CHAPTER 25
SOLUTION FOR PROBLEM 15
(a) After the switches are closed, the potential differences across the capacitors are the same and
the two capacitors are in parallel. The potential difference from a to b is given by Vab = Q/Ceq ,
where Q is the net charge on the combination and Ceq is the equivalent capacitance.
The equivalent capacitance is Ceq = C1 + C2 = 4.0 × 10−6 F. The total charge on the combination
is the net charge on either pair of connected plates. The charge on capacitor 1 is
q1 = C1 V = (1.0 × 10−6 F)(100 V) = 1.0 × 10−4 C
and the charge on capacitor 2 is
q2 = C2 V = (3.0 × 10−6 F)(100 V) = 3.0 × 10−4 C ,
so the net charge on the combination is 3.0×10−4 C−1.0×10−4 C = 2.0×10−4 C. The potential
difference is
2.0 × 10−4 C
Vab =
= 50 V .
4.0 × 10−6 F
(b) The charge on capacitor 1 is now q1 = C1 Vab = (1.0 × 10−6 F)(50 V) = 5.0 × 10−5 C.
(c) The charge on capacitor 2 is now q2 = C2 Vab = (3.0 × 10−6 F)(50 V) = 1.5 × 10−4 C.
CHAPTER 25
SOLUTION FOR PROBLEM 23
(a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitor
is given by 0 A/d, the charge is q = CV = 0 AV /d. After the plates are pulled apart, their
separation is d and the potential difference is V . Then q = 0 AV /d and
V =
d
d 0A
d
8.00 mm
q=
V = V =
(6.00 V) = 16.0 V .
d
3.00 mm
0A
0A d
(b) The initial energy stored in the capacitor is
1
Ui = CV 2 =
2
0 AV
2d
2
=
(8.85 × 10−12 F/m)(8.50 × 10−4 m2 )(6.00 V)
= 4.51 × 10−11 J
−3
2(3.00 × 10 mm)
and the final energy stored is
1
1 0A
(8.85 × 10−12 F/m)(8.50 × 10−4 m2 )(16.0 V)
Uf = C (V )2 =
= 1.20×10−10 J .
(V )2 =
2
2 d
2(8.00 × 10−3 mm)
(c) The work done to pull the plates apart is the difference in the energy: W = Uf − Ui =
1.20 × 10−10 J − 4.51 × 10−11 J = 7.49 × 10−11 J.
CHAPTER 25
SOLUTION FOR PROBLEM 33
(a) The electric field in the region between the plates is given by E = V /d, where V is the
potential difference between the plates and d is the plate separation. The capacitance is given by
C = κ 0 A/d, where A is the plate area and κ is the dielectric constant, so d = κ 0 A/C and
E=
(50 V)(100 × 10−12 F)
VC
=
= 1.0 × 104 V/m .
κ 0 A 5.4(8.85 × 10−12 F/m)(100 × 10−4 m2 )
(b) The free charge on the plates is qf = CV = (100 × 10−12 F)(50 V) = 5.0 × 10−9 C.
(c) The electric field is produced by both the free and induced charge. Since the field of a large
uniform layer of charge is q/2 0 A, the field between the plates is
E=
qf
qf
qi
qi
+
−
−
,
2 0A 2 0A 2 0A 2 0A
where the first term is due to the positive free charge on one plate, the second is due to the
negative free charge on the other plate, the third is due to the positive induced charge on one
dielectric surface, and the fourth is due to the negative induced charge on the other dielectric
surface. Note that the field due to the induced charge is opposite the field due to the free charge,
so the fields tend to cancel. The induced charge is therefore
qi = qf −
0 AE
= 5.0 × 10−9 C − (8.85 × 10−12 F/m)(100 × 10−4 m2 )(1.0 × 104 V/m)
= 4.1 × 10−9 C = 4.1 nC .
CHAPTER 25
HINT FOR PROBLEM 9
C1 and C2 are in parallel and the combination is in series with C3 . First find the equivalent
capacitance C12 of C1 and C2 , then the equivalent capacitance of C12 and C3 .
J
ans: 3.16 µF
o
CHAPTER 25
HINT FOR PROBLEM 14
C3 and C5 are in series and this combination is in parallel with C2 and C4 . C1 and C6 are in
parallel and this combination is in series with the combination of C3 , C5 , C2 , and C4 . The charge
stored by the equivalent capacitor is Ceq V . Find the charge and potential difference of any of the
capacitors by remembering that the potential differences are the same across two capacitors in
parallel and that the charges are the same for two capacitors in series. In addition, the potential
difference for a parallel connection is the same as the potential difference across the equivalent
capacitor and the charge on each capacitor of a series connection is the same as the charge for
the equivalent capacitor.
J
ans: (a) 3.00
o µF; (b) 60 µC; (c) 10 V; (d) 30.0 µC; (e) 10 V; (f) 20.0 µC; (g) 5.00 V;
(h) 20.0 µC
CHAPTER 25
HINT FOR PROBLEM 19
The energy stored in a capacitor is 12 CV 2 , where C is its capacitance and V is the potential
difference across its plates. You need to convert 10 kW · h to joules.
J
ans: 72 F
o