Two charges $q_{1}$ and $q_{2}$ are placed 30cm apart shown in the figure. A third charge $q_{3}$ is moved along the arc of a circle of radius 40cm from C to D. The change in the potential energy of the system is $\frac{q_{3}}{4 π ε_{0}} k$ Where $k$ is

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$8 q_{1}$

$8 q_{2}$

$6 q_{2}$

$6 q_{1}$

answer is A.

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Detailed Solution

Change in potential energy $(Δ U) = U_{f} - U_{i}$

$= Δ U = \frac{1}{4 π ε_{0}} [(\frac{q_{1} q_{3}}{0.4} + \frac{q_{2} q_{3}}{0.1} + \frac{q_{1} q_{2}}{0.3}) - (\frac{q_{1} q_{3}}{0.4} + \frac{q_{2} q_{3}}{0.5} + \frac{q_{1} q_{2}}{0.3})] = \frac{q_{3} q_{2} 8}{4 π ε_{0}}$

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