CAPACITANCE AND DIELECTRICS
24.1.
24.2.
24.3.
24
Q
Vab
SET UP: 1  F  106 F
EXECUTE: Q  CVab  (7.28  106 F)(25.0 V)  1.82  104 C  182  C
EVALUATE: One plate has charge Q and the other has charge Q .
PA
Q
IDENTIFY and SET UP: C  0 , C 
and V  Ed .
d
V
A
0.00122 m 2
(a) C  P0  P0
 3.29 pF
d
0.00328 m
8
Q 4.35  10 C
(b) V  
 13.2 kV
C 3.29  1012 F
V 13.2  103 V
(c) E  
 4.02  106 V/m
d
0.00328 m
EVALUATE: The electric field is uniform between the plates, at points that aren't close to the edges.
IDENTIFY and SET UP: It is a parallel-plate air capacitor, so we can apply the equations of Sections 24.1.
Q
Q 0.148 106 C
EXECUTE: (a) C 
so Vab  
 604 V
Vab
C
245 1012 F
IDENTIFY:
C
12
3
P0 A
Cd  245 10 F 0.328 10 m 
so A 

 9.08 103 m2  90.8 cm2
d
P0
8.854 1012 C2 / N  m2
V
604 V
(c) Vab  Ed so E  ab 
 1.84 106 V/m
d
0.328 103 m

(d) E  so   EP0  1.84  106 V/m 8.854  1012 C2 / N  m 2   1.63  105 C/m 2
P0
Q 0.148  106 C
EVALUATE: We could also calculate  directly as Q/A.   
 1.63  105 C/m2 , which checks.
A 9.08  103 m2
A
IDENTIFY: C  P0 when there is air between the plates.
d
SET UP: A  (3.0  102 m) 2 is the area of each plate.
(b) C 
24.4.
(8.854  1012 F/m)(3.0  102 m)2
 1.59  1012 F  1.59 pF
5.0  103 m
EVALUATE: C increases when A increases and C increases when d decreases.
Q
PA
IDENTIFY: C 
. C 0 .
Vab
d
SET UP: When the capacitor is connected to the battery, Vab  12.0 V .
EXECUTE:
24.5.
C
EXECUTE: (a) Q  CVab  (10.0  106 F)(12.0 V)  1.20  104 C  120 C
(b) When d is doubled C is halved, so Q is halved. Q  60 C .
(c) If r is doubled, A increases by a factor of 4. C increases by a factor of 4 and Q increases by a factor of 4.
Q  480 C.
EVALUATE: When the plates are moved apart, less charge on the plates is required to produce the same potential
difference. With the separation of the plates constant, the electric field must remain constant to produce the same
potential difference. The electric field depends on the surface charge density,  . To produce the same  , more
charge is required when the area increases.
24-1
24-2
24.6.
24.7.
Chapter 24
C
IDENTIFY:
SET UP: When the capacitor is connected to the battery, enough charge flows onto the plates to make Vab  12.0 V.
EXECUTE: (a) 12.0 V
Q
(b) (i) When d is doubled, C is halved. Vab 
and Q is constant, so V doubles. V  24.0 V .
C
(ii) When r is doubled, A increases by a factor of 4. V decreases by a factor of 4 and V  3.0 V .
EVALUATE: The electric field between the plates is E  Q / P0 A . Vab  Ed . When d is doubled E is unchanged and
V doubles. When A is increased by a factor of 4, E decreases by a factor of 4 so V decreases by a factor of 4.
PA
IDENTIFY: C  0 . Solve for d.
d
SET UP: Estimate r  1.0 cm . A   r 2 .
P r 2 P0 (0.010 m)2
P0 A
so d  0

 2.8 mm .
C
1.00  1012 F
d
EVALUATE: The separation between the pennies is nearly a factor of 10 smaller than the diameter of a penny, so it
is a reasonable approximation to treat them as infinite sheets.
Q
PA
INCREASE: C 
. Vab  Ed . C  0 .
Vab
d
C
EXECUTE:
24.8.
Q
PA
. C 0 .
Vab
d
SET UP: We want E  1.00 104 N/C when V  100 V .
V
1.00  102 V
EXECUTE: (a) d  ab 
 1.00  102 m  1.00 cm .
E 1.00  104 N/C
A
Cd (5.00 1012 F)(1.00 102 m)
A

 5.65 103 m2 . A   r 2 so r 
 4.24 102 m  4.24 cm .
12
2
2
P0
8.854 10 C /(N  m )

(b) Q  CVab  (5.00  1012 F)(1.00  102 V)  5.00  1010 C  500 pC
P0 A
. We could have a larger d, along with a larger A, and still achieve the required C without
d
exceeding the maximum allowed E.
IDENTIFY: Apply the results of Example 24.4. C  Q / V .
EVALUATE:
24.9.
SET UP:
C
ra  0.50 mm , rb  5.00 mm
EXECUTE: (a) C 
L 2 P0
(0.180 m)2 P0

 4.35  1012 F .
ln(rb ra ) ln(5.00 0.50)
(b) V  Q / C  (10.0  1012 C) /(4.35  1012 F)  2.30 V
C
 24.2 pF . This value is similar to those in Example 24.4. The capacitance is determined entirely by
L
the dimensions of the cylinders.
IDENTIFY: Capacitance depends on the geometry of the object.
2 P0 L
(a) SET UP: The capacitance of a cylindrical capacitor is C 
. Solving for rb gives rb  ra e2 P0 L / C .
ln  rb / ra 
EXECUTE: Substituting in the numbers for the exponent gives
EVALUATE:
24.10.
2 8.85  1012 C2 /N  m2  (0.120 m)
 0.182
3.67  1011 F
Now use this value to calculate rb: rb = ra e0.182 = (0.250 cm)e0.182 = 0.300 cm
(b) SET UP: For any capacitor, C = Q/V and  = Q/L. Combining these equations and substituting the numbers
gives  = Q/L = CV/L.
EXECUTE: Numerically we get
11
CV  3.67  10 F  125 V 

 3.82  108 C/m = 38.2 nC/m
L
0.120 m
EVALUATE: The distance between the surfaces of the two cylinders would be only 0.050 cm, which is just
0.50 mm. These cylinders would have to be carefully constructed.

Capacitance and Dielectrics
24.11.
24-3
IDENTIFY and SET UP: Use the expression for C/L derived in Example 24.4. Then use Eq.(24.1) to calculate Q.
C
2 P0
EXECUTE: (a) From Example 24.4,

L ln  rb / ra 
12
2
2
C 2 8.854  10 C / N  m 

 6.57  1011 F/m  66 pF/m
L
ln  3.5 mm/1.5 mm 
(b) C   6.57  1011 F/m   2.8 m   1.84  1010 F.
Q  CV  1.84  1010 F  350  103 V   6.4  1011 C  64 pC
24.12.
The conductor at higher potential has the positive charge, so there is +64 pC on the inner conductor and 64 pC on
the outer conductor.
EVALUATE: C depends only on the dimensions of the capacitor. Q and V are proportional.
IDENTIFY: Apply the results of Example 24.3. C  Q / V .
SET UP:
ra  15.0 cm . Solve for rb .
1 r r 
EXECUTE: (a) For two concentric spherical shells, the capacitance is C   a b  . kCrb  kCra  ra rb and
k  rb  ra 
12
kCra
k (116 10 F)(0.150 m)
rb 

 0.175 m .
kC  ra k (116 1012 F)  0.150 m
(b) V  220 V and Q  CV  (116  1012 F)(220 V)  2.55  108 C .
EVALUATE: A parallel-plate capacitor with A  4 ra rb  0.33 m2 and d  rb  ra  2.5  102 m has
P0 A
 117 pF , in excellent agreement with the value of C for the spherical capacitor.
d
IDENTIFY: We can use the definition of capacitance to find the capacitance of the capacitor, and then relate the
capacitance to geometry to find the inner radius.
(a) SET UP: By the definition of capacitance, C = Q/V.
C
24.13.
EXECUTE:
C
Q 3.30  109 C

 1.50  1011 F = 15.0 pF
V 2.20  102 V
(b) SET UP: The capacitance of a spherical capacitor is C  4 P0
ra rb
.
rb  ra
EXECUTE: Solve for ra and evaluate using C = 15.0 pF and rb = 4.00 cm, giving ra = 3.09 cm.
(c) SET UP: We can treat the inner sphere as a point-charge located at its center and use Coulomb’s law,
1 q
E
.
4 P0 r 2
EXECUTE:
24.14.
E
 9.00 10
9
N  m 2 /C 2  3.30  109 C 
 0.0309 m 
2
 3.12  104 N/C
EVALUATE: Outside the capacitor, the electric field is zero because the charges on the spheres are equal in
magnitude but opposite in sign.
IDENTIFY: The capacitors between b and c are in parallel. This combination is in series with the 15 pF capacitor.
SET UP: Let C1  15 pF , C2  9.0 pF and C3  11 pF .
EXECUTE: (a) For capacitors in parallel, Ceq  C1  C2 
so C23  C2  C3  20 pF
(b) C1  15 pF is in series with C23  20 pF . For capacitors in series,
C123 
1
1
1
 

Ceq C1 C2
so
1
1
1


and
C123 C1 C23
C1C23
(15 pF)(20 pF)

 8.6 pF .
C1  C23 15 pF  20 pF
EVALUATE: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors. For
capacitors in series the equivalent capacitance is smaller than any of the individual capacitors.
24-4
Chapter 24
24.15.
IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents. In each equivalent network
apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the
original circuit.
SET UP: Do parts (a) and (b) together. The capacitor network is drawn in Figure 24.15a.
C1  C2  C3  C4  400  F
Vab  28.0 V
Figure 24.15a
EXECUTE: Simplify the circuit by replacing the capacitor combinations by their equivalents: C1 and C2 are in
series and are equivalent to C12 (Figure 24.15b).
1
1
1
 
C12 C1 C2
Figure 24.15b
 4.00 10 F 4.00 106 F  2.00 106 F
C1C2

C1  C2
4.00 106 F  4.00 106 F
C12 and C3 are in parallel and are equivalent to C123 (Figure 24.15c).
6
C12 
C123  C12  C3
C123  2.00  106 F  4.00  106 F
C123  6.00  106 F
Figure 24.15c
C123 and C4 are in series and are equivalent to C1234 (Figure 24.15d).
1
1
1


C1234 C123 C4
Figure 24.15d
 6.00 106 F 4.00 106 F  2.40 106 F
C123C4

C123  C4
6.00 106 F  4.00  106 F
The circuit is equivalent to the circuit shown in Figure 24.15e.
C1234 
V1234  V  28.0 V
Q1234  C1234V   2.40  106 F   28.0 V   67.2  C
Figure 24.15e
Now build back up the original circuit, step by step. C1234 represents C123 and C4 in series (Figure 24.15f).
Q123  Q4  Q1234  67.2 C
(charge same for capacitors in series)
Figure 24.15f
Q123 67.2  C

 11.2 V
C123 6.00  F
Q 67.2  C
V4  4 
 16.8 V
C4 4.00  F
Note that V4  V123  16.8 V  11.2 V  28.0 V, as it should.
Then V123 
Capacitance and Dielectrics
24-5
Next consider the circuit as written in Figure 24.15g.
V3  V12  28.0 V  V4
V3  11.2 V
Q3  C3V3   4.00  F 11.2 V 
Q3  44.8 C
Q12  C12V12   2.00  F 11.2 V 
Q12  22.4 C
Figure 24.15g
Finally, consider the original circuit, as shown in Figure 24.15h.
Q1  Q2  Q12  22.4 C
(charge same for capacitors in series)
Q 22.4  C
V1  1 
 5.6 V
C1 4.00  F
V2 
Q2 22.4  C

 5.6 V
C2 4.00  F
Figure 24.15h
Note that V1  V2  11.2 V, which equals V3 as it should.
Summary: Q1  22.4 C, V1  5.6 V
Q2  22.4 C, V2  5.6 V
Q3  44.8 C, V3  11.2 V
Q4  67.2 C, V4  16.8 V
(c) Vad  V3  11.2 V
EVALUATE:
24.16.
V1  V2  V4  V , or V3  V4  V . Q1  Q2 , Q1  Q3  Q4 and Q4  Q1234 .
IDENTIFY: The two capacitors are in series. The equivalent capacitance is given by
1
1
1
.
 
Ceq C1 C2
SET UP: For capacitors in series the charges are the same and the potentials add to give the potential across the
network.
1
1
1
1
1
EXECUTE: (a)
 


 5.33  105 F1 . Ceq  1.88 106 F . Then
6
Ceq C1 C2 (3.0  10 F) (5.0  106 F)
Q  VCeq  (52.0 V)(1.88 106 F)  9.75 105 C . Each capacitor has charge 9.75 105 C .
(b) V1  Q / C1  9.75  105 C 3.0  106 F  32.5 V .
V2  Q / C2  9.75  105 C 5.0  106 F  19.5 V .
24.17.
EVALUATE: V1  V2  52.0 V , which is equal to the applied potential Vab . The capacitor with the smaller C has the
larger V.
IDENTIFY: The two capacitors are in parallel so the voltage is the same on each, and equal to the applied voltage Vab .
SET UP: Do parts (a) and (b) together. The network is sketched in Figure 24.17.
EXECUTE:
V1  V2  V
V1  52.0 V
V2  52.0 V
Figure 24.17
C  Q /V so Q  CV
Q1  C1V1   3.00  F  52.0 V   156  C. Q2  C2V2   5.00  F  52.0 V   260  C.
EVALUATE: To produce the same potential difference, the capacitor with the larger C has the larger Q.
24-6
Chapter 24
24.18.
IDENTIFY: For capacitors in parallel the voltages are the same and the charges add. For capacitors in series, the
charges are the same and the voltages add. C  Q / V .
SET UP: C1 and C2 are in parallel and C3 is in series with the parallel combination of C1 and C2 .
EXECUTE: (a) C1 and C2 are in parallel and so have the same potential across them:
Q2 40.0 106 C

 13.33 V . Therefore, Q1  V1C1  (13.33 V)(3.00  106 F)  80.0  106 C . Since C3 is
C2 3.00 106 F
in series with the parallel combination of C1 and C2 , its charge must be equal to their combined charge:
V1  V2 
C3  40.0  106 C  80.0  106 C  120.0  106 C .
1
1
1
1
1
(b) The total capacitance is found from
and Ctot  3.21  F .




6
Ctot C12 C3 9.00  10 F 5.00  106 F
Qtot 120.0  106 C

 37.4 V .
Ctot
3.21 106 F
Q 120.0 106 C
EVALUATE: V3  3 
 24.0 V . Vab  V1  V3 .
C3 5.00 106 F
IDENTIFY and SET UP: Use the rules for V for capacitors in series and parallel: for capacitors in parallel the voltages
are the same and for capacitors in series the voltages add.
EXECUTE: V1  Q1 / C1  150  C  /  3.00  F   50 V
Vab 
24.19.
24.20.
C1 and C2 are in parallel, so V2  50 V
V3  120 V  V1  70 V
EVALUATE: Now that we know the voltages, we could also calculate Q for the other two capacitors.
1
1
1
PA
IDENTIFY and SET UP: C  0 . For two capacitors in series,
.
 
Ceq C1 C2
d
1
24.21.
24.22.
1
 d
PA
d 


EXECUTE: Ceq   1  1    1  2   0
. This shows that the combined capacitance for two
C
C
P
A
d
P
A
2 
0
1  d2
 1
 0

capacitors in series is the same as that for a capacitor of area A and separation (d1  d2 ) .
EVALUATE: Ceq is smaller than either C1 or C2 .
PA
IDENTIFY and SET UP: C  0 . For two capacitors in parallel, Ceq  C1  C2 .
d
P0 A1 P0 A2 P0 ( A1  A2 )
EXECUTE: Ceq  C1  C2 
. So the combined capacitance for two capacitors in parallel is


d
d
d
that of a single capacitor of their combined area ( A1  A2 ) and common plate separation d.
EVALUATE: Ceq is larger than either C1 or C2 .
IDENTIFY: Simplify the network by replacing series and parallel combinations of capacitors by their equivalents.
1
1
1
SET UP: For capacitors in series the voltages add and the charges are the same;
 
 For capacitors
Ceq C1 C2
Q
.
V
EXECUTE: (a) The equivalent capacitance of the 5.0  F and 8.0  F capacitors in parallel is 13.0  F. When these
two capacitors are replaced by their equivalent we get the network sketched in Figure 24.22. The equivalent
capacitance of these three capacitors in series is 3.47  F.
(b) Qtot  CtotV  (3.47 F)(50.0 V)  174 C
(c) Qtot is the same as Q for each of the capacitors in the series combination shown in Figure 24.22, so Q for each of
the capacitors is 174  C.
Q
Q
EVALUATE: The voltages across each capacitor in Figure 24.22 are V10  tot  17.4 V , V13  tot  13.4 V and
C10
C13
Qtot
V9 
 19.3 V . V10  V13  V9  17.4 V  13.4 V  19.3 V  50.1 V . The sum of the voltages equals the applied
C9
voltage, apart from a small difference due to rounding.
in parallel the voltages are the same and the charges add; Ceq  C1  C2 
Figure 24.22
C
Capacitance and Dielectrics
24.23.
IDENTIFY: Refer to Figure 24.10b in the textbook. For capacitors in parallel, Ceq  C1  C2 
series,
1
1
1
 

Ceq C1 C2
24-7
. For capacitors in
.
SET UP: The 11  F , 4  F and replacement capacitor are in parallel and this combination is in series with the
9.0  F capacitor.
EXECUTE:
24.24.

1
1
1
1 



 . (15  x) F  72 F and x  57 F .
Ceq 8.0  F  (11  4.0  x)  F 9.0  F 
EVALUATE: Increasing the capacitance of the one capacitor by a large amount makes a small increase in the
equivalent capacitance of the network.
PA
IDENTIFY: Apply C  Q / V . C  0 . The work done to double the separation equals the change in the stored
d
energy.
1
Q2
SET UP: U  CV 2 
.
2
2C
EXECUTE: (a) V  Q / C  (2.55  C) (920  1012 F)  2770 V
P0 A
says that since the charge is kept constant while the separation doubles, that means that the capacitance
d
halves and the voltage doubles to 5540 V.
Q2 (2.55 106 C)2
(c) U 

 3.53 103 J . When if the separation is doubled while Q stays the same, the
2C 2(920 1012 F)
capacitance halves, and the energy stored doubles. So the amount of work done to move the plates equals the
difference in energy stored in the capacitor, which is 3.53  103 J.
EVALUATE: The oppositely charged plates attract each other and positive work must be done by an external force to
pull them farther apart.
IDENTIFY and SET UP: The energy density is given by Eq.(24.11): u  12 P0 E 2 . Use V  Ed to solve for E.
(b) C 
24.25.
EXECUTE: Calculate E : E 
V
400 V

 8.00 104 V/m.
d 5.00 103 m
Then u  12 P0 E 2  12 8.854 1012 C2 / N  m2 8.00 104 V/m   0.0283 J/m3
2
24.26.
EVALUATE: E is smaller than the value in Example 24.8 by about a factor of 6 so u is smaller by about a factor of
62  36.
Q
PA
IDENTIFY: C 
. C  0 . Vab  Ed . The stored energy is 12 QV .
Vab
d
SET UP:
d  1.50 103 m . 1 C  106 C
0.0180  106 C
 9.00  1011 F  90.0 pF
200 V
PA
Cd (9.00 1011 F)(1.50 103 m)
(b) C  0 so A 

 0.0152 m2 .
d
P0
8.854 1012 C2 /(N  m2 )
EXECUTE: (a) C 
(c) V  Ed  (3.0  106 V/m)(1.50  103 m)  4.5  103 V
(d) Energy  12 QV  12 (0.0180  106 C)(200 V)  1.80  106 J  1.80  J
EVALUATE: We could also calculate the stored energy as
24.27.
IDENTIFY: The energy stored in a charged capacitor is
SET UP:
EXECUTE:
1
2
Q2 (0.0180 106 C)2

 1.80  J .
2C
2(9.00 1011 F)
CV 2 .
1  F  106 F
1
2
CV 2  12 (450  106 F)(295 V) 2  19.6 J
EVALUATE: Thermal energy is generated in the wire at the rate I 2 R , where I is the current in the wire. When the
capacitor discharges there is a flow of charge that corresponds to current in the wire.
24-8
Chapter 24
24.28.
IDENTIFY: After the two capacitors are connected they must have equal potential difference, and their combined
charge must add up to the original charge.
Q2 1
SET UP: C  Q / V . The stored energy is U 
 CV 2
2C 2
EXECUTE: (a) Q  CV0 .
Q Q
Q
Q2
C
Q
3
(b) V  1  2 and also Q1  Q2  Q  CV0 . C1  C and C2 
so 1 
and Q2  1 . Q  Q1 .
C1 C2
C (C 2)
2
2
2
2
Q1 2 Q 2
Q1  Q and V 

 V0 .
3
C 3C 3
1  Q 2 Q 2  1  ( 2 Q) 2 2( 13 Q) 2  1 Q 2 1

 CV0 2
(c) U   1  2    3

2  C1 C2  2  C
C  3C 3
24.29.
1
(d) The original U was U  12 CV0 2 , so U   CV02 .
6
(e) Thermal energy of capacitor, wires, etc., and electromagnetic radiation.
EVALUATE: The original charge of the charged capacitor must distribute between the two capacitors to make the
potential the same across each capacitor. The voltage V for each after they are connected is less than the original
voltage V0 of the charged capacitor.
IDENTIFY and SET UP: Combine Eqs. (24.9) and (24.2) to write the stored energy in terms of the separation
between the plates.
Q2
PA
xQ2
EXECUTE: (a) U 
; C  0 so U 
2C
x
2P0 A
x  dx  Q 2

(b) x  x  dx gives U 
2P0 A
dU 
 x  dx  Q 2 
2P0 A
xQ 2  Q 2 

 dx
2P0 A  2P0 A 
Q2
2P0 A
(d) EVALUATE: The capacitor plates and the field between the plates are shown in Figure 24.29a.

Q
E 
P0 P0 A
F  12 QE, not QE
Figure 24.29a
(c) dW  F dx  dU , so F 
The reason for the difference is that E is the field due to both plates. If we consider the positive plate only and
calculate its electric field using Gauss’s law (Figure 24.29b):
Q
ъE  dA  Pencl
0
A
2 EA 
P0

Q
E

2P0 2P0 A
Figure 24.29b
The force this field exerts on the other plate, that has charge Q, is F 
24.30.
IDENTIFY:
C
0 A
. The stored energy can be expressed either as
Q2
.
2P0 A
Q2
CV 2
or as
, whichever is more convenient
2
2C
d
for the calculation.
SET UP: Since d is halved, C doubles.
EXECUTE: (a) If the separation distance is halved while the charge is kept fixed, then the capacitance increases and
the stored energy, which was 8.38 J, decreases since U  Q 2 2C. Therefore the new energy is 4.19 J.
(b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance
leads to a doubling of the stored energy to 16.8 J, using U  CV 2 2 , when V is held constant throughout.
Capacitance and Dielectrics
24.31.
24-9
EVALUATE: When the capacitor is disconnected, the stored energy decreases because of the positive work done by
the attractive force between the plates. When the capacitor remains connected to the battery, Q  CV tells us that the
charge on the plates increases. The increased stored energy comes from the battery when it puts more charge onto the
plates.
Q
IDENTIFY and SET UP: C  . U  12 CV 2 .
V
EXECUTE: (a) Q  CV  (5.0 F)(1.5 V)  7.5 C . U  12 CV 2  12 (5.0  F)(1.5 V) 2  5.62  J
(b) U  12 CV 2  12 C (Q / C ) 2  Q 2 / 2C . Q  2CU  2(5.0 106 F)(1.0 J)  3.2 103 C .
Q 3.2  103 C

 640 V .
C 5.0  106 F
EVALUATE: The stored energy is proportional to Q 2 and to V 2 .
V
24.32.
IDENTIFY: The two capacitors are in series.
1
1
1
 

Ceq C1 C2
C
Q
. U  12 CV 2 .
V
SET UP:
For capacitors in series the voltages add and the charges are the same.
CC
(150 nF)(120 nF)
1
1
1
 66.7 nF .
EXECUTE: (a)
so Ceq  1 2 
 
C1  C2 150 nF  120 nF
Ceq C1 C2
Q  CV  (66.7 nF)(36 V)  2.4  106 C  2.4  C
(b) Q  2.4 C for each capacitor.
(c) U  12 CeqV 2  12 (66.7 109 F)(36 V)2  43.2 J
(d) We know C and Q for each capacitor so rewrite U in terms of these quantities. U  12 CV 2  12 C (Q / C ) 2  Q 2 / 2C
(2.4 106 C)2
(2.4 106 C) 2
 19.2 J ; 120 nF: U 
 24.0  J
9
2(150 10 F)
2(120 109 F)
Note that 19.2  J  24.0  J  43.2  J , the total stored energy calculated in part (c).
150 nF: U 
Q 2.4  106 C
Q 2.4  106 C

 16 V ; 120 nF: V  
 20 V
9
C 150  10 F
C 120  109 F
Note that these two voltages sum to 36 V, the voltage applied across the network.
EVALUATE: Since Q is the same the capacitor with smaller C stores more energy ( U  Q 2 / 2C ) and has a larger
voltage ( V  Q / C ).
(e) 150 nF: V 
24.33.
Q
. U  12 CV 2 .
V
SET UP: For capacitors in parallel, the voltages are the same and the charges add.
EXECUTE: (a) Ceq  C1  C2  35 nF  75 nF  110 nF . Qtot  CeqV  (110 109 F)(220 V)  24.2  C
IDENTIFY: The two capacitors are in parallel. Ceq  C1  C2 . C 
(b) V  220 V for each capacitor.
35 nF: Q35  C35V  (35  109 F)(220 V)  7.7  C ; 75 nF: Q75  C75V  (75  109 F)(220 V)  16.5  C . Note that
Q35  Q75  Qtot .
(c) U tot  12 CeqV 2  12 (110 109 F)(220 V)2  2.66 mJ
(d) 35 nF: U 35  12 C35V 2  12 (35  109 F)(220 V)2  0.85 mJ ;
24.34.
75 nF: U 75  12 C75V 2  12 (75  109 F)(220 V) 2  1.81 mJ . Since V is the same the capacitor with larger C stores more
energy.
(e) 220 V for each capacitor.
EVALUATE: The capacitor with the larger C has the larger Q.
IDENTIFY: Capacitance depends on the geometry of the object.

ln  rb / ra  . Solving for the linear charge
(a) SET UP: The potential difference between the core and tube is V 
2 P0
density gives  
2 P0V
4 P0V
.

ln  rb / ra  2 ln  rb / ra 
EXECUTE: Using the given values gives  
6.00 V
 2.00 
2  9.00  10 N  m /C  ln 

 1.20 
9
2
2
 6.53  1010 C/m
24-10
Chapter 24
(b) SET UP:
Q  L
EXECUTE: Q  (6.53  1010 C/m)(0.350 m) = 2.29  1010 C
(c) SET UP: The definition of capacitance is C  Q / V .
2.29  1010 C
 3.81 1011 F
6.00 V
(d) SET UP: The energy stored in a capacitor is U  12 CV 2 .
EXECUTE:
24.35.
C
EXECUTE: U  12 (3.81 1011 F)(6.00 V) 2  6.85  1010 J
EVALUATE: The stored energy could be converted to heat or other forms of energy.
IDENTIFY: U  12 QV . Solve for Q. C  Q / V .
SET UP: Example 24.4 shows that for a cylindrical capacitor,
EXECUTE: (a) U  12 QV gives Q 
C
2 P0

.
L ln(rb /ra )
2U 2(3.20 109 J)

 1.60 109 C.
V
4.00 V
C
2 P0
r

. b  exp(2 P0 L/C )  exp(2 P0 LV/Q )  exp(2 P0 (15.0 m)(4.00 V) (1.60 109 C))  8.05.
L ln(rb ra ) ra
The radius of the outer conductor is 8.05 times the radius of the inner conductor.
EVALUATE: When the ratio rb / ra increases, C / L decreases and less charge is stored for a given potential
difference.
IDENTIFY: Apply Eq.(24.11).
 rr 
Q
Q
  a b Vab .
SET UP: Example 24.3 shows that E 
between the conducting shells and that
2
4 P0  rb  ra 
4 P0 r
(b)
24.36.
 r r V
 [0.125 m][0.148 m]  120 V 96.5 V  m
E   a b  ab2  
 2 
r

r
r
r2
 0.148 m  0.125 m  r
 b a
(a) For r  0.126 m , E  6.08 103 V/m . u  12 P0 E 2  1.64  104 J/m3 .
EXECUTE:
24.37.
(b) For r  0.147 m , E  4.47 103 V/m . u  12 P0 E 2  8.85  105 J/m3 .
EVALUATE: (c) No, the results of parts (a) and (b) show that the energy density is not uniform in the region between
the plates. E decreases as r increases, so u decreases also.
IDENTIFY: Use the rules for series and for parallel capacitors to express the voltage for each capacitor in terms of
the applied voltage. Express U, Q, and E in terms of the capacitor voltage.
SET UP: Le the applied voltage be V. Let each capacitor have capacitance C. U  12 CV 2 for a single capacitor with
voltage V.
EXECUTE: (a) series
Voltage across each capacitor is V/2. The total energy stored is Us  2  12 C[V/2]2   14 CV 2
parallel
Voltage across each capacitor is V. The total energy stored is Up  2  12 CV 2   CV 2
Up  4Us
(b) Q  CV for a single capacitor with voltage V. Qs  2  C[V/2]  CV ; Qp  2(CV )  2CV ; Q p  2Qs
(c) E  V/d for a capacitor with voltage V. Es  V/2d ; Ep  V/d ; Ep  2 Es
24.38.
EVALUATE: The parallel combination stores more energy and more charge since the voltage for each capacitor is
larger for parallel. More energy stored and larger voltage for parallel means larger electric field in the parallel case.
IDENTIFY: V  Ed and C  Q / V . With the dielectric present, C  KC0 .
SET UP: V  Ed holds both with and without the dielectric.
EXECUTE: (a) V  Ed  (3.00  104 V/m)(1.50  103 m)  45.0 V .
Q  C0V  (5.00  1012 F)(45.0 V)  2.25 1010 C .
(b) With the dielectric, C  KC0  (2.70)(5.00 pF)  13.5 pF . V is still 45.0 V, so
Q  CV  (13.5  1012 F)(45.0 V)  6.08  1010 C .
EVALUATE: The presence of the dielectric increases the amount of charge that can be stored for a given potential
difference and electric field between the plates. Q increases by a factor of K.
Capacitance and Dielectrics
24.39.
24-11
IDENTIFY and SET UP: Q is constant so we can apply Eq.(24.14). The charge density on each surface of the
dielectric is given by Eq.(24.16).
E
E
3.20  105 V/m
EXECUTE: E  0 so K  0 
 1.28
K
E 2.50  105 V/m
(a)  i   (1  1/ K )
  P0 E0  (8.854  1012 C2 /N  m2 )(3.20  105 N/C)  2.833  106 C/m2
24.40.
 i  (2.833  106 C/m2 )(1  1/1.28)  6.20  107 C/m2
(b) As calculated above, K  1.28.
EVALUATE: The surface charges on the dielectric produce an electric field that partially cancels the electric field
produced by the charges on the capacitor plates.
IDENTIFY: Capacitance depends on geometry, and the introduction of a dielectric increases the capacitance.
SET UP: For a parallel-plate capacitor, C  K P0 A/d .
EXECUTE: (a) Solving for d gives
d
K P0 A (3.0)(8.85 1012 C2 /N  m 2 )(0.22 m)(0.28 m)

 1.64 103 m = 1.64 mm .
C
1.0 109 F
1.64 mm
 8 sheets .
0.20 mm/sheet
Cd
(1.0 109 F)(0.012 m)
(b) Solving for the area of the plates gives A 

 0.45 m2 .
K P0 (3.0)(8.85 1012 C2/N  m2 )
(c) Teflon has a smaller dielectric constant (2.1) than the posterboard, so she will need more area to achieve the same
capacitance.
EVALUATE: The use of dielectric makes it possible to construct reasonable-sized capacitors since the dielectric
increases the capacitance by a factor of K.
IDENTIFY and SET UP: For a parallel-plate capacitor with a dielectric we can use the equation C  KP0 A / d .
Dividing this result by the thickness of a sheet of paper gives
24.41.
24.42.
24.43.
Minimum A means smallest possible d. d is limited by the requirement that E be less than 1.60 107 V/m when V is
as large as 5500 V.
V
5500 V
EXECUTE: V  Ed so d  
 3.44 104 m
E 1.60 107 V/m
Cd
(1.25 109 F)(3.44 104 m)
Then A 

 0.0135 m2 .
K P0 (3.60)(8.854 1012 C2 / N  m2 )
EVALUATE: The relation V = Ed applies with or without a dielectric present. A would have to be larger if there were
no dielectric.
IDENTIFY and SET UP: Adapt the derivation of Eq.(24.1) to the situation where a dielectric is present.
EXECUTE: Placing a dielectric between the plates just results in the replacement of P for P0 in the derivation of
Equation (24.20). One can follow exactly the procedure as shown for Equation (24.11).
EVALUATE: The presence of the dielectric increases the energy density for a given electric field.
IDENTIFY: The permittivity P of a material is related to its dielectric constant by P K P0 . The maximum voltage is
E0


, where
K K P0
 is the surface charge density on each plate. The induced surface charge density on the surface of the dielectric is
given by  i   (1  1/ K ) .
SET UP: From Table 24.2, for polystyrene K  2.6 and the dielectric strength (maximum allowed electric field) is
2 107 V/m .
EXECUTE: (a) P K P0  (2.6)P0  2.3  1011 C2 N  m 2
related to the maximum possible electric field before dielectric breakdown by Vmax  Emax d . E 
(b) Vmax  Emax d  (2.0  107 V m)(2.0  103 m)  4.0  104 V
(c) E 



K P0
 i   1 
and   PE  (2.3  1011 C2 /N  m 2 )(2.0  107 V/m)  0.46  103 C/m 2 .
1
3
2
4
2
  (0.46 10 C/m )(1  1/ 2.6)  2.8 10 C/m .
K
EVALUATE: The net surface charge density is  net     i  1.8  104 C/m 2 and the electric field between the
plates is E   net / P0 .
24-12
Chapter 24
24.44.
IDENTIFY: C  Q / V . C  KC0 . V  Ed .
SET UP: Table 24.1 gives K  3.1 for mylar.
EXECUTE: (a) Q  Q  Q0  ( K  1)Q0  ( K  1)C0V0  (2.1)(2.5  107 F)(12 V)  6.3  106 C .
24.45.
(b)  i   (1  1/ K ) so Qi  Q(1  1/K )  (9.3  106 C)(1  1/ 3.1)  6.3  106 C .
(c) The addition of the mylar doesn’t affect the electric field since the induced charge cancels the additional charge
drawn to the plates.
EVALUATE: E  V / d and V is constant so E doesn't change when the dielectric is inserted.
(a) IDENTIFY and SET UP: Since the capacitor remains connected to the power supply the potential difference
doesn’t change when the dielectric is inserted. Use Eq.(24.9) to calculate V and combine it with Eq.(24.12) to obtain a
relation between the stored energies and the dielectric constant and use this to calculate K.
EXECUTE: Before the dielectric is inserted U 0  12 C0V 2 so V 
2 1.85 105 J 
2U 0

 10.1 V
C0
360 109 F
(b) K  C/C0
U 0  12 C0V 2 , U  12 CV 2 so C/C0  U / U 0
K
24.46.
U 1.85 105 J  2.32 105 J

 2.25
U0
1.85 105 J
EVALUATE: K increases the capacitance and then from U  12 CV 2 , with V constant an increase in C gives an
increase in U.
IDENTIFY: C  KC0 . C  Q / V . V  Ed .
SET UP: Since the capacitor remains connected to the battery the potential between the plates of the capacitor
doesn't change.
EXECUTE: (a) The capacitance changes by a factor of K when the dielectric is inserted. Since V is unchanged (the
C
Q
45.0 pC
 K  1.80 .
battery is still connected), after  after 
Cbefore Qbefore 25.0 pC
(b) The area of the plates is  r 2   (0.0300 m) 2  2.827  103 m2 and the separation between them is thus
P0 A (8.85  1012 C2 N  m2 )(2.827  103 m2 )
PA Q

 2.00  103 m . Before the dielectric is inserted, C  0 
C
12.5  1012 F
d V
Qd
(25.0 1012 C)(2.00 103 m)
and V 

 2.00 V . The battery remains connected, so the potential
P0 A (8.85 1012 C2/N  m2 )(2.827 103 m 2 )
difference is unchanged after the dielectric is inserted.
Q
25.0 1012 C
(c) Before the dielectric is inserted, E 

 1000 N/C
P0 A (8.85 1012 C2/N  m2 )(2.827 103 m2 )
Again, since the voltage is unchanged after the dielectric is inserted, the electric field is also unchanged.
V
2.00 V
EVALUATE: E  
 1000 N/C , whether or not the dielectric is present. This agrees with the result
d 2.00 103 m
in part (c). The electric field has this value at any point between the plates. We need d to calculate E because V is the
potential difference between points separated by distance d.
IDENTIFY: C  KC0 . U  12 CV 2 .
d
24.47.
SET UP: C0  12.5 F is the value of the capacitance without the dielectric present.
EXECUTE: (a) With the dielectric, C  (3.75)(12.5 F)  46.9 F .
before: U  12 C0V 2  12 (12.5  106 F)(24.0 V)2  3.60 mJ
24.48.
after: U  12 CV 2  12 (46.9  106 F)(24.0 V) 2  13.5 mJ
(b) U  13.5 mJ  3.6 mJ  9.9 mJ . The energy increased.
EVALUATE: The power supply must put additional charge on the plates to maintain the same potential difference
when the dielectric is inserted. U  12 QV , so the stored energy increases.
IDENTIFY: Gauss’s law in dielectrics has the same form as in vacuum except that the electric field is multiplied by a
factor of K and the charge enclosed by the Gaussian surface is the free charge. The capacitance of an object depends
on its geometry.
(a) SET UP: The capacitance of a parallel-plate capacitor is C  KP0 A/d and the charge on its plates is Q = CV.
Capacitance and Dielectrics
EXECUTE:
24-13
First find the capacitance:
K P0 A (2.1)(8.85  1012 C2 /N  m2 )(0.0225 m2 )

 4.18  1010 F .
d
1.00  103 m
C
Now find the charge on the plates: Q  CV  (4.18  1010 F)(12.0 V) = 5.02  109 C .
(b) SET UP: Gauss’s law within the dielectric gives KEA  Qfree /P0 .
EXECUTE: Solving for E gives
E
Qfree
5.02 109 C

 1.20 104 N/C
KAP0 (2.1)(0.0225 m2 )(8.85 1012 C2/N  m 2 )
(c) SET UP: Without the Teflon and the voltage source, the charge is unchanged but the potential increases, so
C  P0 A/d and Gauss’s law now gives EA  Q/P0 .
EXECUTE: First find the capacitance:
P0 A (8.85  1012 C2 /N  m 2 )(0.0225 m 2 )

 1.99  1010 F.
d
1.00  103 m
C
The potential difference is V 
E
24.49.
Q 5.02  109 C

 25.2 V. From Gauss’s law, the electric field is
C 1.99  1010 F
Q
5.02 109 C

 2.52 104 N/C.
12
P0 A (8.85 10 C2 /N  m2 )(0.0225 m2 )
EVALUATE: The dielectric reduces the electric field inside the capacitor because the electric field due to the dipoles
of the dielectric is opposite to the external field due to the free charge on the plates.
IDENTIFY: Apply Eq.(24.23) to calculate E. V = Ed and C = Q/V apply whether there is a dielectric between the
plates or not.
(a) SET UP: Apply Eq.(24.23) to the dashed surface in Figure 24.49:
Q
EXECUTE: ъKE  dA  encl-free
P0
ъKE  dA  KEA
since E = 0 outside the plates
Qencl-free   A   Q / A  A
Figure 24.49
Thus KEA 
 Q/A A
and E 
Q
P0 AK
P0
Qd
(b) V  Ed 
P0 AK
Q
Q
PA
 K 0  KC0 .
(c) C  
V (Qd/P0 AK )
d
24.50.
EVALUATE: Our result shows that K  C/C0 , which is Eq.(24.12).
PA
IDENTIFY: C  0 . C  Q / V . V  Ed . U  12 CV 2 .
d
SET UP: With the battery disconnected, Q is constant. When the separation d is doubled, C is halved.
P A P (0.16 m)2
EXECUTE: (a) C  0  0
 4.8  1011 F
d
4.7  103 m
(b) Q  CV  (4.8  1011 F)(12 V)  0.58  109 C
(c) E  V/d  (12 V) /(4.7  103 m)  2550 V/m
(d) U  12 CV 2  12 (4.8  1011 F)(12 V) 2  3.46  109 J
(e) If the battery is disconnected, so the charge remains constant, and the plates are pulled further apart to 0.0094 m, then
the calculations above can be carried out just as before, and we find: (a) C  2.411011 F
(b) Q  0.58 109 C
Q2 (0.58 109 C)2

 6.91109 J
2C 2(2.411011 F)
Q
EVALUATE: Q is unchanged. E 
so E is unchanged. U doubles because C is halved. The additional stored
P0 A
energy comes from the work done by the force that pulled the plates apart.
(c) E  2550 V m
(d) U 
24-14
Chapter 24
24.51.
IDENTIFY and SET UP: If the capacitor remains connected to the battery, the battery keeps the potential difference
between the plates constant by changing the charge on the plates.
PA
EXECUTE: (a) C  0
d
12
2
(8.854  10 C / N  m2 )(0.16 m) 2
C
 2.4  1011 F  24 pF
9.4  103 m
(b) Remains connected to the battery says that V stays 12 V. Q  CV  (2.4  1011 F)(12 V)  2.9  1010 C
V
12 V

 1.3 103 V/m
d 9.4 103 m
(d) U  12 QV  12 (2.9  1010 C)(12.0 V)  1.7  109 J
EVALUATE: Increasing the separation decreases C. With V constant, this means that Q decreases and U decreases.
Q decreases and E  Q / P0 A so E decreases. We come to the same conclusion from E  V / d .
(c) E 
24.52.
24.53.
24.54.
24.55.
IDENTIFY:
C  KC0  K P0
A
. V  Ed for a parallel plate capacitor; this equation applies whether or not a dielectric
d
is present.
SET UP: A  1.0 cm2  1.0 104 m2 .
(8.85  1012 F/m)(1.0  104 m2 )
EXECUTE: (a) C  (10)
 1.18  F per cm2.
7.5  109 m
V
85 mV
(b) E  
 1.13 107 V/m .
K 7.5 109 m
EVALUATE: The dielectric material increases the capacitance. If the dielectric were not present, the same charge
density on the faces of the membrane would produce a larger potential difference across the membrane.
IDENTIFY: P  E/t , where E is the total light energy output. The energy stored in the capacitor is U  12 CV 2 .
SET UP: E  0.95U
EXECUTE: (a) The power output is 600 W, and 95% of the original energy is converted, so
E  Pt  (2.70  105 W)(1.48  103 s)  400 J . E0  400 J  421 J .
0.95
2U 2(421 J)
2
1
 0.054 F .
(b) U  2 CV so C  2 
V
(125 V) 2
EVALUATE: For a given V, the stored energy increases linearly with C.
PA
IDENTIFY: C  0
d
SET UP: A  4.2 105 m2 . The original separation between the plates is d  0.700 103 m . d  is the separation
between the plates at the new value of C.
AP (4.20  105 m2 )P0
EXECUTE: C0  0 
 5.31 1013 F . The new value of C is C  C0  0.25 pF  7.81  1013 F .
d
7.00  104 m
AP (4.20  105 m 2 )P0
AP
But C  0 , so d   0 
 4.76  104 m . Therefore the key must be depressed by a distance of
C
7.81 1013 F
d
7.00  104 m  4.76  104 m  0.224 mm .
EVALUATE: When the key is depressed, d decreases and C increases.
2 P0 L
IDENTIFY: Example 24.4 shows that C 
for a cylindrical capacitor.
ln( rb / ra )
SET UP:
ln(1  x)  x when x is small. The area of each conductor is approximately A  2 ra L .
2 P0 L
2 P0 L
2 P0 L
2 ra LP0 P0 A




ln( rb ra ) ln((d  ra ) ra ) ln(1  d ra )
d
d
(b) At the scale of part (a) the cylinders appear to be flat, and so the capacitance should appear like that
EXECUTE: (a) d  ra : C 
24.56.
EVALUATE:
of flat plates.
IDENTIFY: Initially the capacitors are connected in parallel to the source and we can calculate the charges Q1 and
Q2
.
2C
SET UP: After they are reconnected, the charges add and the voltages are the same, so Ceq  C1  C2 , as for
Q2 on each. After they are reconnected to each other the total charge is Q  Q2  Q1 . U  12 CV 2 
capacitors in parallel.
Capacitance and Dielectrics
24-15
EXECUTE: Originally Q1  C1V1  (9.0  F) (28 V)  2.52  104 C and Q2  C2V2  (4.0  F)(28 V)  1.12  104 C .
Ceq  C1  C2  13.0  F . The original energy stored is U  12 CeqV 2  12 (13.0 106 F)(28 V)2  5.10 103 J .
Disconnect and flip the capacitors, so now the total charge is Q  Q2  Q1  1.4  104 C and the equivalent capacitance
is still the same, Ceq  13.0  F . The new energy stored is U 
24.57.
Q2
(1.4  104 C) 2

 7.54  104 J . The change in
2Ceq 2(13.0  106 F)
stored energy is U  7.45  104 J  5.10  103 J  4.35  103 J .
EVALUATE: When they are reconnected, charge flows and thermal energy is generated and energy is radiated as
electromagnetic waves.
IDENTIFY: Simplify the network by replacing series and parallel combinations by their equivalent. The stored
energy in a capacitor is U  12 CV 2 .
SET UP:
For capacitors in series the voltages add and the charges are the same;
1
1
1
 

Ceq C1 C2
. For capacitors
Q
. U  12 CV 2 .
V
EXECUTE: (a) Find Ceq for the network by replacing each series or parallel combination by its equivalent. The
successive simplified circuits are shown in Figure 24.57a–c.
U tot  12 CeqV 2  12 (2.19 106 F)(12.0 V)2  1.58 104 J  158 J
in parallel the voltages are the same and the charges add; Ceq  C1  C2 
C
(b) From Figure 24.57c, Qtot  CeqV  (2.19 106 F)(12.0 V)  2.63105 C. From Figure 24.57b, Qtot  2.63 105 C.
Q4.8 2.63 105 C

 5.48 V . U 4.8  12 CV 2  12 (4.80  106 F)(5.48 V) 2  7.21 105 J  72.1  J
C4.8 4.80 106 F
This one capacitor stores nearly half the total stored energy.
Q2
EVALUATE: U 
. For capacitors in series the capacitor with the smallest C stores the greatest amount of
2C
energy.
V4.8 
Figure 24.57
24.58.
IDENTIFY: Apply the rules for combining capacitors in series and parallel. For capacitors in series the voltages add
and in parallel the voltages are the same.
SET UP: When a capacitor is a moderately good conductor it can be replaced by a wire and the potential across it is zero.
EXECUTE: (a) A network that has the desired properties is sketched in Figure 24.58a. Ceq  C  C  C . The total
2 2
capacitance is the same as each individual capacitor, and the voltage is spilt over each so that V  480 V.
(b) If one capacitor is a moderately good conductor, then it can be treated as a “short” and thus removed from the
circuit, and one capacitor will have greater than 600 V across it.
EVALUATE: An alternative solution is two in parallel in series with two in parallel, as sketched in Figure 24.58b.
Figure 24.58
24-16
Chapter 24
24.59.
(a) IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents.
SET UP: The network is sketched in Figure 24.59a.
C1  C5  8.4  F
C2  C3  C4  4.2 F
Figure 24.59a
EXECUTE: Simplify the circuit by replacing the capacitor combinations by their equivalents: C3 and C4 are in
series and can be replaced by C34 (Figure 24.59b):
1
1
1


C34 C3 C4
1
C  C4
 3
C34
C3C4
Figure 24.59b
C3C4
 4.2 F 4.2 F  2.1 F

C3  C4
4.2  F  4.2  F
C2 and C34 are in parallel and can be replaced by their equivalent (Figure 24.59c):
C34 
C234  C2  C34
C234  4.2 F  2.1 F
C234  6.3  F
Figure 24.59c
C1 , C5 and C234 are in series and can be replaced by Ceq (Figure 24.59d):
1
1
1
1
 

Ceq C1 C5 C234
1
2
1


Ceq 8.4  F 6.3  F
Ceq  2.5  F
Figure 24.59d
EVALUATE: For capacitors in series the equivalent capacitor is smaller than any of those in series. For capacitors in
parallel the equivalent capacitance is larger than any of those in parallel.
(b) IDENTIFY and SET UP: In each equivalent network apply the rules for Q and V for capacitors in series and
parallel; start with the simplest network and work back to the original circuit.
EXECUTE: The equivalent circuit is drawn in Figure 24.59e.
Qeq  CeqV
Qeq   2.5  F  220 V   550  C
Figure 24.59e
Q1  Q5  Q234  550 C (capacitors in series have same charge)
V1 
Q1 550  C

 65 V
C1 8.4  F
V5 
Q5 550  C

 65 V
C5 8.4  F
V234 
Q234 550  C

 87 V
C234 6.3  F
Capacitance and Dielectrics
24-17
Now draw the network as in Figure 24.59f.
V2  V34  V234  87 V
capacitors in parallel have the same potential
Figure 24.59f
Q2  C2V2   4.2  F 87 V   370 C
Q34  C34V34   2.1  F  87 V   180 C
Finally, consider the original circuit (Figure 24.59g).
Q3  Q4  Q34  180 C
capacitors in series have the same charge
Figure 24.59g
V3 
Q3 180  C

 43 V
C3 4.2  F
V4 
Q4 180  C

 43 V
C4 4.2  F
Summary: Q1  550 C, V1  65 V
Q2  370 C, V2  87 V
Q3  180 C, V3  43 V
Q4  180 C, V4  43 V
Q5  550 C, V5  65 V
EVALUATE:
24.60.
V3  V4  V2 and V1  V2  V5  220 V (apart from some small rounding error)
Q1  Q2  Q3 and Q5  Q2  Q4
IDENTIFY: Apply the rules for combining capacitors in series and in parallel.
SET UP: With the switch open each pair of 3.00  F and 6.00  F capacitors are in series with each other and each
pair is in parallel with the other pair. When the switch is closed each pair of 3.00  F and 6.00  F capacitors are in
parallel with each other and the two pairs are in series.
1
1



 
EXECUTE: (a) With the switch open Ceq    1  1    1  1    4.00  F .
  3 F 6 F 
 3  F 6  F  

Qtotal  CeqV  (4.00 F) (210 V)  8.40 104 C . By symmetry, each capacitor carries 4.20 104 C. The
voltages are then calculated via V  Q / C . This gives Vad  Q / C3  140 V and Vac  Q / C6  70 V .
Vcd  Vad  Vac  70 V .
(b) When the switch is closed, the points c and d must be at the same potential, so the equivalent capacitance is
1


1
1
4
Ceq  

  4.5  F . Qtotal  CeqV  (4.50  F)(210 V)  9.5 10 C , and each
 (3.00  6.00)  F (3.00  6.00)  F 
capacitor has the same potential difference of 105 V (again, by symmetry).
(c) The only way for the sum of the positive charge on one plate of C2 and the negative charge on one plate of
C1 to change is for charge to flow through the switch. That is, the quantity of charge that flows through the
switch is equal to the change in Q2  Q1 . With the switch open, Q1  Q2 and Q2  Q1  0. After the switch is
closed, Q2  Q1  315 C , so 315 C of charge flowed through the switch.
EVALUATE: When the switch is closed the charge must redistribute to make points c and d be at the same
potential.
24-18
Chapter 24
24.61.
(a) IDENTIFY: Replace the three capacitors in series by their equivalent. The charge on the equivalent capacitor
equals the charge on each of the original capacitors.
SET UP: The three capacitors can be replaced by their equivalent as shown in Figure 24.61a.
Figure 24.61a
EXECUTE:
C3  C1 / 2 so
1
1
1
1
4
and Ceq  8.4  F/4  2.1  F
 


Ceq C1 C2 C3 8.4  F
Q  CeqV   2.1  F  36 V   76  C
The three capacitors are in series so they each have the same charge: Q1  Q2  Q3  76 C
EVALUATE: The equivalent capacitance for capacitors in series is smaller than each of the original capacitors.
(b) IDENTIFY and SET UP: Use U  12 QV . We know each Q and we know that V1  V2  V3  36 V.
EXECUTE:
1
1
U  12 QV
1 1  2 Q2V2  2 Q3V3
But Q1  Q2  Q3  Q so U  12 Q(V1  V2  V3 )
But also V1  V2  V3  V  36 V, so U  12 QV  12 (76  C)(36 V)  1.4  103 J.
EVALUATE: We could also use U  Q 2 / 2C and calculate U for each capacitor.
(c) IDENTIFY: The charges on the plates redistribute to make the potentials across each capacitor the same.
SET UP: The capacitors before and after they are connected are sketched in Figure 24.61b.
Figure 24.61b
EXECUTE: The total positive charge that is available to be distributed on the upper plates of the three capacitors is
Q0  Q01  Q02  Q03  3 76  C   228  C. Thus Q1  Q2  Q3  228 C. After the circuit is completed the charge
distributes to make V1  V2  V3. V  Q / C and V1  V2 so Q1 / C1  Q2 / C2 and then C1  C2 says Q1  Q2. V1  V3 says
Q1 / C1  Q3 / C3 and Q1  Q3  C1 / C3   Q3 8.4  F/4.2  F   2Q3
Using Q2  Q1 and Q1  2Q3 in the above equation gives 2Q3  2Q3  Q3  228 C.
5Q3  228 C and Q3  45.6 C, Q1  Q2  91.2 C
Q1 91.2  C
Q 91.2  C
Q 45.6  C

 11 V, V2  2 
 11 V, and V3  3 
 11 V.
C1 8.4  F
C2
8.4  F
C3
4.2  F
The voltage across each capacitor in the parallel combination is 11 V.
1
1
(d) U  12 QV
1 1  2 Q2V2  2 Q3V3 .
Then V1 
But V1  V2  V3 so U  12 V1 Q1  Q2  Q3   12 11 V 228 C  1.3 103 J.
24.62.
EVALUATE: This is less than the original energy of 1.4 103 J. The stored energy has decreased, as in
Example 24.7.
PA
Q
IDENTIFY: C  0 . C  . V  Ed . U  12 QV .
d
V
SET UP: d  3.0 103 m . A   r 2 , with r  1.0 103 m .
P A (8.854  1012 C2 /N  m2 ) (1.0 103 m) 2
EXECUTE: (a) C  0 
 9.3  109 F .
d
3.0  103 m
Q
20 C
(b) V  
 2.2 109 V
C 9.3 109 F
V 2.2  109 V
(c) E  
 7.3  105 V/m
d 3.0  103 m
(d) U  12 QV  12 (20 C)(2.2  109 V)  2.2  1010 J
EVALUATE: Thunderclouds involve very large potential differences and large amounts of stored energy.
Capacitance and Dielectrics
24.63.
24-19
IDENTIFY: Replace series and parallel combinations of capacitors by their equivalents. In each equivalent network
apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the
original circuit.
(a) SET UP: The network is sketched in Figure 24.63a.
C1  6.9  F
C2  4.6  F
Figure 24.63a
EXECUTE: Simplify the network by replacing the capacitor combinations by their equivalents. Make the
replacement shown in Figure 24.63b.
1
3

Ceq C1
Ceq 
C1 6.9  F

 2.3  F
3
3
Figure 24.63b
Next make the replacement shown in Figure 24.63c.
Ceq  2.3  F  C2
Ceq  2.3  F  4.6  F  6.9  F
Figure 24.63c
Make the replacement shown in Figure 24.63d.
1
2
1
3
 

Ceq C1 6.9  F 6.9  F
Ceq  2.3  F
Figure 24.63d
Make the replacement shown in Figure 24.63e.
Ceq  C2  2.3  F  4.6  F  2.3  F
Ceq  6.9  F
Figure 24.63e
Make the replacement shown in Figure 24.63f.
1
2
1
3
 

Ceq C1 6.9  F 6.9  F
Ceq  2.3  F
Figure 24.63f
(b) Consider the network as drawn in Figure 24.63g.
From part (a) 2.3  F is the equivalent
capacitance of the rest of the network.
Figure 24.63g
24-20
Chapter 24
The equivalent network is shown in Figure 24.63h.
The capacitors are in series,
so all three capacitors have
the same Q.
Figure 24.63h
But here all three have the same C, so by V = Q/C all three must have the same V. The three voltages must add
to 420 V, so each capacitor has V = 140 V. The 6.9  F to the right is the equivalent of C2 and the 2.3  F
capacitor in parallel, so V2  140 V. (Capacitors in parallel have the same potential difference.) Hence
Q1  C1V1  (6.9  F)(140 V)  9.7  104 C and Q2  C2V2  (4.6  F)(140 V)  6.4  104 C.
(c) From the potentials deduced in part (b) we have the situation shown in Figure 24.63i.
From part (a) 6.9  F is the
equivalent capacitance of the
rest of the network.
Figure 24.63i
24.64.
The three right-most capacitors are in series and therefore have the same charge. But their capacitances are also equal,
so by V = Q/C they each have the same potential difference. Their potentials must sum to 140 V, so the potential
across each is 47 V and Vcd  47 V.
EVALUATE: In each capacitor network the rules for combining V for capacitors in series and parallel are obeyed.
Note that Vcd  V , in fact V  2(140 V)  2(47 V)  Vcd .
IDENTIFY: Find the total charge on the capacitor network when it is connected to the battery. This is the amount of
charge that flows through the signal device when the switch is closed.
SET UP: For capacitors in parallel, Ceq  C1  C2  C3 
EXECUTE:
24.65.
Cequiv  C1  C2  C3  60.0  F . Q  CV  (60.0 F)(120 V)  7200  C .
EVALUATE: More charge is stored by the three capacitors in parallel than would be stored in each capacitor used
alone.
(a) IDENTIFY and SET UP: Q is constant. C  KC0 ; use Eq.(24.1) to relate the dielectric constant K to the ratio of
the voltages without and with the dielectric.
EXECUTE: With the dielectric: V  Q / C  Q /  KC0 
without the dielectric: V0  Q / C0
V0 / V  K , so K   45.0 V  / 11.5 V   3.91
EVALUATE: Our analysis agrees with Eq.(24.13).
(b) IDENTIFY: The capacitor can be treated as equivalent to two capacitors C1 and C2 in parallel, one with
area 2A/3 and air between the plates and one with area A/3 and dielectric between the plates.
SET UP: The equivalent network is shown in Figure 24.65.
Figure 24.65
EXECUTE: Let C0  P0 A / d be the capacitance with only air between the plates. C1  KC0 / 3, C2  2C0 / 3;
Ceq  C1  C2  (C0 / 3)( K  2)
V
Q
Q 3 
 3 
 3 
 
  V0 
   45.0 V  
  22.8 V
Ceq C0  K  2 
K

2


 5.91 
EVALUATE: The voltage is reduced by the dielectric. The voltage reduction is less when the dielectric doesn’t
completely fill the volume between the plates.
Capacitance and Dielectrics
24.66.
24-21
IDENTIFY: This situation is analogous to having two capacitors C1 in series, each with separation 12 (d  a).
SET UP:
For capacitors in series,
1
1
1
.
 
Ceq C1 C2
1
24.67.
P0 A
PA


EXECUTE: (a) C   1  1   12 C1  12
 0
C
C
(
d

a
)
2
d
a
1
 1
PA PA d
d
(b) C  0  0
 C0
d a d d a
d a
(c) As a  0 , C  C0 . The metal slab has no effect if it is very thin. And as a  d , C   . V  Q / C . V  Ey is
the potential difference between two points separated by a distance y parallel to a uniform electric field. When the
distance is very small, it takes a very large field and hence a large Q on the plates for a given potential difference.
Since Q  CV this corresponds to a very large C.
(a) IDENTIFY: The conductor can be at some potential V, where V = 0 far from the conductor. This potential
depends on the charge Q on the conductor so we can define C = Q/V where C will not depend on V or Q.
(b) SET UP: Use the expression for the potential at the surface of the sphere in the analysis in part (a).
EXECUTE: For any point on a solid conducting sphere V  Q / 4 P0 R if V  0 at r  .
C
 4 P0 R 
Q
 Q
  4 P0 R
V
 Q 
(c) C  4 P0 R  4 8.854  1012 F/m  6.38  106 m   7.10  10 4 F  710  F.
24.68.
EVALUATE: The capacitance of the earth is about seven times larger than the largest capacitances in this range. The
capacitance of the earth is quite small, in view of its large size.
Q2
IDENTIFY: The electric field energy density is 12 P0 E 2 . For a capacitor U 
.
2C
Q
SET UP: For a solid conducting sphere of radius R, E  0 for r  R and E 
for r  R .
4 P0 r 2
EXECUTE: (a) r  R : u  12 P0 E 2  0.
2
 Q 
Q2

.
(b) r  R : u  PE  P 
2 
2
4
 4 P0 r  32 P0 r
1
2 0
2

1
2 0
(c) U   udV  4  r 2udr 
R
24.69.

Q 2 dr
Q2

.
2

8 P0 R r
8 P0 R
Q2
(d) This energy is equal to 1
which is just the energy required to assemble all the charge into a spherical
2 4 P0 R
distribution. (Note that being aware of double counting gives the factor of 1/2 in front of the familiar potential energy
formula for a charge Q a distance R from another charge Q.)
Q2
Q2
EVALUATE: (e) From Equation (24.9), U 
.U
from part (c) , C  4 P0 R , as in Problem (24.67).
2C
8 P0 R
IDENTIFY: We model the earth as a spherical capacitor.
rr
SET UP: The capacitance of the earth is C  4 P0 a b and, the charge on it is Q = CV, and its stored energy is
rb  ra
U  12 CV 2 .
EXECUTE: (a) C 
 6.38 106 m  6.45 106 m   6.5 102 F
1
9.00  109 N  m 2 /C2 6.45 106 m  6.38 106 m
(b) Q  CV   6.54  102 F  (350,000 V) = 2.3  104 C
(c) U  12 CV 2  12  6.54  102 F  (350,000 V)2  4.0  109 J
EVALUATE: While the capacitance of the earth is larger than ordinary laboratory capacitors, capacitors much larger
than this, such as 1 F, are readily available.
24-22
Chapter 24
24.70.
IDENTIFY: The electric field energy density is u  12 P0 E 2 . U 
SET UP:
For this charge distribution, E  0 for r  ra , E 
Example 24.4 shows that
Q2
.
2C

for ra  r  rb and E  0 for r  rb .
2 P0 r
U
2 P0

for a cylindrical capacitor.
L ln( rb / ra )
2
  
2
EXECUTE: (a) u  12 P0 E 2  12 P0 
  2 2
 2 P0 r  8 P0 r
r
(b) U   udV  2 L  urdr 
U
2
L 2 b dr
and

ln(rb / ra ) .
4 P0 ra r
L 4 P0
Q2
Q2
 2L

ln(rb / ra ) 
ln(rb / ra ) . This agrees with the result of part (b).
2C 4 P0 L
4 P0
Q2
EVALUATE: We could have used the results of part (b) and U 
to calculate U / L and would obtain the same
2C
result as in Example 24.4.
IDENTIFY: C  Q /V , so we need to calculate the effect of the dielectrics on the potential difference between the
plates.
SET UP: Let the potential of the positive plate be Va , the potential of the negative plate be Vc , and the potential
(c) Using Equation (24.9), U 
24.71.
midway between the plates where the dielectrics meet be Vb , as shown in Figure 24.71.
C
Q
Q

.
Va  Vc Vac
Vac  Vab  Vbc .
Figure 24.71
EXECUTE: The electric field in the absence of any dielectric is E0 
reduced to E1 
E2 
Q
. In the first dielectric the electric field is
P0 A
E0
Q
Qd
d 

and Vab  E1   
. In the second dielectric the electric field is reduced to
K1 K1P0 A
2
K
 
1 2P
0A
Qd
Qd
Qd  1
1 
E0
Q
Qd
d 



and Vbc  E2   
. Thus Vac  Vab  Vbc 
 
.
K1 2P0 A K 2 2P0 A 2P0 A  K1 K 2 
K2 K2P0 A
 2  K2 2P0 A
 2P A   K K  2P A  K K 
Qd  K1  K 2 
Q
 Q  0   1 2   0  1 2 .

 . This gives C 
2P0 A  K1K 2 
Vac
d  K1  K 2 
 Qd  K1  K 2 
EVALUATE: An equivalent way to calculate C is to consider the capacitor to be two in series, one with dielectric
constant K1 and the other with dielectric constant K 2 and both with plate separation d/2. (Can imagine inserting a
thin conducting plate between the dielectric slabs.)
PA
PA
C1  K1 0  2K1 0
d /2
d
P0 A
P0 A
C2  K 2
 2K 2
d /2
d
1 1
1
CC
2P A  K K 
 
so C  1 2  0  1 2 
Since they are in series the total capacitance C is given by
C C1 C2
C1  C2
d  K1  K 2 
Vac 
24.72.
IDENTIFY: This situation is analogous to having two capacitors in parallel, each with an area A/2.
SET UP:
For capacitors in parallel, Ceq  C1  C2 . For a parallel-plate capacitor with plates of area A/2, C 
EXECUTE:
EVALUATE:
P0 A / 2 P0 A / 2 P0 A


( K1  K2 )
d
d
2d
PA
If K1  K 2 , Ceq  K 0 , which is Eq.(24.19).
d
Ceq  C1  C2 
P0 (A/2)
.
d
Capacitance and Dielectrics
24.73.
24-23
IDENTIFY and SET UP: Show the transformation from one circuit to the other:
Figure 24.73a
EXECUTE: (a) Consider the two networks shown in Figure 24.73a. From Circuit 1: Vac 
q  q3
q1  q3
and Vbc  2
.
Cx
Cy
q
CxC yCz  q1 q2 
q3 q1  q3 q2  q3
q 
. This gives q3 
  K 1  2 .






Cx  C y  Cz  C y Cx 
Cz
Cy
Cx
 C y Cx 
 1
 1
q q q
1 
1
q
q q
1
1 
From Circuit 2: Vac  1  1 2  q1     q2
and Vbc  2  1 2  q1  q2    . Setting the
C1
C3
C3
C2
C3
C3
 C1 C3 
 C2 C3 
coefficients of the charges equal to each other in matching potential equations from the two circuits results in three
1
1 
1
1 
independent equations relating the two sets of capacitances. The set of equations are


1 
,
C1 C y  KC y KCx 
1
1
1
1 
1
1 
. From these, subbing in the expression for K , we get
 1 


 and

C2 Cx  KC y KCx 
C3 KC yCx
C1  (CxC y  C yCz  CzCx ) Cx , C2  (CxC y  C yCz  CzCx ) C y and C3  (CxC y  C yCz  CzCx ) Cz .
q3 is derived from Vab : Vab 
(b) Using the transformation of part (a) we have the equivalent networks shown in Figure 24.73b:
Figure 24.73b
C1  126  F , C2  28  F , C3  42  F , C4  42  F , C5  147  F and C6  32  F . The total equivalent capacitance
 1
1
1
1
1 
is Ceq  





72

F
126

F
34.8

F
147

F
72
F 

1
1
 1
 1
1 
1  
34.8  F   




  .
  42  F 32  F 
 28  F 42  F  

1
 14.0  F, where the 34.8 F comes from
24-24
Chapter 24
(c) The circuit diagram can be redrawn as shown in Figure 25.73c. The overall charge is given by
Q  CeqV  (14.0 F)(36 V)  5.04 104 C . And this is also the charge on the 72 F capacitors, so
V72 
5.04 104 C
 7.0 V .
72 106 F
Figure 24.73c
Next we will find the voltage over the numbered capacitors, and their associated voltages. Then those voltages
will be changed back into voltage of the original capacitors, and then their charges. QC1  QC5  Q72  5.04 104 C .
VC5 
5.04  104 C
5.04 104 C
 3.43 V and VC1 
 4.00 V . Therefore,
6
147  10 F
126 106 F
1
 1
1 
VC2C4  VC3C8  (36.0  7.00  7.00  4.00  3.43) V  14.6 V . But Ceq (C2C4 )  
   16.8  F and
C
C
4 
 2
1
 1
1 
Ceq (C3C6 )      18.2  F , so QC2  QC4  VC2C4 Ceq(C2C4 )  2.45 104 C and
 C3 C6 
QC
QC
QC
QC3  QC6  VC3C6 Ceq(C3C6 )  2.64 104 C . Then VC2  2  8.8 V , VC3  3  6.3 V , VC4  4  5.8 V and
C2
C3
C4
QC
VC6  6  8.3 V . Vac  VC1  VC2  V18  13 V and Q18  C18V18  2.3  104 C . Vab  VC1  VC3  V27  10 V and
C6
Q27  C27V27  2.8  104 C . Vcd  VC4  VC5  V28  9 V and Q28  C28V28  2.6  104 C . Vbd  VC5  VC6  V21  12 V
and Q21  C21V21  2.5  104 C . Vbc  VC3  VC2  V6  2.5 V and Q6  C6V6  1.5  105 C .
24.74.
EVALUATE: Note that 2V72  V18  V28  2(7.0 V)  13 V  9 V  36 V, as it should.
IDENTIFY: The force on one plate is due to the electric field of the other plate. The electrostatic force must be
balanced by the forces from the springs.

SET UP: The electric field due to one plate is E 
. The force exerted by a spring compressed a distance
2P0
z0  z from equilibrium is k ( z0  z ) .
q
q2
(CV )2 P02 A2 V 2
P AV 2


 2
 0 2 .
2P0 2P0 A 2P0 A
z 2P0 A
2z
(b) When V  0, the separation is just z0 . When V  0 , the total force from the four springs must equal the
EXECUTE: (a) The force between the two parallel plates is F  qE 
P0 AV 2
P AV 2
0.
and 2 z 3  2 z 3 z0  0
2
2z
4k
(c) For A  0.300 m 2 , z0  1.2  103 m , k  25 N/m and V  120 V , so 2z 3  (2.4  103 m)z 2  3.82  1010 m3  0 .
The physical solutions to this equation are z  0.537 mm and 1.014 mm.
EVALUATE: (d) Stable equilibrium occurs if a slight displacement from equilibrium yields a force back toward the
equilibrium point. If one evaluates the forces at small displacements from the equilibrium positions above, the
1.014 mm separation is seen to be stable, but not the 0.537 mm separation.
electrostatic force calculated in part (a). F4 springs  4k ( z0  z ) 
Capacitance and Dielectrics
24.75.
24-25
IDENTIFY: The system can be considered to be two capacitors in parallel, one with plate area L( L  x) and air
between the plates and one with area Lx and dielectric filling the space between the plates.
KP A
SET UP: C  0 for a parallel-plate capacitor with plate area A.
d
P
PL
EXECUTE: (a) C  0 (( L  x) L  xKL)  0 ( L  ( K  1) x)
D
D
P0 L
PL
2
1
(b) dU  2 (dC )V , where C  C0 
(dx  dxK ) , with C0  0 ( L  ( K  1) x) . This gives
D
D
2
P
Ldx
(
K

1)
P
V
L


0
dU  12  0
( K  1) V 2 
dx .
2D
 D

C
PLV
(c) If the charge is kept constant on the plates, then Q  0
( L  ( K  1) x) and U  12 CV 2  12 C0V 2   .
D
 C0 
U

C0V 2 
P0 L
( K  1)P0V 2 L
( K  1)dx  and U  U  U 0  
dx .
1 
2  DC0
2D

( K  1)P0V 2 L
dx , the force is in the opposite direction to the motion dx, meaning that the
2D
slab feels a force pushing it out.
EVALUATE: (e) When the plates are connected to the battery, the plates plus slab are not an isolated system. In
addition to the work done on the slab by the charges on the plates, energy is also transferred between the battery and
( K  1)P0V 2 L
the plates. Comparing the results for dU in part (c) to dU   Fdx gives F 
.
2D
IDENTIFY: C  Q / V . Apply Gauss's law and the relation between potential difference and electric field.
(d) Since dU   Fdx  
24.76.
r
r
r0
ra
SET UP: Each conductor is an equipotential surface. Va  Vb   b EU  d r   b EL  d r , so EU  EL , where these
are the fields between the upper and lower hemispheres. The electric field is the same in the air space as in the
dielectric.
 rr 
EXECUTE: (a) For a normal spherical capacitor with air between the plates, C0  4 P0  a b  . The capacitor in
 rb  ra 
this problem is equivalent to two parallel capacitors, C L and C U , each with half the plate area of the normal
capacitor. CL 
 rr 
 rr 
 rr 
KC0
C
 2 K P0  a b  and CU  0  2 P0  a b  . C  CU  CL  2 P0 (1  K )  a b  .
2
2
 rb  ra 
 rb  ra 
 rb  ra 
(b) Using a hemispherical Gaussian surface for each respective half, EL
QL
4 r 2 QL
, so EL 
, and

2

K P0 r 2
2
K P0
QU
VC K
4 r 2 QU
, so EU 
. But QL  VCL and QU  VCU . Also, QL  QU  Q . Therefore, QL  0  KQU

2
2 P0 r
2
2
P0
KQ
1
2
Q
Q
KQ

and QU 
, QL 
. This gives EL 
and
1  K 2 K P0 r 2 1  K 4 P0 r 2
1 K
1 K
Q
1
2
Q
EU 

. We do find that EU  EL .
2
1  K 2 K P0 r
1  K 4 K P0 r 2
EU
(c) The free charge density on upper and lower hemispheres are: ( f,ra ) U 
( f , rb ) U 
QU
Q

and
2
2
2 ra
2 ra (1  K )
QU
Q
Q
KQ
Q
KQ

.
; ( f,ra ) L  L 2 
and ( f,rb ) L  L 2 
2
2
2
2 rb
2 rb (1  K )
2 ra
2 ra (1  K )
2 rb
2 rb 2 (1  K )
 ( K  1)  Q  K   K  1  Q
(d)  i,ra   f,ra (1  1 K )  



2
2
 K  2 ra  K  1   K  1  2 ra
 ( K  1)  Q  K   K  1  Q
 i,rb   f,rb (1  1 K )  



2
2
 K  2 rb  K  1   K  1  2 rb
(e) There is zero bound charge on the flat surface of the dielectric-air interface, or else that would imply a
circumferential electric field, or that the electric field changed as we went around the sphere.
EVALUATE: The charge is not equally distributed over the surface of each conductor. There must be more charge on
the lower half, by a factor of K, because the polarization of the dielectric means more free charge is needed on the
lower half to produce the same electric field.
24-26
Chapter 24
24.77.
IDENTIFY: The object is equivalent to two identical capacitors in parallel, where each has the same area A, plate
separation d and dielectric with dielectric constant K.
PA
SET UP: For each capacitor in the parallel combination, C  0 .
d
EXECUTE: (a) The charge distribution on the plates is shown in Figure 24.77.
2
 P A  2(4.2)P0 (0.120 m)
(b) C  2  0  
 2.38 109 F .
4.5 104 m
 d 
EVALUATE:
If two of the plates are separated by both sheets of paper to form a capacitor, C 
P0 A 2.38  109 F
,

2d
4
smaller by a factor of 4 compared to the capacitor in the problem.
Figure 24.77
24.78.
IDENTIFY: As in Problem 24.72, the system is equivalent to two capacitors in parallel. One of the capacitors has
plate separation d, plate area w( L  h) and air between the plates. The other has the same plate separation d, plate area
wh and dielectric constant K.
K PA
SET UP: Define K eff by Ceq  eff 0 , where A  wL . For two capacitors in parallel, Ceq  C1  C2 .
d
Pw( L  h) K P0wh P0wL  Kh h 
EXECUTE: (a) The capacitors are in parallel, so C  0


  . This gives
1 
d
d
d 
L L
 Kh h 
Keff  1 
 .
L L

(b) For gasoline, with K  1.95:
1
1
L
L


full: Keff  h    1.24 ; full: Keff  h    1.48 ;
4
2
4
2



3
3L 

full: Keff  h 
  1.71.
4
4 

1
1
3
3L 
L
L



full: Keff  h    9 ;
full: Keff  h    17 ; full: Keff  h 
  25.
4
2
4
4 
4
2



(d) This kind of fuel tank sensor will work best for methanol since it has the greater range of K eff values.
(c) For methanol, with K  33:
EVALUATE: When h  0 , Keff  1 . When h  L , Keff  K .