A parallel plate capacitor `(C=50 muF. d=4mm)` is charged to `200V` and then charging battery is removed. Now a dielectric slab `(K=4)` of thickness `2mm` is placed between the plates. Find new potential difference across capacitor.

Explanation

To solve the problem, we need to find the new potential difference across a parallel plate capacitor after inserting a dielectric slab. Here’s a step-by-step solution:

### Step 1: Calculate the initial charge on the capacitor
The initial capacitance \( C \) is given as \( 50 \, \mu F \) and the initial voltage \( V \) is \( 200 \, V \).

The charge \( Q \) stored in the capacitor can be calculated using the formula:
\[
Q = C \times V
...