A particle of charge q and mass m is projected from a large distance towards another identical charged particle at rest with velocity v0. The distance of closest approach will be (Assume negligible gravitational attraction)

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$\frac{2 q^{2}}{{πε}_{0} {mv}_{0}^{2}}$

$\frac{q^{2}}{{πε}_{0} {mv}_{0}^{2}}$

$\frac{q^{2}}{2 {πε}_{0} {mv}_{0}^{2}}$

$\frac{q^{2}}{4 {πε}_{0} {mv}_{0}^{2}}$

answer is B.

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Detailed Solution

Applying conservation of momentum : 2mv = mv₀ . .. (i)
conservation of energy : $2 \times \frac{1}{2} {mv}^{2} + \frac{1}{4 {πε}_{0}} \frac{q^{2}}{d} = \frac{1}{2} {mv}_{0}^{2} . . . . (ii)$
$(I) \Rightarrow V = v_{0} / 2$ ,putting in (2), we get,
$d = \frac{q^{2}}{{πε}_{0} {mv}_{0}^{2}}$