When air is replaced by a dielectric medium of constant $$K,$$ the maximum force of attraction between two charges, separated by a distance

Solution :
According to Coulomb's law, force between two charges is directly proportional to product of charges and inversely proportional to square of distance between them. Thus,
$$F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}}\,\,......\left( {\text{i}} \right)$$
where, $$\frac{1}{{4\pi {\varepsilon _0}}} = $$   proportionality constant.
If a dielectric medium of constant $$K$$ is placed between them, then new force between them,
$${F_1} = \frac{1}{{4\pi {\varepsilon _0}K}} \cdot \frac{{{q_1}{q_2}}}{{{r^2}}}\,\,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we have
$$\eqalign{ & \frac{{{F_1}}}{F} = \frac{1}{K} \cr & {\text{or}}\,\,F' = \frac{F}{K} \cr} $$
Thus, new force decreases $$K$$ times.