Question

Two charges each $Q$ are released when the distance between them is $d$. Then the velocity of each charge of mass $m$ each when the distance between them is $2d$ is :

• A. $\frac{Q}{\sqrt{8\pi {\epsilon }_{0}dm}}$
• B. $\frac{Q}{\sqrt{4\pi {\epsilon }_{0}dm}}$
• C. $\frac{Q}{4\sqrt{\pi {\epsilon }_{0}dm}}$
• D. $\frac{Q}{\sqrt{2\pi {\epsilon }_{0}dm}}$

Answer 1

The correct option is A $\frac{Q}{\sqrt{8\pi {\epsilon }_{0}dm}}$

we have P.E between two charges =$\frac{k{q}_1{q}_2}{d_1}$
applying law of conservation of energy
$\Rightarrow \frac{kQQ}{d}=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+\frac{kQQ}{2d}$

$\Rightarrow \frac{k{Q}^2}{2d}=m{v}^2$

$\Rightarrow \frac{Q^2}{8\pi {ϵ}_{0}dm}=v^2$

$\Rightarrow v=\frac{Q}{\sqrt{8\pi {ϵ}_{0}dm}}$