The plates of a parallel plate capacitor have surface area ‘ $A$ ’ and are initially separated by a distance ‘ $d$ ’. They are connected to a battery of voltage $v_{0}$ . Now, the plates of the capacitor are pulled apart to a separation $2 d$ . Then, increase in the energy of the battery is

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$\frac{\in_{0} A {V_{0}}^{2}}{8 d}$

$\frac{\in_{0} A {V_{0}}^{2}}{2 d}$

$\frac{\in_{0} A {V_{0}}^{2}}{4 d}$

there will be no change in the energy of the battery

answer is B.

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Detailed Solution

The force of attraction acting between the plates $F = \frac{q^{2}}{2 \in_{0} A}$
$F = \frac{C^{2} V^{2}}{2 \in_{0} A} = \frac{\in_{0} A V^{2}}{2 d^{2}}$ ; $W = \int_{d}^{2 d} F d x = \frac{\in_{0} A V^{2}}{4 d}$
When charge $Δ Q$ leaves the plates of the capacitor it increases the energy of the battery by $Δ Q V = Δ C V^{2} = \frac{\in_{0} A V^{2}}{2 d}$ .