A charged particle q is shot towards another charged particle Q which is fixed, with a speed $v$ . It approaches Q upto a closest distance r and then returns. If q were given a speed $2v^{2}$ , the closest distances of approach would be
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Charge q will momentarily come to rest at a distance r from charge Q when all it's kinetic energy converted to potential energy i.e. $\frac{1}{2}mv^{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{qQ}{r}$
Therefore the distance of closest approach is given by
$r = \frac{qQ}{4 \pi \epsilon_0} \cdot \frac{2}{mv^2}$ ⇒ ⇒ $r \propto \frac{1}{v^{2}}$
Hence if v is doubled, r becomes one fourth.
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