a
According to conservation principal of mechanical energy.
$\mathrm{U}_{\mathrm{i}}+\mathrm{T}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{T}_{\mathrm{f}}$
$\mathrm{qV}+\frac{1}{2} \mathrm{mv}_{0}^{2}=\frac{\mathrm{qQ}}{4 \pi \in_{0} \mathrm{a}}+0$ $....(1)$
where $v_{0}$ is velocity of at point $'p'$
the potential at the point $p$ due to $+Q$ is
$v=\frac{Q}{4 \pi \in_{0}(4 a)}$
From eqn $(1)$
$\mathrm{q} \mathrm{v}+\frac{1}{2} \mathrm{mv}_{0}^{2}=4 \mathrm{qV}$
$\frac{1}{2} \mathrm{mv}_{0}^{2}=3 \mathrm{qV}$
$\mathrm{v}_{0}^{2}=\frac{6 \mathrm{qV}}{\mathrm{m}}$
$\Rightarrow \quad v_{0}=\sqrt{\frac{6 q V}{m}}$
