d
(d) 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. \(\frac{1}{2}m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{qQ}}{r}\)
Therefore the distance of closest approach is given by
\(r = \frac{{qQ}}{{4\pi {\varepsilon _0}}}.\frac{2}{{m{v^2}}}\) ==> \(r \propto \frac{1}{{{v^2}}}\)
Hence if \(v\) is doubled, \(r\) becomes one fourth.