A parallel plate capacitor of capacity 100µF is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joules is

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12.5 × 10^–3

0.125 × 10^–3

1.25 × 10^–3

62.5 × 10^–3

answer is A.

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Detailed Solution

$\begin{array}{l} U_{i} = \frac{1}{2} C_{1} V^{2} = \frac{1}{2} \times 100 \times 10^{- 6} \times 50 \times 50 \\ U_{i} = 12 .5 \times 10^{- 2} J \end{array}$
If distance is doubled, capacity reduced to half
$(C = \frac{{Aε}_{0}}{d})$
Question Image
$\begin{array}{l} U_{f} = \frac{1}{2} C_{2} V^{2} \\ = \frac{1}{2} \times 50 \times 10^{- 6} \times 50 \times 50 \\ = 6 .25 \times 10^{- 2} J \end{array}$
Required energy $= U_{i} - U_{f}$
$= (12.5 - 6.25) \times 10^{- 2} = 62.5 \times 10^{- 3} J$