we are testing is that the capacitance increases by less than 10%, or,

′<

CC

1 10. .

Substituting the expressions for C and ′C from Example 26.2, we have,

′= = <

CC

k

k

eb

a

eba

bab

a

2

2

1 101.10

1.10

ln

ln

ln

ln.

c h

c h

c hc h .

This becomes,

ln . ln.

. ln . ln.

. ln . ln .ba

ba

ba

ba

FHGIKJ <

FHGIKJ =

FHGIKJ +

FHGIKJ =

FHGIKJ −1 10

1 101 10 1 10

11 10

1 10 1 10 1 10a f .

We can rewrite this as,

− FHGIKJ < −

FHGIKJ > =

0 10 1 10 1 10

11 0 1 10 1 10 11.0

. ln . ln .

ln . ln . ln .

ba

ba

a f

a f a f

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

ba> =1 10 2 8511.0. .a f .

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

Chapter 26 103

ANSWERS TO EVEN PROBLEMS

P26.2 (a) 1 00. Fµ ; (b) 100 VP26.40 (a) C V∆a f2 ; (b)

43∆V

; (c) 43

2

CV∆a f

;

P26.4 (a) 8 99. mm ; (b) 0 222. pF ; (c) 22 2. pC (d) Positive work is done on the system bythe agent pulling the plates apart.

P26.6 11 1. nF ; 26 6. C

P26.42 (a) qR Q

R R11

1 2=

+ and q

R QR R2

2

1 2=

+;

P26.8 3 10. nm(b) see the solution

P26.102 1 0

2N Rd

− ∈ −a f a fπ θP26.44 (a) 13 3. nC ; (b) 272 nC

P26.46 ~10 6− F and ~102 V for two 40 cm by100 cm sheets of aluminum foilsandwiching a thin sheet of plastic.

P26.12 2 13 1016. × m3

P26.14mgd

qtanθ

P26.48 (a) 1 53. nF ; (b) 18 4. nC ; (c) 184 C m2µ

free; 183 C m2µ induced; (d) 694 V mP26.16 (a) 17 0. Fµ ; (b) 9 00. V ;

(c) 45 0. Cµ and 108 CµP26.50 (a) − + ⋅9 10 8 40. .i je j pC m ;

(b) − ⋅20 9. nN mk ; (c) 112 nJ ; (d) 228 nJP26.18 1 83. C

P26.20C C

C Cp pp s2 4

2

+ − and C C

C Cp pp s2 4

2

− −P26.52 579 V

P26.54 (a) 3 33. Fµ ;(b) ∆V3 60 0= . V ; ∆V6 30 0= . V ;∆V2 60 0= . V ; ∆V4 30 0= . V ;

P26.22 (a) 2C ; (b) Q Q Q1 3 2> > ;(c) ∆ ∆ ∆V V V1 2 3> = ;

(c) Q Q3 6 180= = Cµ ; Q Q2 4 120= = Cµ ;(d) Q Q3 1 and increase and Q2 decreases

(d) 13 4. mJ

P26.24 (a) 398 Fµ in series; (b) 2 20. Fµ in parallel P26.56 189 kV

P26.58 (a) 40 0. Jµ ; (b) 500 VP26.26 19 8. Cµ

P26.60 yes; 1 00. Mm sP26.28 83.6 Cµ

P26.62 23 3. V ; 26 7. VP26.30 3 1

20−e jC

P26.64 (a) ∈ + −0

2 1x

d

κa f;

P26.32 4 47. kV

(b) ∈ + −0

2 2 1

2

∆V x

d

a f a fκ;P26.34 energy doubles

(c) ∈ −0

2 12

∆Vd

a f a fκ to the left ;P26.36 2 51 10 2 513. .× =− m L3

(d) 1 55. mN leftP26.38 (a) 400 Cµ ; (b) 2 50. kN m


P26.66 Gasoline has 194 times the specific energycontent of the battery, and 727 000 timesthat of the capacitor.

P26.72 3 00. Fµ

P26.74 see the solution

P26.68 see the solution; 45 V P26.76 see the solution

P26.7023

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