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

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Aicharya · Accepted Answer

Change in potential energy $(\Delta U)=U_{f}-U_{i}$
$\Rightarrow \quad \Delta U=\frac{1}{4 \pi \varepsilon_{0}}\left[\left(\frac{q_{1} q_{3}}{0.4}+\frac{q_{2} q_{3}}{0.1}\right)-\left(\frac{q_{1} q_{3}}{0.4}+\frac{q_{2} q_{3}}{0.5}\right)\right]$
$\Rightarrow \quad \Delta U=\frac{1}{4 \pi \varepsilon_{0}}\left[8 q_{2} q_{3}\right]=\frac{q_{3}}{4 \pi \varepsilon_{0}}\left(8 q_{2}\right)$
$\therefore \quad k=8 q_{2}$
$\therefore \quad$ (a)

Two charges $q_{1}$ and $q_{2}$ are placed $30 \mathrm{~cm}$ apart as shown. A third charge $q_{3}$ is moved…

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