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Question

According to conservation principal of mechanical energy.

$\mathrm{U}_{\mathrm{i}}+\mathrm{T}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{T}_{\mat

Vedclass · Accepted Answer

$\sqrt {\frac{{6qV}}{m}} $

Vedclass · Answer

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

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