A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is now introduced between the two plates to occupy the space completely. State the effect on the following:
(i) the capacitance of the capacitor.
(ii) potential difference between the plates.
(iii) the energy stored in the capacitor.

Explanation

To solve the problem, we will analyze the effects of introducing a dielectric slab into a parallel plate capacitor that has been charged and then disconnected from the battery. We will examine the changes in capacitance, potential difference, and energy stored in the capacitor.

### Step-by-Step Solution:

1. **Initial Setup**:
- Let the initial capacitance of the capacitor (without the dielectric) be \( C_0 \).
- The potential difference across the capacitor when connected to the battery is \( V_0 \).
- The charge stored on the capacitor is \( Q_0 = C_0 V_0 \).
...