A parallel plate capacitor of plate area A and plate separation $d$ charged by a battery of voltage V. The battery is then disconnected. The work needed to pull the plates to separation $2 d$ is

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$\frac{2 A V^{2} ε_{0}}{d}$

$\frac{3 A V^{2} ε_{0}}{d}$

$\frac{A V^{2} ε_{0}}{d}$

$\frac{A V^{2} ε_{0}}{2 d}$

answer is C.

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Detailed Solution

Energy stored initially is $U_{i} = \frac{Q^{2}}{2 C} .$ if d is doubled, C becomes $C / 2$ .Hence, energy stored when d is doubled is $U_{f} = \frac{Q^{2}}{C}$
$∴$ Work needed is
$W = U_{f} - U_{i} = \frac{Q^{2}}{2 (\frac{C}{2})} - \frac{Q^{2}}{2 C} = \frac{Q^{2}}{2 C} = \frac{1}{2} C V^{2}$
$(∵ Q = C V)$
Now $C = \frac{A ε_{0}}{d}$ . Hence
$W = \frac{A ε_{0} V^{2}}{2 d}$ .