Two charges $q_{1}$ and $q_{2}$ are placed $30 cm$ apart, as shown in the figure. A third charge $q_{3}$ is moved along the arc of a circle of radius $40 cm$ 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

8 q_{2}

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6 q_{2}

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8 q_{1}

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6 q_{1}

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Solution

The correct option is A $8 q_{2}$

Initial potential energy
$U_{i} = \frac{1}{4 π ϵ_{0}} [\frac{q_{1} q_{3}}{(0.4)} + \frac{q_{1} q_{2}}{(0.3)} + \frac{q_{2} q_{3}}{(0.5)}]$
Final potential energy
$U_{f} = \frac{1}{4 π ϵ_{0}} [\frac{q_{1} q_{3}}{(0.4)} + \frac{q_{1} q_{2}}{(0.3)} + \frac{q_{2} q_{3}}{(0.1)}]$
Change in potential energy
$Δ U = U_{f} - U_{i} = \frac{q_{2} q_{3}}{4 π ϵ_{0}} (\frac{1}{0.1} - \frac{1}{0.5})$
$Δ U = \frac{q_{3}}{4 π ϵ_{0}} (8 q_{2})$
So, $k = 8 q_{2}$

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Two charges $q_{1}$ and $q_{2}$ are placed 30cm apart, as shown in figure. A third charges $q_{3}$ is moved along the arc of a circle of radius 40 cm 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