Two charges Q₁ C and Q₂ C are shown in fig. A third charge Q₃ coulomb is moved from point R to S along a circular path with P as centre. Change in potential energy is

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$\frac{Q_{1} Q_{2}}{{πϵ}_{0}}$

$\frac{4 Q_{1} Q_{2}}{{πϵ}_{0}}$

$\frac{Q_{1} Q_{2} Q_{3}}{4 {πϵ}_{0}}$

$\frac{Q_{2} Q_{3}}{{πϵ}_{0}}$

answer is C.

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Detailed Solution

$\begin{array}{l} ΔU = U_{f} - U_{i} = \frac{1}{4 {πε}_{0}} [\frac{Q_{1} Q_{2}}{0 .6} + \frac{Q_{1} Q_{3}}{0 .8} + \frac{Q_{3} Q_{2}}{0 .2}] \\ - \frac{1}{4 {πε}_{0}} [\frac{Q_{1} Q_{2}}{0 .6} + \frac{Q_{1} Q_{3}}{0 .8} + \frac{Q_{3} Q_{2}}{1}] \\ = \frac{Q_{2} Q_{3}}{4 {πε}_{0}} [5 - 1] = \frac{Q_{2} Q_{3}}{{πε}_{0}} \end{array}$