A variable capacitor is kept connected to a 10V battery. If the capacitance of the capacitor is changed from7µF to 3µF , the work done is

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8 × 10^–4 J

2 × 10^–4 J

4 × 10^–4 J

6 × 10^–4 J

answer is B.

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Detailed Solution

$\begin{array}{l} W = \frac{1}{2} C_{2} V^{2} - \frac{1}{2} C_{1} V^{2} \\ = \frac{1}{2} \times (4 \times 10^{- 6}) \times 100 = 2 \times 10^{- 4} J \end{array}$

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