A parallel plate capacitor having plate area | vedclass

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(A) Initial capacitance $C_i = \frac{\epsilon_0 A}{d}$.Initial charge $Q = C_i V = \frac{\epsilon_0 A V}{d}$.Since the battery is disconnected, the ch

Vedclass · Accepted Answer

$\frac{3 \epsilon_0 A V^2}{2 d}$

Vedclass · Answer

(A) Initial capacitance $C_i = \frac{\epsilon_0 A}{d}$.Initial charge $Q = C_i V = \frac{\epsilon_0 A V}{d}$.Since the battery is disconnected, the charge $Q$ remains constant.Final separation $d_f = 4d$.Final capacitance $C_f = \frac{\epsilon_0 A}{4d} = \frac{C_i}{4}$.Initial energy $U_i = \frac{Q^2}{2C_i} = \frac{1}{2} C_i V^2 = \frac{\epsilon_0 A V^2}{2d}$.Final energy $U_f = \frac{Q^2}{2C_f} = \frac{Q^2}{2(C_i/4)} = 4 \left( \frac{Q^2}{2C_i} \right) = 4 U_i$.Work done $W = U_f - U_i = 4 U_i - U_i = 3 U_i$.Substituting $U_i$, $W = 3 \left( \frac{\epsilon_0 A V^2}{2d} \right) = \frac{3 \epsilon_0 A V^2}{2d}$.

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