Solution:
When the battery is disconnected, the charge \(Q\) remains constant. Decreasing the plate separation \(d\) increases capacitance \(C = \frac{\epsilon_0 A}{d}\). This in turn decreases potential difference \(V = \frac{Q}{C}\) and decreases energy stored \(U = \frac{Q^2}{2C}\). Thus, statements B, C, and D are correct.
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