# Solution manual fundamentals of physics extended, 8th editionch31

1. (a) All the energy in the circuit resides in the capacitor when it has its maximum
charge. The current is then zero. If Q is the maximum charge on the capacitor, then the
total energy is

c

h

2

2.90 × 10−6 C
Q2
U=
=
= 117
. × 10−6 J.
−6
2C 2 3.60 × 10 F

c

h

(b) When the capacitor is fully discharged, the current is a maximum and all the energy
resides in the inductor. If I is the maximum current, then U = LI2/2 leads to

c

h

2 1168
.
× 10−6 J
2U
. × 10−3 A .
I=
=
= 558
−3
75 × 10 H
L

2. According to U = 21 LI 2 = 21 Q 2 C , the current amplitude is
I=

Q
=
LC

3.00 × 10−6 C

c1.10 × 10 Hhc4.00 × 10 Fh
−3

−6

= 4.52 × 10−2 A .

3. We find the capacitance from U = 21 Q 2 C :

2

c

h
h

160
. × 10−6 C
Q2
C=
=
= 9.14 × 10−9 F.
2U 2 140 × 10−6 J

c

4. (a) The period is T = 4(1.50 µs) = 6.00 µs.
(b) The frequency is the reciprocal of the period:
f =

1
1
=
= 167
. × 105 Hz.
T 6.00µs

(c) The magnetic energy does not depend on the direction of the current (since UB ∝ i2),
so this will occur after one-half of a period, or 3.00 µs.

5. (a) We recall the fact that the period is the reciprocal of the frequency. It is helpful to
refer also to Fig. 31-1. The values of t when plate A will again have maximum positive
charge are multiples of the period:
t A = nT =

n
n
=
= n 5.00 µs ,
f 2.00 × 103 Hz

b

g

where n = 1, 2, 3, 4, ! . The earliest time is (n=1) t A = 5.00 µ s.
(b) We note that it takes t = 21 T for the charge on the other plate to reach its maximum
positive value for the first time (compare steps a and e in Fig. 31-1). This is when plate A
acquires its most negative charge. From that time onward, this situation will repeat once
every period. Consequently,

( 2n −1) = ( 2n −1) = 2n −1 2.50 µ s ,
1
1
t = T + (n −1)T = ( 2n −1) T =
(
)(
)
2
2
2f
2 ( 2 ×103 Hz )
where n = 1, 2, 3, 4, ! . The earliest time is (n=1) t = 2.50 µ s.
(c) At t = 41 T , the current and the magnetic field in the inductor reach maximum values
for the first time (compare steps a and c in Fig. 31-1). Later this will repeat every halfperiod (compare steps c and g in Fig. 31-1). Therefore,
tL =

T (n −1)T
T
+
= ( 2n − 1) = ( 2n −1)(1.25 µ s ) ,
4
2
4

where n = 1, 2, 3, 4, ! . The earliest time is (n=1) t = 1.25 µ s.

6. (a) The angular frequency is

ω=

k
=
m

F x
=
m

c

8.0 N
2.0 × 10 m 0.50 kg
−13

hb

g

(b) The period is 1/f and f = ω/2π. Therefore,
T=

ω

=

= 7.0 × 10−2 s.

(c) From ω = (LC)–1/2, we obtain
C=

1

ω L
2

=

1

2

= 2.5 × 10−5 F.

7. (a) The mass m corresponds to the inductance, so m = 1.25 kg.
(b) The spring constant k corresponds to the reciprocal of the capacitance. Since the total
energy is given by U = Q2/2C, where Q is the maximum charge on the capacitor and C is
the capacitance,

c

h

2

175 × 10−6 C
Q2
C=
=
= 2.69 × 10−3 F
−6
2U 2 5.70 × 10 J

c

h

and
k=

(c) The maximum
xmax = 1.75 ×10−4 m.

1
= 372 N / m.
2.69 × 10−3 m / N

displacement

corresponds

to

the

maximum

charge,

so

(d) The maximum speed vmax corresponds to the maximum current. The maximum
current is
I = Qω =

Q
=
LC

175 × 10−6 C
. H gc2.69 × 10 Fh
b125

Consequently, vmax = 3.02 × 10–3 m/s.

−3

= 3.02 × 10−3 A.

8. We apply the loop rule to the entire circuit:

(

)

§
¨

ε total = ε L + ε C + ε R + " = ¦ ε L + ε C + ε R = ¦ ¨ L j
1

1

1

j

=L

di q
+ + iR
dt C

j

j

j

with L = ¦ L j ,
j

j

·
di q
+ + iR j ¸
¸
dt C j
¹

1
1
= ¦ , R = ¦ Rj
C
j Cj
j

where we require εtotal = 0. This is equivalent to the simple LRC circuit shown in Fig. 3124(b).

9. The time required is t = T/4, where the period is given by T = 2 π / ω = 2 π LC.
Consequently,
T 2 π LC 2 π
t= =
=
4
4

b0.050 Hgc4.0 × 10 Fh = 7.0 × 10
−6

4

−4

s.

10. We find the inductance from f = ω / 2 π = 2 π LC

d

L=

1
1
=
2
4 π f C 4 π 2 10 × 103 Hz
2

c

2

i

−1

.

h c6.7 × 10 Fh
−6

= 38
. × 10−5 H.

11. (a) Q = CVmax = (1.0 × 10–9 F)(3.0 V) = 3.0 × 10–9 C.
(b) From U = 21 LI 2 = 21 Q 2 / C we get
I=

Q
=
LC

3.0 × 10−9 C

c3.0 × 10 Hhc10. × 10 Fh
−3

−9

= 17
. × 10−3 A.

(c) When the current is at a maximum, the magnetic field is at maximum:
U B,max =

1 2 1
LI = 3.0 × 10−3 H 17
. × 10−3 A
2
2

c

hc

h

2

= 4.5 × 10−9 J.

12. The capacitors C1 and C2 can be used in four different ways: (1) C1 only; (2) C2 only;
(3) C1 and C2 in parallel; and (4) C1 and C2 in series.
(a) The smallest oscillation frequency is
f3 =

1

b

2 π L C1 + C2

g

=

1

c

−2

hc

−6

−6

h

. × 10 H 2.0 × 10 F + 5.0 × 10 F
2 π 10

= 6.0 × 102 Hz

(b) The second smallest oscillation frequency is
f1 =

1
1
=
= 7.1×102 Hz
2π LC1 2π (1.0 ×10−2 H )( 5.0 ×10−6 F )

(c) The second largest oscillation frequency is
f2 =

1
1
=
= 1.1×103 Hz
2
6

2π LC2 2π (1.0 ×10 H )( 2.0 ×10 F )

(d) The largest oscillation frequency is
f4 =

1
2π LC1C2 / ( C1 + C2 )

=

1

2.0 ×10−6 F+5.0 ×10−6 F
= 1.3 ×103 Hz
−2
−6
−6
1.0
×
10
H
2.0
×
10
F
5.0
×
10
F
(
)(
)(
)

13. (a) After the switch is thrown to position b the circuit is an LC circuit. The angular
frequency of oscillation is ω = 1/ LC . Consequently,
f =

ω

=

1
2 π LC

=

1

c

hc

h

2 π 54.0 × 10−3 H 6.20 × 10−6 F

= 275 Hz.

(b) When the switch is thrown, the capacitor is charged to V = 34.0 V and the current is
zero. Thus, the maximum charge on the capacitor is Q = VC = (34.0 V)(6.20 × 10–6 F) =
2.11 × 10–4 C. The current amplitude is

b

gc

h

I = ωQ = 2 πfQ = 2 π 275 Hz 2.11 × 10−4 C = 0.365 A.

14. For the first circuit ω = (L1C1)–1/2, and for the second one ω = (L2C2)–1/2. When the
two circuits are connected in series, the new frequency is

ω′ =
=

1
=
Leq Ceq
1
L1C1

where we use ω −1 =

1

( L1 + L2 ) C1C2 / ( C1 + C2 )
1

( C1 + C2 ) / ( C1 + C2 )
L1C1 =

L2 C2 .

=ω,

=

1

( L1C1C2 + L2C2C1 ) / ( C1 + C2 )

15. (a) Since the frequency of oscillation f is related to the inductance L and capacitance
C by f = 1 / 2 π LC , the smaller value of C gives the larger value of f. Consequently,
f max = 1 / 2 π LCmin , f min = 1 / 2 π LCmax , and
Cmax
365 pF
f max
=
=
= 6.0.
f min
Cmin
10 pF

(b) An additional capacitance C is chosen so the ratio of the frequencies is
r=

160
. MHz
= 2.96.
0.54 MHz

Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds
to that of the tuning capacitor. If C is in picofarads, then
C + 365 pF
C + 10 pF

= 2.96.

The solution for C is

b365 pFg − b2.96g b10 pFg = 36 pF.
C=
b2.96g − 1
2

2

(c) We solve f = 1 / 2 π LC for L. For the minimum frequency C = 365 pF + 36 pF =
401 pF and f = 0.54 MHz. Thus
L=

1

=

1

b2πg Cf b2πg c401 × 10 Fhc0.54 × 10 Hzh
2

2

2

−12

6

2

= 2.2 × 10−4 H.

16. The linear relationship between θ (the knob angle in degrees) and frequency f is

FG
H

f = f0 1+

IJ  θ = 180° FG f − 1IJ
180° K
Hf K
θ

0

where f0 = 2 × 105 Hz. Since f = ω/2π = 1/2π
θ:
C=

1

b

4 π Lf 1 + 180°
2

2
0

θ

g

2

=

LC , we are able to solve for C in terms of

81

b

400000π 2 180°+θ

g

2

with SI units understood. After multiplying by 1012 (to convert to picofarads), this is
plotted, below.

17. (a) The total energy U is the sum of the energies in the inductor and capacitor:
−6
−3
−3
q 2 i 2 L ( 3.80 ×10 C ) ( 9.20 ×10 A ) ( 25.0 ×10 H )
U =U E +U B =
+
=
+
= 1.98 ×10−6 J.
2C 2 2 ( 7.80 ×10−6 F )
2
2

2

(b) We solve U = Q2/2C for the maximum charge:

c

hc

h

Q = 2CU = 2 7.80 × 10−6 F 198
. × 10−6 J = 556
. × 10−6 C.
(c) From U = I2L/2, we find the maximum current:
I=

c

h

2 198
. × 10−6 J
2U
=
= 126
. × 10−2 A.
−3
25.0 × 10 H
L

(d) If q0 is the charge on the capacitor at time t = 0, then q0 = Q cos φ and

φ = cos

−1

FG q IJ = cos FG 380
. × 10 C I
J = ±46.9° .
. × 10 C K
H QK
H 556
−1

−6

−6

For φ = +46.9° the charge on the capacitor is decreasing, for φ = –46.9° it is increasing.
To check this, we calculate the derivative of q with respect to time, evaluated for t = 0.
We obtain –ωQ sin φ, which we wish to be positive. Since sin(+46.9°) is positive and
sin(–46.9°) is negative, the correct value for increasing charge is φ = –46.9°.
(e) Now we want the derivative to be negative and sin φ to be positive. Thus, we take
φ = +46.9°.

18. (a) Since the percentage of energy stored in the electric field of the capacitor is
(1 − 75.0%) = 25.0% , then

U E q 2 / 2C
= 2
= 25.0%
U
Q / 2C
which leads to q / Q = 0.250 = 0.500.
(b) From
U B Li 2 / 2
= 2
= 75.0%,
U
LI / 2
we find i / I = 0.750 = 0.866.

19. (a) The charge (as a function of time) is given by q = Q sin ωt, where Q is the
maximum charge on the capacitor and ω is the angular frequency of oscillation. A sine
function was chosen so that q = 0 at time t = 0. The current (as a function of time) is
i=

dq
= ωQ cosωt ,
dt

and at t = 0, it is I = ωQ. Since ω = 1/ LC ,

b

Q = I LC = 2.00 A

. × 10
g c3.00 × 10 Hhc2.70 × 10 Fh = 180
−3

−6

−4

C.

(b) The energy stored in the capacitor is given by
UE =

q 2 Q 2 sin 2 ωt
=
2C
2C

and its rate of change is
dU E Q 2ω sin ωt cos ωt
=
dt
C

b g

We use the trigonometric identity cos ωt sin ωt = 21 sin 2ωt to write this as
dU E ωQ 2
sin 2ωt .
=
dt
2C

b g

The greatest rate of change occurs when sin(2ωt) = 1 or 2ωt = π/2 rad. This means
t=

π π
π
=
LC =
4ω 4
4

( 3.00 ×10

−3

H )( 2.70 ×10−6 F ) = 7.07 ×10−5 s.

(c) Substituting ω = 2π/T and sin(2ωt) = 1 into dUE/dt = (ωQ2/2C) sin(2ωt), we obtain

FG dU IJ
H dt K
E

Now T = 2 π LC = 2 π

max

2 πQ 2 πQ 2
=
=
.
TC
2TC

c3.00 × 10 Hhc2.70 × 10 Fh = 5.655 × 10
−3

−6

−4

s, so

FG dU IJ
H dt K

=

E

max

c

π 180
. × 10−4 C

h

2

c5.655 × 10 shc2.70 × 10 Fh = 66.7 W.
−4

−6

We note that this is a positive result, indicating that the energy in the capacitor is indeed
increasing at t = T/8.

20. (a) We use U = 21 LI 2 = 21 Q 2 / C to solve for L:
2

2

2

1 § Q · 1 § CV ·
§V ·
L = ¨ ¸ = ¨ max ¸ = C ¨ max ¸ = ( 4.00 ×10−6 F )

2

§ 1.50V ·
−3
¨
¸ = 3.60 ×10 H.
−3

(b) Since f = ω/2π, the frequency is
f =

1
2 π LC

=

1

c3.60 × 10 Hhc4.00 × 10 Fh
−3

−6

. × 103 Hz.
= 133

(c) Referring to Fig. 31-1, we see that the required time is one-fourth of a period (where
the period is the reciprocal of the frequency). Consequently,
t=

1
1
1
T=
=
= 188
. × 10−4 s.
3
4
4 f 4 133
. × 10 Hz

e

j

21. (a) We compare this expression for the current with i = I sin(ωt+φ0). Setting (ωt+φ) =
2500t + 0.680 = π/2, we obtain t = 3.56 × 10–4 s.
(b) Since ω = 2500 rad/s = (LC)–1/2,
L=

1
1
=
= 2.50 × 10−3 H.
2
2
ω C 2500 rad / s 64.0 × 10−6 F

b

gc

h

(c) The energy is
U=

1 2 1
LI = 2.50 × 10−3 H 160
. A
2
2

c

hb

g

2

= 3.20 × 10−3 J.

22. (a) From V = IXC we find ω = I/CV. The period is then T = 2π/ω = 2πCV/I = 46.1 µs.
2

(b) 12 CV = 6.88 nJ.
(c) The answer is again 6.88 nJ (see Fig. 31-4).
(d) We apply Eq. 30-35 as V = L(di/dt)max . We can substitute L = CV2/I2 (combining
what we found in part (a) with Eq. 31-4) into Eq. 30-35 (as written above) and solve for
(di/dt)max . Our result is 1.02 × 103 A/s.
(e) The derivative of U = 12 Li2 leads to dU/dt = LI2ω sin(ωt)cos(ωt) = 12 LI2ω sin(2ωt).
Therefore, (dU/dt)max =

1 2
2 LI ω

= 12 IV = 0.938 mW.

23. The loop rule, for just two devices in the loop, reduces to the statement that the
magnitude of the voltage across one of them must equal the magnitude of the voltage
across the other. Consider that the capacitor has charge q and a voltage (which we’ll
consider positive in this discussion) V = q/C. Consider at this moment that the current in
the inductor at this moment is directed in such a way that the capacitor charge is
increasing (so i = +dq/dt). Eq. 30-35 then produces a positive result equal to the V across
the capacitor: V = −L(di/dt), and we interpret the fact that −di/dt > 0 in this discussion to
mean that d(dq/dt)/dt = d2q/dt2 < 0 represents a “deceleration” of the charge-buildup
process on the capacitor (since it is approaching its maximum value of charge). In this
way we can “check” the signs in Eq. 31-11 (which states q/C = − L d2q/dt2) to make sure
we have implemented the loop rule correctly.

24. The charge q after N cycles is obtained by substituting t = NT = 2πN/ω' into Eq.
31-25:
q = Qe − Rt / 2 L cos (ω ′t + φ ) = Qe − RNT / 2 L cos ª¬ω ′ ( 2πN / ω ′ ) + φ º¼
= Qe

(

)

− RN 2 π L / C / 2 L

= Qe − N πR

C/L

cos ( 2πN + φ )

cos φ .

We note that the initial charge (setting N = 0 in the above expression) is q0 = Q cos φ,
where q0 = 6.2 µC is given (with 3 significant figures understood). Consequently, we
write the above result as q N = q0e − NπR C / L .
(a) For N = 5,
q5 = ( 6.2 µ C ) e

−5 π( 7.2 Ω ) 0.0000032 F/12 H

q10 = ( 6.2 µ C ) e

−10 π( 7.2 Ω ) 0.0000032 F/12 H

= 5.85 µ C.

(b) For N = 10,
= 5.52 µ C.

(c) For N = 100,
q100 = ( 6.2 µ C ) e

−100 π( 7.2 Ω ) 0.0000032 F/12 H

= 1.93 µ C.

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