Parallel Plate Capacitor: Practice Problem & Solution
A parallel plate capacitor is charged and then the battery is removed. If the plates are now moved close to each other then, (A) The charge stored on it, increases. (B) The energy stored in it, decreases. (C) Its capacitance increases. (D) The potential difference between the plates decreases. The correct statement(s) is/are
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:
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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