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(D) Initial capacitance $C_i = \frac{\varepsilon_0 A}{d}$.Initial energy $U_i = \frac{1}{2} C_i V_0^2 = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \

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$\frac{\varepsilon_0 A V_0^2}{d}$

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(D) Initial capacitance $C_i = \frac{\varepsilon_0 A}{d}$.Initial energy $U_i = \frac{1}{2} C_i V_0^2 = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) V_0^2$.Since the battery is disconnected,the charge $Q$ on the plates remains constant.$Q = C_i V_0 = \left( \frac{\varepsilon_0 A}{d} \right) V_0$.When the separation becomes $d' = 3d$,the new capacitance is $C_f = \frac{\varepsilon_0 A}{3d} = \frac{C_i}{3}$.The new energy $U_f = \frac{Q^2}{2 C_f} = \frac{(C_i V_0)^2}{2 (C_i / 3)} = \frac{3}{2} C_i V_0^2$.Work done $W = U_f - U_i = \frac{3}{2} C_i V_0^2 - \frac{1}{2} C_i V_0^2 = C_i V_0^2$.Substituting $C_i = \frac{\varepsilon_0 A}{d}$,we get $W = \frac{\varepsilon_0 A V_0^2}{d}$.

$A$ parallel plate capacitor has plate area $A$ and separation $d$. It is charged to a potential difference $V_0$. The charging battery is disconnected and the plates are pulled apart to $3$ times the initial separation. The work required to separate the plates is

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