A parallel plate capacitor with a dielectric | vedclass

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Question

(A) Let the initial capacitance be $C = \frac{K \epsilon_0 A}{d}$. The charge on the capacitor is $Q = CV = \frac{K \epsilon_0 A V}{d}$.When the diele

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Zero

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(A) Let the initial capacitance be $C = \frac{K \epsilon_0 A}{d}$. The charge on the capacitor is $Q = CV = \frac{K \epsilon_0 A V}{d}$.When the dielectric is removed,the capacitance becomes $C_0 = \frac{\epsilon_0 A}{d} = \frac{C}{K}$.Since the capacitor is disconnected from the battery (implied by the process of removing/reinserting),the charge $Q$ remains constant.The initial potential energy is $U_i = \frac{Q^2}{2C}$.When the dielectric is removed,the potential energy becomes $U_{removed} = \frac{Q^2}{2C_0} = \frac{Q^2}{2(C/K)} = K \frac{Q^2}{2C} = K U_i$.When the dielectric is reinserted,the capacitance returns to $C$,and the potential energy returns to $U_f = \frac{Q^2}{2C} = U_i$.The net work done by the system is the change in potential energy: $W = -(U_f - U_i) = -(U_i - U_i) = 0$.Thus,the net work done by the system in this process is zero.

$A$ parallel plate capacitor with a dielectric of dielectric constant $K$ between the plates has a capacitance $C$ and is charged to a potential $V \ volt$. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is

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