Question

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 $\frac{q_{3}}{4 \pi \varepsilon_{o}} k$, where $k$ is

• A. $8 q_{1}$
• B. $6 q_{1}$
• C. $8 q_{2}$
• D. $6 q_{2}$

Answer 1

Answer: (C)

The potential energy when $q_{3}$ is at point $C$
$U_{1}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1} q_{3}}{0.40}+\frac{q_{2} q_{3}}{\sqrt{(0.40)^{2}+(0.30)^{2}}}\right]$
The potential energy when $q_{3}$ is at point D
$U_{2}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1} q_{3}}{0.40}+\frac{q_{2} q_{3}}{0.10}\right]$
Thus charge in potential energy is $\Delta U=U_{2}-U_{1}$
$\Rightarrow \frac{q_{3}}{4 \pi \varepsilon_{0}} k=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1} q_{3}}{0.40}+\frac{q_{2} q_{3}}{0.10}-\frac{q_{1} q_{3}}{0.40}-\frac{q_{2} q_{3}}{0.50}\right] $
$\Rightarrow k=\frac{5 q_{2}-q_{2}}{0.50}=\frac{4 q_{2}}{0.50}=8 q_{2}$