Question

When the battery across the plates of a charged condenser is disconnected and a dielectric slab is introduced between its plates, then the energy stored

• A. becomes infinity
• B. does not change
• C. increases
• D. decreases

Hint:

Always track the state of the battery! If the battery stays connected, voltage remains constant ($V = \text{constant}$), and energy increases ($U = \frac{1}{2}CV^2 \propto C$). But if the battery is disconnected, charge remains constant ($Q = \text{constant}$), and energy decreases ($U = \frac{Q^2}{2C} \propto \frac{1}{C}$). Keeping this distinction clear avoids a common trap!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Distinguish how the energy stored in a capacitor changes when a dielectric is inserted, depending on whether the battery remains connected or is disconnected.

Step 2: Key Formula or Approach:
Battery connected → V constant → U = ½CV² ∝ C (energy increases). Battery disconnected → Q constant → U = Q²/(2C) ∝ 1/C (energy decreases).

Step 3: Detailed Explanation:
With the battery connected, inserting a dielectric raises capacitance, and the constant voltage forces more charge onto the plates, increasing stored energy. With the battery disconnected, the charge is trapped; the increased capacitance lowers the voltage, reducing stored energy. Confusing these two scenarios is a classic exam pitfall—tracking the battery state avoids it completely.

Step 4: Final Answer:
Energy increases with battery connected; decreases with battery disconnected.