The battary remains connected to a parallel plate capacitor and a dielectric slab is inserted between the plates. What will be the effect on its (i) capacity (ii) charge , (iiI) potential difference (iv) electric field, (v) energy stored ? Justify your answer.

Explanation

To solve the problem step by step, we will analyze the effects of inserting a dielectric slab into a parallel plate capacitor that is connected to a battery.

### Step-by-Step Solution:

1. **Understanding Capacitance with Dielectric**:
- The capacitance \( C \) of a parallel plate capacitor without a dielectric is given by:
\[
C = \frac{A \epsilon_0}{d}
...