If two charges $$q_1$$ and $$q_2$$ are separated with distance $$d$$ and placed in a medium of dielectric constant $$k$$. What will be the equivalent distance between charges in air for the same electrostatic force?

In a medium of dielectric constant $$k$$, the electrostatic force between charges $$q_1$$ and $$q_2$$ separated by distance $$d$$ is given by

$$F = \frac{q_1 q_2}{4\pi\epsilon_0 k d^2}$$

Similarly, if the same force acts in air at an equivalent distance $$d'$$, it can be expressed as

$$F = \frac{q_1 q_2}{4\pi\epsilon_0 d'^2}$$

Equating these two expressions yields $$\frac{1}{k d^2} = \frac{1}{d'^2}$$, which implies $$d'^2 = k d^2$$ and therefore $$d' = d\sqrt{k}$$, showing that the equivalent distance in air is $$d\sqrt{k}$$.