Two identical particles of mass $m$ each carry a charge $Q$ each. One is at rest on a horizontal surface and the other is projected on the plane directly towards the first from a large distance with a speed $v$ . The closest distance of approach is

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$\frac{1}{4 π ε_{o}} \frac{Q^{2}}{v^{2}}$

$\frac{1}{4 π ε_{o}} \frac{Q^{2}}{2m}$

$\frac{1}{2 π ε_{o}} \frac{Q^{2}}{{mv}^{2}}$

$\frac{1}{4 π ε_{o}} \frac{Q^{2}}{{mv}^{2}}$

answer is D.

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Detailed Solution

The kinetic energy of the second particle is equal to the electrostatic potential energy between the two particles which are separated by a distance $x$

$\frac{1}{2} {mv}^{2} = \frac{1}{4 π ε_{o}} \frac{Q^{2}}{x} \Rightarrow x = \frac{1}{2 π ε_{o}} \times \frac{Q^{2}}{{mv}^{2}}$

So, the correct answer is (3)

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