A parallel plate capacitor each with plate area A and separation ‘d’ is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant K is now placed between the plates. What change if any, will take place in
(i) charge on the plates
(ii) electric field intensity between the plates,
(iii) capacitance of the capacitor
Justify your answer in each case

Explanation

To solve the problem step by step, we will analyze the situation of a parallel plate capacitor before and after inserting a dielectric slab.

### Given:
- Plate area of the capacitor = A
- Separation between plates = d
- Initial potential difference = V
- Dielectric slab thickness = d
- Dielectric constant of the slab = K
...