b
(b)
At distance of closest approach, Total initial $KE =$ Total final PE
$\therefore \quad \frac{1}{2} m v^2+\frac{1}{2} m v^2=\frac{k q^2}{r}$
$\Rightarrow \quad m v^2=\frac{q^2}{4 \pi \varepsilon_0 r}$
$\Rightarrow \quad r=\frac{q^2}{4 \pi \varepsilon_0 m v^2}$