Write the expression for the Coulombian force acting between two point charges kept in a medium.

(N/A) The Coulombian force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ in a medium with permittivity $\epsilon$ is given by:
$F = \frac{1}{4\pi\epsilon} \frac{q_1 q_2}{r^2}$
Where $\epsilon = \epsilon_0 \epsilon_r$ (or $\epsilon = \epsilon_0 K$),$\epsilon_0$ is the permittivity of free space,and $\epsilon_r$ (or $K$) is the relative permittivity or dielectric constant of the medium.
Thus,the expression can also be written as:
$F = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{r^2}$