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(A) The capacitance of a parallel plate capacitor is given by $C_0 = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the sep

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$4$

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(A) The capacitance of a parallel plate capacitor is given by $C_0 = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the separation.Initially,the charge stored is $Q = C_0 V_0$.$(i)$ When the battery is disconnected,the charge $Q$ remains constant. If the separation is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{C_0}{2}$. The energy stored is $E_1 = \frac{Q^2}{2C'} = \frac{(C_0 V_0)^2}{2(C_0 / 2)} = C_0 V_0^2$.(ii) When the battery remains connected,the potential $V_0$ remains constant. If the separation is doubled,the new capacitance is $C' = \frac{C_0}{2}$. The energy stored is $E_2 = \frac{1}{2} C' V_0^2 = \frac{1}{2} (\frac{C_0}{2}) V_0^2 = \frac{1}{4} C_0 V_0^2$.Therefore,the ratio is $\frac{E_1}{E_2} = \frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2} = 4$.

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