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 .