26 Capacitance 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 (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 C 1 2 3 , , Parallel C C C C C C C C C 1 2 3 1 2 1 3 2 3 + + + + + , , , Series-Parallel 1 1 1 2 1 3 C C C + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + − , 1 1 1 3 1 2 C C C + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + − , 1 1 2 3 1 1 C 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 Volume is proportional to radius cubed. Increasing the radius by a factor of 3 1/3 will triple the volume. Capacitance is proportional to radius, so it increases by a factor of 3 1/3 . Answer (d). *Q26.4 Let C 2 = NC 1 be the capacitance of the large capacitor and C 1 that of the small one. The equivalent capacitance is C eq = C C NC N NC N N C eq = + = + = + 1 1 1 1 1 1 1 1 1 1 / / ( )/ This is slightly less than C 1 , answer (d). *Q26.5 We find 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 = Q 2 /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
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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.
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).
Q26.11 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.
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
)) = ×( ) ×( ) =−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) 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
� �p a i j= = ×( ) − +( ) × =− −2 3 50 10 2 60 2 40 109 3q . . ˆ . ˆC m −− +( ) × ⋅−9 10 8 40 10 12. ˆ . ˆi j C m
(b) � � �ττ = × = − +( ) × ⋅⎡
⎣⎤⎦ ×−p E i j9 10 8 40 10 7 8012. ˆ . ˆ .C m ˆ . ˆi 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 . ˆi 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
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
(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.
*P26.57 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.
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
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 � .
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.