CHAPTER 24 – Capacitance, Dielectrics, Electric Energy Storage
5.
From Q = CV, we see that
ΔQ = C ΔV;
16 μC = C(48 V – 28 V), which gives C =
0.80 μF.
13. The uniform electric field between the plates is related to the potential difference across the plates:
E = V/d.
For a parallel-plate capacitor, we have
Qmax = CVmax = (ε0A/d)(Emaxd) = ε0AEmax
= (8.85 × 10–12 C2/N · m2)(8.5 × 10–4 m2)(3.0 × 106 V/m) = 2.3 × 10–8 C = 23 nC.
17. We find the potential difference across the plates from
Q = CV;
72 μC = (0.80 μF)V, which gives V = 90 V.
We find the uniform electric field between the plates from
4.5 × 104 V/m.
E = V/d = (90 V)/(2.0 × 10–3 m) =
20. We find the equivalent capacitance for a parallel connection from
10.8 μF.
Cparallel = ∑Ci = 6C1 = 6(1.8 μF) =
When the capacitors are connected in series, we find the equivalent capacitance from
1/Cseries = ∑(1/Ci ) = 6/C1 = 6/(1.8 μF), which gives Cseries =
0.30 μF.
.
(a) From the circuit, we see that C2 and C3 are in series
and find their equivalent capacitance from
1/C4 = (1/C2) + (1/C3), which gives C4 = C2C3/(C2 + C3).
From the new circuit, we see that C1 and C4 are in parallel,
with an equivalent capacitance
Ceq = C1 + C4 = C1 + [C2C3/(C2 + C3)]
=
(C1C2 + C1C3 + C2C3)/(C2 + C3).
(b) Because V is across C1 , we have
Q1 = C1V = (14.0 μF)(25.0 V) =
350 μC.
Because C2 and C3 are in series, the charge on each is the
charge on their equivalent capacitance:
Q2 = Q3 = C4V = [C2C3/(C2 + C3)]V
= [(14.0 μF)(7.00 μF)/(14.0 μF + 7.00 μF)](25.0 V) = 117 μC.
23. (a) From the circuit, we see that C2 and C3 are in series
and find their equivalent capacitance from
1/C4 = (1/C2) + (1/C3), which gives C4 = C2C3/(C2 + C3).
From the new circuit, we see that C1 and C4 are in parallel,
with an equivalent capacitance
Ceq = C1 + C4 = C1 + [C2C3/(C2 + C3)]
=
(C1C2 + C1C3 + C2C3)/(C2 + C3).
(b) Because V is across C1 , we have
Q1 = C1V = (14.0 μF)(25.0 V) =
350 μC.
Because C2 and C3 are in series, the charge on each is the
charge on their equivalent capacitance:
Q2 = Q3 = C4V = [C2C3/(C2 + C3)]V
= [(14.0 μF)(7.00 μF)/(14.0 μF + 7.00 μF)](25.0 V) =
C1
C2
a
C3
c
b
V
C1
117 μC.
C4
a
27. The capacitance increases with a parallel connection, so the maximum capacitance
is
Cmax = C1 + C2 + C3
V
b
= 3000 pF + 5000 pF + 0.010 μF = 3.0 nF + 5.0 nF + 10 nF =
18 nF (parallel).
The capacitance decreases with a series connection, so we find the minimum capacitance from
1/Cmin = (1/C1) + (1/C2) + (1/C3) = [1/(3000 pF)] + [1/(5000 pF)] + [1/(0.010 μF)]
1.6 nF (series).
= [1/(3.0 nF)] + [1/(5.0 nF)] + [1/(10 nF)], which gives Cmin =
29. (a) From the circuit, we see that C1 and C2 are in series
and find their equivalent capacitance from
1/C5 = (1/C1) + (1/C2) = (1/C) + (1/C), which gives C5 = C/2.
From the new circuit, we see that C3 and C5 are in parallel,
with an equivalent capacitance
C6 = C3 + C5 = C + C/2 = 3C/2.
From the new circuit, we see that C4 and C6 are in series
and find their equivalent capacitance from
1/Ceq = (1/C4) + (1/C6)
3C/5.
= (1/C) + [1/(3C/2)], which gives Ceq =
(b) The charge on the equivalent capacitor is also the
charge on C4 and C6 :
Qeq = Q4 = Q6 = CeqVab = (3C/5)V = 3CV/5.
We find the potential difference between c and b from
Vcb = Q6/C6 = (3CV/5)/(3C/2) = 2V/5.
The charge on C5 is also the charge on C1 and C2 :
Q5 = Q1 = Q2 = C5Vcb = (C/2)(2V/5) = CV/5.
We find the potential differences from
Vcd = Q1/C1 = (CV/5)/(C) = V/5;
Vdb = Q2/C2 = (CV/5)/(C) = V/5;
Vac = Q4/C4 = (3CV/5)/(C) = 3V/5.
The charge on C3 is
Q3 = C3Vcb = (C)(2V/5) = 2CV/5.
Thus we have
Q1 = Q2 = CV/5, Q3 = 2CV/5, Q4 = 3CV/5;
V1 = V2 = V/5, V3 = 2V/5, V4 = 3V/5.
33. Because C3 and C4 are in parallel, we have
Vcb = V3 = V4 = Q3 /C3 = 30 μC/8.0 μF = 3.75 V.
Thus we have
Q4 = C4V4 = (16.0 μF)(3.75 V) = 60 μC.
From the diagram we see that
Q5 = Q6 = Q3 + Q4 = 30 μC + 60 μC = 90 μC.
We see that C1 and C2 are in parallel:
C5 = C1 + C2 = 8.0 μF + 16.0 μF = 24.0 μF, so we have
V1 = V2 = V5 = Vac = Q5/C5 = 90 μC/24.0 μF = 3.75 V.
We can now find the charges on C1 and C2 :
Q1 = C1V1 = (8.0 μF)(3.75 V) = 30 μC;
Q2 = C2V2 = (16.0 μF)(3.75 V) = 60 μC.
As expected from the symmetry we have
(a) Q1 = Q3 = 30 μC; Q2 = Q4 = 60 μC.
(b) V1 = V2 = V3 = V4 = 3.75 V.
(c) Vab = Vac + Vcb = 3.75 V + 3.75 V = 7.5 V.
C1
C2
d
C3
c
C4
a
b
C5
c
C3
a
C4
C6
c
a
C4
c
C2
C5
a
b
C3
C1
a
b
b
C4
C6
c
b
36. We find the equivalent capacitance for the series combination from
1/Cseries = (1/C1) + (1/C2) = [1/(3200 pF)] + [1/(2200 pF)],
which gives Cseries = 1300 pF.
The charge on the equivalent capacitor is the initial charge on each capacitor:
Q1 = Q2 = Qseries = CseriesV0 = (1300 pF)(12.0 V) = 1.56 × 104 pC.
When the capacitors are connected with positive plates together, some charge will flow until the potential
difference across the two capacitors is the same:
V1 = V2 = V.
Because charge is conserved, we have
Q = Q1′ + Q2′ = Q1 + Q2 = 1.56 × 104 pC + 1.56 × 104 pC = 3.12 × 104 pC.
For the two capacitors we have
V = Q1′/C1 = Q2′/C2 ;
Q1′ = 1.85 × 104 pC.
Q1′/3200 pF = (3.12 × 104 pC – Q1′)/2200 pF, which gives
Thus
Q2′ = 3.12 × 104 pC – 1.85 × 104 pC =
1.27 × 104 pC.
42. (a) From C = ε0A/d, we see that separating the plates will change C. For the stored energy we have
U = ½CV 2 = ½Q 2/C.
Because the charge is constant, for the two conditions we have
2.
U2/U1 = C1/C2 = d2/d1 =
(b) The work changes the energy stored in the capacitor:
W = U2 – U1 = ½Q 2(1/C2 – 1/C1) = (½Q 2/ε0A)(d2 – d1) =
Q 2d/2ε0A.
.47 (a) Because there is no stored energy on the uncharged 5.0-μF capacitor, the total stored energy is
Ua = ½C1V02 = ½(3.0 × 10–6 F)(12 V)2 =
2.2 × 10–4 J.
(b) We find the initial charge on the 3.0-μF capacitor when it is connected to the battery;
Q = C1V0 = (3.0 μF)(12 V) = 36 μC.
When the capacitors are connected, some charge will flow from C1 to C2 until the potential
difference across the two capacitors is the same; so the two capacitors are connected in parallel:
Ceq = C1 + C2 = 3.0 μF + 5.0 μF = 8.0 μF.
For the stored energy we have
Ub = ½CeqV2 = ½Q 2 /Ceq = ½(36 μC)2/(8.0 μF) = 81 μJ =
8.1 × 10–5 J.
(c) The change in stored energy is
ΔU = Ub – Ua = 8.1 × 10–5 J – 2.2 × 10–4 J = – 1.4 × 10–4 J.
(d) The stored potential energy is not conserved. During the flow of charge before the final
steady state, some of the stored energy is dissipated as thermal and radiant energy.
53. The mica will change the capacitance. The potential difference is constant, so we have
ΔQ = Q2 – Q1 = (C2 – C1)V = (K – 1)C1V
0.46 μC.
= (7 – 1)(3500 × 10–12 F)(22 V) = 4.6 × 10–7 C =
56. The potential difference must be the same on each
half of the capacitor, so we can treat the system
as two capacitors in parallel:
C = C1 + C2 = [K1ε 0(½A)/d] + [K2ε 0(½A)/d]
= (ε0½A/d)(K1 + K2) = ½(K1 + K2)(ε0A/d)
=
ε0A(K1 + K2)/2d .
57. If we think of a layer of equal and opposite charges on the
interface between the two dielectrics, we see that they
are in series. For the equivalent capacitance, we have
1/C = (1/C1) + (1/C2) = (½d/K1ε0A) + (½d/K2ε0A)
= (d/2ε0A)[(1/K1) + (1/K2)]
= (d/2ε0A)[(K1 + K2)/K1K2],
which gives
C = 2ε0AK1K2/d(K1 + K2).
d
K1
d
K2
K1
K2
75. Because the capacitor is isolated, the charge will not change. The initial stored energy is
U1 = ½C1V12 = ½Q 2/C1 , with C1 = ε0A/d1 .
The changes will change the capacitance:
C2 = Kε0A/d2 .
For the ratio of stored energies, we have
U2/U1 = C1/C2 = (ε0A/d1)/(Kε0A/d2) = d2/Kd1 = ½/K =
1/2K.
The stored energy decreases from two factors. Because the plates attract each other, when the separation is
halved, work is done by the field, so the energy decreases. When the dielectric is inserted, the induced charges
on the dielectric are attracted to the plates; again work is done by the field and the energy decreases.
The uniform electric field between the plates is related to the potential difference across the plates:
E = V/d.
For a parallel-plate capacitor, we have
Q = C1V1 = C1E1d1 = C2E2d2 , or
E2/E1 = C1d1/C2d2 = ε0A/Kε0A =
1/K.