The capacitance of a parallel-plate capacitor is when the region between the pla

The capacitor remains connected to a battery of emf \( \varepsilon \). Initial capacitance with dielectric is \( K C_0 \), so initial charge \( Q_i = K C_0 \varepsilon \) and initial energy \( U_i = \frac{1}{2} K C_0 \varepsilon^2 \). After removing the dielectric, capacitance becomes \( C_0 \), so final charge \( Q_f = C_0 \varepsilon \) and final energy \( U_f = \frac{1}{2} C_0 \varepsilon^2 \). The charge that flows through the battery is \( Q_i - Q_f = C_0 \varepsilon (K-1) \), and the battery absorbs energy \( \varepsilon \cdot \Delta Q = \varepsilon^2 C_0 (K-1) \). By conservation of energy, the work done by the external agent \( W_{\text{ext}} \) satisfies: \( U_i + W_{\text{ext}} = U_f + \text{energy absorbed by battery} \). Substituting values gives \( W_{\text{ext}} = \frac{1}{2} \varepsilon^2 C_0 (K-1) \). Options A and B are also correct, but D is the work done by the external agent, which is a common focus in such problems.