What is the limit of achieving maximum velocity for two charges of same mass $m$ and having unequal charges $q_{1}$ and $q_{2}$ in a gravity free space where they are at present distance ' $a$ ' apart which is almost negligible compared to the distance where their velocity will be maximum :

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${(\frac{q_{1} q_{2}}{2 π ε_{0} m a})}^{\frac{1}{2}}$

${(\frac{q_{1} q_{2}}{4 π ε_{0} m a})}^{\frac{1}{2}}$

${(\frac{q_{1} q_{2}}{π ε_{0} m a})}^{\frac{1}{2}}$

${(\frac{q_{1} q_{2}}{4 \sqrt{2} π ε_{0} m a})}^{\frac{1}{2}}$

answer is B.

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Detailed Solution

At distance $' a'$ apart, kinetic energy of the particles is assumed zero and at infinity their kinetic energy will be maximum.
$\begin{array}{l} \frac{q_{1} q_{2}}{4 π ε_{0} m a} = \frac{1}{2} m {V_{\max}}^{2} \times 2 \\ V_{\max} = \sqrt{\frac{q_{1} q_{2}}{4 π ε_{0} m a}} \end{array}$