A parallel plate capacitor of capacity $C_{0}$ is charged to a potential $V_{0}$ .
(1) The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_{1}$ .
(2) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_{2} .$ Then $E_{1} / E_{2}$

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$\frac{1}{2}$

$\frac{3}{2}$

answer is A.

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Detailed Solution

When battery is disconnected, $U_{0} = \frac{Q^{2}}{2 c}$

Q is same when battery is kept connected $U_{0} = \frac{1}{2} C V^{2}$

$V is same E_{1} = \frac{Q^{2}}{2 \times \frac{C_{0}}{2}} = \frac{Q^{2}}{C_{0}} = C_{0} V_{0}^{2}$

$E_{2} = \frac{1}{2} \frac{C_{0}}{2} V_{0}^{2} = \frac{C_{0} V_{0}^{2}}{4} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{4}{1}$

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