p33 016

background image

16.

(a) Since the percentage of energy stored in the electric field of the capacitor is (1

75.0%) = 25.0%,

then

U

E

U

=

q

2

/2C

Q

2

/2C

= 25.0%

which leads to q =

0.250 Q = 0.500Q.

(b) From

U

B

U

=

Li

2

/2

LI

2

/2

= 75.0% ,

we find i =

0.750 I = 0.866I.


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