The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection, if the plates of the capacitor are separated, so that the distance between them is twice the original distance, then the electrostatic energy becomes

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } E=\frac{1}{2} C V^2 \Rightarrow C \propto \frac{1}{\mathrm{~d}} \ & \Rightarrow \frac{C_1}{C_2}=\frac{d_1}{d_2}=\frac{2 d}{d}=2\left[\because C=\frac{A \in_0}{d}ight] \ & 	ext { Now, } \frac{E_1}{E_2}=\frac{C_1}{C_2}=\frac{C}{C / 2}=\frac{2}{1}\end{aligned}$ $\Rightarrow \frac{E}{E_2}=\frac{2}{1} \Rightarrow E_2=\frac{E}{2}$

The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in…

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