There is an air capacitor with capacitance C, charged to a potential difference of V. If the capacitor is isolated from the battery after charging, and then a dielectric slab is inserted in between the plates of capacitor. How will the energy stored in this capacitor change?

Explanation

To solve the problem step by step, let's analyze the situation involving the capacitor, the dielectric slab, and the energy stored in the capacitor.

### Step 1: Understand the Initial Conditions
- The capacitor has a capacitance \( C \) and is charged to a potential difference \( V \).
- The initial energy stored in the capacitor can be calculated using the formula:
\[ U_i = \frac{1}{2} C V^2 \]
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