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(D) $1$. Initial capacitance $C_0 = \frac{\epsilon_0 A}{d}$.$2$. Initial charge $Q = C_0 V = \frac{\epsilon_0 A V}{d}$.$3$. Initial energy $U_i = \fra

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$- \frac{1}{4} \frac{\epsilon_0 A}{d} V^2$

Vedclass · Answer

(D) $1$. Initial capacitance $C_0 = \frac{\epsilon_0 A}{d}$.$2$. Initial charge $Q = C_0 V = \frac{\epsilon_0 A V}{d}$.$3$. Initial energy $U_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} \frac{\epsilon_0 A}{d} V^2$.$4$. After the battery is disconnected,the charge $Q$ remains constant.$5$. When a metal slab of thickness $t = d/2$ is inserted,the new capacitance $C_f$ is given by $C_f = \frac{\epsilon_0 A}{d - t} = \frac{\epsilon_0 A}{d - d/2} = \frac{2 \epsilon_0 A}{d} = 2 C_0$.$6$. The final energy $U_f = \frac{Q^2}{2 C_f} = \frac{(C_0 V)^2}{2(2 C_0)} = \frac{C_0^2 V^2}{4 C_0} = \frac{1}{4} C_0 V^2 = \frac{1}{4} \frac{\epsilon_0 A}{d} V^2$.$7$. Work done by the external agent $W = U_f - U_i = \frac{1}{4} C_0 V^2 - \frac{1}{2} C_0 V^2 = - \frac{1}{4} C_0 V^2 = - \frac{1}{4} \frac{\epsilon_0 A}{d} V^2$.

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