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 $\frac{q_{3}}{4 \pi \, \in _{o}} k$ where $k$ is -

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

Answer 1

Answer: (C)

$U_{\text{1}}\text{ =}\frac{\textit{kq}_{\text{1}} \textit{q}_{\text{3}}}{4 \pi \epsilon _{0} \times \text{0.4}}\text{+}\frac{\textit{kq}_{\text{1}} \textit{q}_{\text{2}}}{4 \pi \epsilon _{0} \times \text{0.3}}\text{+}\frac{\textit{kq}_{\text{2}} \textit{q}_{\text{3}}}{4 \pi \epsilon _{0} \times \text{0.5}}$
$U_{\text{2}}\text{=}\frac{\textit{kq}_{\text{1}} \textit{q}_{\text{2}}}{4 \pi \epsilon _{0} \times \text{0.3}}\text{+}\frac{\textit{kq}_{\text{2}} \textit{q}_{\text{3}}}{4 \pi \epsilon _{0} \times \text{0.1}}\text{+}\frac{\textit{kq}_{\text{1}} \textit{q}_{\text{3}}}{4 \pi \epsilon _{0} \times \text{0.4}}$
$\therefore \Delta U=\frac{8 k q_{2} q_{3}}{4 \pi \left(\epsilon \right)_{0}}=8q_{2}\left(\frac{q_{3}}{4 \pi \left(\epsilon \right)_{0}}\right)\Rightarrow k=8q_{2}$