Explanation

(d) Let a particle of change $q\ $having velocity v approaches Q upto a closest distance $r$ and if the velocity becomes $2v$, the closest distance will be $r$.$^{’}$
The law of conservation of energy yields,
Kinetic energy of particle=electric potential energy between them at closest distance of approach.
Or $\frac{1}{2}{mv}^{2} = \frac{1}{4\pi\varepsilon_{0}}\frac{Q_{q}}{r}\ $
Or $\frac{1}{2}{mv}^{2} = k\frac{Qq}{r}$ …(i)
$\left( k = constant = \ \frac{1}{4\pi\varepsilon_{0}} \right)$
and $\frac{1}{2}m(2v)^{2}\ = \ k\frac{Qq}{r'}$ …(ii)
Dividing Eq. (i) by Eq.(ii),
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\frac{1}{2}{mv}^{2}}{\frac{1}{2}{m(2v)}^{2}}$ =$\frac{\frac{kQq}{r}}{\frac{kQq}{r^{'}}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{4} = \frac{r^{'}}{r}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r’\ = \frac{r}{4}$