Explanation

(d) From Coulomb’s law, the force between two points charges $q_{1}$ and $q_{2}$ separated by the distance $r$ is given by
$\ \ \ \ \ F = \frac{1}{4\pi\varepsilon_{0}}.\frac{q_{1}q_{2}}{r^{2}}$
Since, $q_{1} = q_{2} = q$
$\therefore\ \ \ \ \ \ \ \ F = \frac{1}{4\pi\varepsilon_{0}}.\frac{q^{2}}{r^{2}}\ \ \ldots(i)$
In second case when
$q_{1} = q_{2} = \frac{q}{2}$ and $r^{'} = 2r$
We have
$F^{'} = \frac{1}{4\pi\varepsilon_{0}}\frac{(q/2)^{2}}{(2r)^{2}} = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{16r^{2}}\ \ \ \ ..(ii)$
Dividing Eq. (ii) by Eq. (i), we have
$\ \ \ \ \ \ \ \ \ \frac{F^{'}}{F} = \frac{q^{2}}{16r^{2}} \times \frac{r^{2}}{q^{2}} = \frac{1}{16}$
$\Longrightarrow \ \ F^{'} = \frac{F}{16}$