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Q.

A 60 $p$ F capacitor has charge on each plate 3 x 10^-8 coulomb, then the energy stored is

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a

$1 .2 \times 10^{- 4} J$

b

$7 \cdot 5 \times 10^{- 6} J$

c

$15 \times 10^{- 6} J$

d

$2 .4 \times 10^{- 4} J$

answer is B.

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Detailed Solution

$E= \frac{1}{2} \times \frac{q^{2}}{C} = \frac{1}{2} \times \frac{{(3 \times 10^{- 8})}^{2}}{60 \times 10^{- 12}} = 7 \cdot 5 \times 10^{- 6} J$

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