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Four particles, each of mass m and charge q, are held at the vertices of a square of side 'a', as shown in Fig. 2.132. They are released at t = 0 and move under mutual repulsive forces . Speed of any particle when its.distance from the centre of square doubles, is :

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${[\frac{1}{4 {πε}_{0}} \frac{q^{2}}{ma} (1 + \frac{1}{2 \sqrt{2}})]}^{\frac{1}{2}}$

${[\frac{1}{4 {πε}_{0}} \frac{q^{2}}{ma}]}^{\frac{1}{2}}$

${[\frac{1}{4 {πε}_{0}} \frac{2 q^{2}}{{ma}^{2}} (1 + \frac{1}{2 \sqrt{2}})]}^{\frac{1}{2}}$

${[\frac{1}{4 {πε}_{0}} \frac{q^{2}}{{ma}^{2}}]}^{\frac{1}{2}}$

answer is A.

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Detailed Solution

Loss in potential energy

$= [\frac{4 {kq}^{2}}{a} + \frac{2 {kq}^{2}}{a \sqrt{2}}] - [\frac{4 {kq}^{2}}{(2 a)} + \frac{2 {kq}^{2}}{(2 a \sqrt{2})}] = (2 + \frac{1}{\sqrt{2}}) \frac{{kq}^{2}}{a}$

As gain in kinetic energy = loss in potential energy

So, $4 x \frac{1}{2} {mv}^{2} = (2 + \frac{1}{\sqrt{2}}) \frac{{kq}^{2}}{a}$

$\Rightarrow v = \sqrt{(1 + \frac{1}{2 \sqrt{2}}) \frac{q^{2}}{4 {πε}_{0} ma}}$

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