Question

The force of repulsion between two identical positive charges when kept with the separation are in air is f half the gap between the two charges if filled by a dielectrib slab of dielectri constant 4

Answer 1

it is given that, force of repulsion between two identical positive charge $q_1$ and $q_2$ when kept with separation r in air is f.

e.g., f = $\frac{kq_1q_2}{r^2}$.....(1)

thickness of slab is filled between the charges , t = r/2

dielectric constant of slab, k = 4

force of repulsion when a slab is filled between charges , F = $\frac{kq_1q_2}{(r-t+t\sqrt{k})^2}$

= $\frac{kq_1q_2}{(r-r/2+r)^2}$

= $\frac{4kq_1q_2}{9r^2}$

from equation (1),

F = 4f/9