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Assertion (A): A parallel plate capacitor connected to a battery has a dielectric slab of constant K inserted between its plates. The energy stored in the capacitor becomes K times the original value. Reason (R): The surface charge density on the plates remains unchanged when the dielectric is inserted.

1. Both A and R are true but R is not the correct explanation of A.
2. A is true but R is false.
3. A is false but R is true.
4. Both A and R are true and R is the correct explanation of A.

Correct answer: Both A and R are true and R is the correct explanation of A.

Solution

When the battery remains connected, the potential V across the capacitor stays constant. Capacitance C = K*epsilon₀*A/d increases by K. Energy U = (1/2)*K*C*V² = K*(1/2)*C*V² becomes K times. The electric field E = V/d is unchanged; sigma_free = K*epsilon₀*(V/d) = K*sigma₀ — actually sigma INCREASES by K. So R is FALSE. A is true (energy becomes K times with battery), but R is false. Answer: A is true but R is false.

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