A parallel plate capacitor has plate area A a | vedclass

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(D) Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d}$.Initial charge $Q = C_0 V_0$.Since the battery is disconnected,the charge $Q$ remains consta

Vedclass · Accepted Answer

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

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(D) Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d}$.Initial charge $Q = C_0 V_0$.Since the battery is disconnected,the charge $Q$ remains constant.Final separation $d' = 3d$.Final capacitance $C' = \frac{\varepsilon_0 A}{3d} = \frac{C_0}{3}$.Initial potential energy $U_i = \frac{Q^2}{2C_0} = \frac{1}{2} C_0 V_0^2$.Final potential energy $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(C_0/3)} = \frac{3Q^2}{2C_0} = \frac{3}{2} C_0 V_0^2$.Work done $W = U_f - U_i = \frac{3}{2} C_0 V_0^2 - \frac{1}{2} C_0 V_0^2 = C_0 V_0^2$.Substituting $C_0 = \frac{\varepsilon_0 A}{d}$,we get $W = \frac{\varepsilon_0 A V_0^2}{d}$.

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