(i) Battery remains connected
(ii) Battery is disconnected
Mark the CORRECT statement.
Solution:
Initial capacitance \(C = \frac{\epsilon_0 A}{d}\). Doubling separation gives \(C' = C/2\).
Case (i) Battery connected: \(V\) is constant. Energy \(U = \frac{1}{2} C V^2\). \(U' = \frac{1}{2} (C/2) V^2 = U/2\). Energy decreases. Work by battery \(W_{batt} = V \Delta Q = V(C'V - CV) = V(-CV/2) = -QV/2\), which is negative.
Case (ii) Battery disconnected: \(Q\) is constant. Energy \(U = \frac{Q^2}{2C}\). \(U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/2)} = 2U\). Energy increases. Potential difference \(V = Q/C\), \(V' = Q/C' = Q/(C/2) = 2V\). Potential difference increases. Thus, statement C is correct.
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