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Halliday/Resnick/Walker 7e
Chapter 25 – Capacitance
- (a) The capacitance of the system is
C
q V
pC V
. pF.
(b) The capacitance is independent of q; it is still 3.5 pF.
(c) The potential difference becomes
∆V
q C
pC pF
V.
3. (a) The capacitance of a parallel-plate capacitor is given by C = ε 0 A/d, where A is the area of
each plate and d is the plate separation. Since the plates are circular, the plate area is A = πR^2 , where R is the radius of a plate. Thus,
( ) ( ) 2 12 2 2 0 10 3
8.85 10 F m 8.2 10 m 1.44 10 F 144pF. 1.3 10 m
R
C
d
ε π −^ π − −
−
× ×
= = = × =
×
(b) The charge on the positive plate is given by q = CV, where V is the potential difference across the plates. Thus,
q = (1.44 × 10 –10^ F)(120 V) = 1.73 × 10 –8^ C = 17.3 nC.
- Assuming conservation of volume, we find the radius of the combined spheres, then use C =
4 π ε 0 R to find the capacitance. When the drops combine, the volume is doubled. It is then V =
2(4π/3)R^3. The new radius R' is given by
( )
R ′^ = R ⇒
p p R ′ = 2 1 3^ R.
The new capacitance is 1 3
C ′^ = 4 p ε 0 R ′= 4 pε 0 2 R =5.04 pε 0 R.
With R = 2.00 mm, we obtain C = 5.04 π (^) ( 8.85 × 10 −^12 F m (^) )( 2.00 × 10 −^3 m (^) )= 2.80 × 10 −^13 F.
- The equivalent capacitance is given by Ceq = q/V, where q is the total charge on all the capacitors and V is the potential difference across any one of them. For N identical capacitors in parallel, Ceq = NC, where C is the capacitance of one of them. Thus, NC = q V/ and
3 6
1 00C
110V 1 00 10 F
q. N.. VC (^). −
= = = ×
×
- The equivalent capacitance is
eq 1 2 3
10.0 F 5.00 F 4.00 F
3.16 F.
10.0 F 5.00 F 4.00 F
C C C
C
C C C
- The charge that passes through meter A is
q = C Veq = 3 CV = 3 25 0b. μF gb 4200 V g=0 315. C.
- (a) and (b) The original potential difference V 1 across C 1 is
eq^ (^ )(^ )
1 1 2
3.16 F 100.0 V
21.1V.
10.0 F 5.00 F
C V
V
C C
Thus ∆V 1 = 100.0 V – 21.1 V = 78.9 V and
∆q 1 = C 1 ∆V 1 = (10.0 μF)(78.9 V) = 7.89 × 10 –4^ C.
- (a) The potential difference across C 1 is V 1 = 10.0 V. Thus,
q 1 = C 1 V 1 = (10.0 μF)(10.0 V) = 1.00 × 10 –4^ C.
(b) Let C = 10.0 μF. We first consider the three-capacitor combination consisting of C 2 and its
two closest neighbors, each of capacitance C. The equivalent capacitance of this combination is
2 eq 2
C C
C C. C.
C C
Also, the voltage drop across this combination is
1 1 0 40 1 eq^ 1 50
CV CV
V. V.
C C C. C
Since this voltage difference is divided equally between C 2 and the one connected in series with it, the voltage difference across C 2 satisfies V 2 = V/2 = V 1 /5. Thus
2 2 2 (^ )^5
10 0V
10 0 F 2 00 10 C.
q C V. μ. −
= = = ×
(e) The potential difference across C 2 is given by V 2 = V – V 1 = 20.0 V – 10.0 V = 10.0 V.
(f) The charge carried by C 2 is q 2 = C 2 V 2 = (2.00 μF)(10.0 V) = 2.00 × 10 –5^ C.
(g) Since this voltage difference V 2 is divided equally between C 3 and the other 4.00- μF
capacitors connected in series with it, the voltage difference across C 3 is given by V 3 = V 2 /2 = 10.0 V/2 = 5.00 V.
(h) Thus, q 3 = C 3 V 3 = (4.00 μF)(5.00 V) = 2.00 × 10 –5^ C.
- Eq. 23-14 applies to each of these capacitors. Bearing in mind that σ = q/A, we find the total charge to be
qtotal = q 1 + q 2 = σ 1 A 1 + σ 2 A 2 = εo E 1 A 1 + εo E 2 A 2 = 3.6 pC
where we have been careful to convert cm^2 to m^2 by dividing by 10^4.
- The charges on capacitors 2 and 3 are the same, so these capacitors may be replaced by an equivalent capacitance determined from
2 3
2 3 C C C 2 3
C C
eq C C
Thus, Ceq = C 2 C 3 /(C 2 + C 3 ). The charge on the equivalent capacitor is the same as the charge on either of the two capacitors in the combination and the potential difference across the equivalent capacitor is given by q 2 /Ceq. The potential difference across capacitor 1 is q 1 /C 1 , where q 1 is the charge on this capacitor. The potential difference across the combination of capacitors 2 and 3 must be the same as the potential difference across capacitor 1, so q 1 /C 1 = q 2 /Ceq. Now some of the charge originally on capacitor 1 flows to the combination of 2 and 3. If q 0 is the original charge, conservation of charge yields q 1 + q 2 = q 0 = C 1 V 0 , where V 0 is the original potential difference across capacitor 1.
(a) Solving the two equations
q C
q C
(^1) q q C V 1
2 = 1 + 2 = 1 0 eq
and
for q 1 and q 2 , we obtain
12 0 12 0 12 (^2 3 )^0 1 eq 1 2 3 1 2 1 3 2 3 1 2 3
C V C V C C C V
q C C C C C C C C C C C C C
With V 0 = 12.0 V, C 1 = 4.00 μF, C 2 = 6.00 μF and C 3 =3.00 μF, we find Ceq = 2.00 μF and q 1 =
32.0 μC.
(b) The charge on capacitors 2 is
q 2 = C V 1 0 − q 1 = (4.00 μF)(12.0V) − 32.0 μF =16.0 μF
(c) The charge on capacitor 3 is the same as that on capacitor 2:
q 3 = C V 1 0 − q 1 = (4.00 μF)(12.0V) − 32.0 μF =16.0 μF
- Let V = 1.00 m^3. Using Eq. 25-25, the energy stored is
( ) (^) ( )
(^2 ) 2 12 3 8 (^0 )
1 1 C
8.85 10 150 V m 1.00 m 9.96 10 J. 2 2 N m
U u ε E −^ −
= = = × = ×
V V
- The total energy is the sum of the energies stored in the individual capacitors. Since they are connected in parallel, the potential difference V across the capacitors is the same and the total energy is
( ) (^) ( )( ) 2 6 6 2 1 2
2.0 10 F 4.0 10 F 300 V 0.27 J.
U = C + C V = × −^ + × − =
- (a) The potential difference across C 1 (the same as across C 2 ) is given by
3 (^ )^ (^ ) 1 2 1 2 3
15.0 F 100V
50.0V.
10.0 F+5.00 F+15.0 F
C V
V V
C C C
Also, V 3 = V – V 1 = V – V 2 = 100 V – 50.0 V = 50.0 V. Thus,
( ) ( ) ( ) ( )
4 1 1 1 4 2 2 2 4 4 4 3 1 2
10.0 F 50.0V 5.00 10 C
5.00 F 50.0V 2.50 10 C
5.00 10 C 2.50 10 C=7.50 10 C.
q C V q C V q q q
− − − − −
= = = ×
= = = ×
= + = × + × ×
(b) The potential difference V 3 was found in the course of solving for the charges in part (a). Its value is V 3 = 50.0 V.
(c) The energy stored in C 3 is
( ) ( ) 2 2 2 3 3 3
15.0 F 50.0V 1.88 10 J.
U = C V = μ = × −
W = Uf – Ui = 7.52 × 10 −^11 J.
- If the original capacitance is given by C = ε 0 A/d, then the new capacitance is C ' = ε κ 0 A / 2d.
Thus C'/C = κ/2 or
κ = 2C'/C = 2(2.6 pF/1.3 pF) = 4.0.
- The capacitance with the dielectric in place is given by C = κC 0 , where C 0 is the capacitance before the dielectric is inserted. The energy stored is given by U = 12 CV 2 =^12 κC V 0 2 , so
6 2 12 2 0
2 2(7.4 10 J)
(7.4 10 F)(652V)
U
C V
κ
− −
×
×
According to Table 25-1, you should use Pyrex.
- Each capacitor has 12.0 V across it, so Eq. 25-1 yields the charge values once we know C 1 and C 2. From Eq. 25-9,
C 2 =
ε 0 A d = 2.21^ ×^10
− 11 F ,
and from Eq. 25-27,
C 1 =
κε 0 A d = 6.64^ ×^10
− 11 F.
This leads to q 1 = C 1 V 1 = 8.00 × 10 −^10 C and q 2 = C 2 V 2 = 2.66 × 10 −^10 C. The addition of these gives the desired result: qtot = 1.06 × 10 −^9 C. Alternatively, the circuit could be reduced to find the qtot.
- The capacitance is given by C = κC 0 = κε 0 A/d, where C 0 is the capacitance without the dielectric, κ is the dielectric constant, A is the plate area, and d is the plate separation. The electric field between the plates is given by E = V/d, where V is the potential difference between the plates. Thus, d = V/E and C = κε 0 AE/V. Thus,
A
CV
E
κε (^0)
For the area to be a minimum, the electric field must be the greatest it can be without breakdown occurring. That is,
A =
× ×
× ×
− −
8 12
F)(4.0 10 V) 2
F / m)(18 10 V / m)
m
3 6
- We assume there is charge q on one plate and charge –q on the other. The electric field in the lower half of the region between the plates is
E
q A 1 1 0
κ ε
where A is the plate area. The electric field in the upper half is
E
q (^22 0) A
κ ε
Let d/2 be the thickness of each dielectric. Since the field is uniform in each region, the potential difference between the plates is
V
E d E d qd A
qd A
L
N
M
O
Q
P =^
0 1 2 0
1 2 (^2 2 21 )
so
C
q V
A
d
1 2
ε κ κ κ κ
This expression is exactly the same as that for Ceq of two capacitors in series, one with dielectric constant κ 1 and the other with dielectric constant κ 2. Each has plate area A and plate separation d/2. Also we note that if κ 1 = κ 2 , the expression reduces to C = κ 1 ε 0 A/d, the correct result for a parallel-plate capacitor with plate area A, plate separation d, and dielectric constant κ 1.
With A = 7.89 × 10 −^4 m^2 , d = 4.62 × 10 −^3 m, κ 1 =11.0 and κ 2 = 12.0, the capacitance is, (in SI units)
12 4 11 3
1.73 10 F.
C
− − − −
× ×
= = ×
× +
51.One way to approach this is to note that – since they are identical – the voltage is evenly divided between them. That is, the voltage across each capacitor is V = (10/n) volt. With C =
2.0 × 10 −^6 F, the electric energy stored by each capacitor is 12 CV^2. The total energy stored by the
capacitors is n times that value, and the problem requires the total be equal to 25 × 10 −^6 J. Thus,
n 2 (2.0^ ×^10
n
2 = 25 × 10 −^6
Using the relation N = nAd, we obtain d = 1.1 × 10 −^12 m, a remarkably small distance!
- (a) Put five such capacitors in series. Then, the equivalent capacitance is 2.0 μF/5 = 0.40 μF. With each capacitor taking a 200-V potential difference, the equivalent capacitor can withstand 1000 V.
(b) As one possibility, you can take three identical arrays of capacitors, each array being a five- capacitor combination described in part (a) above, and hook up the arrays in parallel. The equivalent capacitance is now Ceq = 3(0.40 μF) = 1.2 μF. With each capacitor taking a 200-V potential difference the equivalent capacitor can withstand 1000 V.
- We do not employ energy conservation since, in reaching equilibrium, some energy is dissipated either as heat or radio waves. Charge is conserved; therefore, if Q = C 1 Vbat = 100 μC, and q 1 , q 2 and q 3 are the charges on C 1 , C 2 and C 3 after the switch is thrown to the right and equilibrium is reached, then
Q = q 1 + q 2 + q 3.
Since the parallel pair C 2 and C 3 are identical, it is clear that q 2 = q 3. They are in parallel with C 1 so that V 1 =V 3 , or
q 1 C 1 =^
q 3 C 3
which leads to q 1 = q 3 /2. Therefore,
Q =
2 q^3 +^ q^3 +q^3
which yields q 3 = 40 μC and consequently q 1 = 20 μC.
- The pair C 3 and C 4 are in parallel and consequently equivalent to 30 μF. Since this numerical value is identical to that of the others (with which it is in series, with the battery), we observe that each has one-third the battery voltage across it. Hence, 3.0 V is across C 4 , producing a charge
q 4 = C 4 V 4 = (15 μF)(3.0 V) = 45 μC.
