Consider a capacitor as shown in Fig. 1. If we pull the plates of capacitor apart to a final position as shown in Fig. 2, then we must perform some work against electric force. For this situation, mark out the correct statement. [take areas of plates as A.]

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Work done is $\frac{Q^{2}}{2 ε_{0} A} (d_{2} - d_{1})$ and is stored in volume A(d₂ – d₁)

Work done is $+ \frac{ε_{0} {AV}^{2}}{2} (\frac{d_{2} - d_{1}}{d_{1} d_{2}})$ and is stored in volume A(d₂ – d₁)

Work done is $\frac{ε_{0} {AV}^{2}}{2} (\frac{d_{2} - d_{1}}{d_{1} d_{2}})$ and is stored in volume Ad₂.

Work done is $+ \frac{Q^{1}}{2 ε_{0} A} (d_{2} - d_{1})$ and is stored in volume Ad₂ .

answer is C.

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Detailed Solution

As the battery remains connected the potential difference across the capacitor remains constant equal to V and the battery does some work:

$\begin{array}{l} U_{1} = \frac{ε_{0} A}{d_{1}} \times \frac{V^{2}}{2}; U_{2} = \frac{ε_{0} A}{d_{2}} \times \frac{V^{2}}{2} \\ dU = W_{ext} + W_{by battery} \\ \Rightarrow \frac{ε_{0} {AV}^{2}}{2} [\frac{1}{d_{2}} - \frac{1}{d_{1}}] = W_{ext} + [\frac{ε_{0} AV}{d_{2}} - \frac{ε_{0} AV}{d_{1}}] V \\ \Rightarrow W_{ext} = \frac{ε_{0} {AV}^{2}}{2} [\frac{1}{d_{1}} - \frac{1}{d_{2}}] = \frac{ε_{0} {AV}^{2}}{2} [\frac{d_{2} - d_{1}}{d_{1} d_{2}}] \end{array}$