12.
(a) We use U =
1
2
LI
2
=
1
2
Q
2
/C to solve for L:
L
=
1
C
Q
I
2
=
1
C
CV
max
I
2
=
C
V
max
I
2
=
(4.00
× 10
−6
F)
1.50 V
50.0
× 10
−3
A
2
=
3.60
× 10
−3
H .
(b) Since f = ω/2π, the frequency is
f =
1
2π
√
LC
=
1
2π
(3.60
× 10
−3
H)(4.00
× 10
−6
F)
= 1.33
× 10
3
Hz .
(c) Referring to Fig. 33-1, we see that the required time is one-fourth of a period (where the period is
the reciprocal of the frequency). Consequently,
t =
1
4
T =
1
4f
=
1
4(1.33
× 10
3
Hz)
= 1.88
× 10
−4
s .