Capacitor With Dielectrics: Practice Problem & Solution
Assertion (A): A parallel plate capacitor is connected across battery through a key. A dielectric slab of dielectric constant \(K\) is introduced between the plates. The energy which is stored becomes \(K\) times. Reason (R): The surface density of charge on the plate remains constant or unchanged.
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Capacitor With Dielectrics. Understanding the underlying formula is key to arriving at the correct answer below:
A: True. When connected to a battery, potential \(V\) is constant. Energy \(U = \frac{1}{2}CV^2\). As dielectric \(K\) is inserted, \(C\) becomes \(KC_0\), so \(U\) becomes \(KU_0\).\nR: False. Charge \(Q = CV\). Since \(C\) increases by \(K\) and \(V\) is constant, \(Q\) also increases by \(K\). Thus, surface charge density \(\sigma = Q/A\) also increases. Therefore, (A) is true and (R) is false.
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