Two charge $q_1$ and $q_2$ are placed $30 \mathrm{~cm}$ apart, as shown in the figure. A third charge $q_3$ is moved along the are of a circle of radius $40 \mathrm{~cm}$ from $\mathrm{C}$ to $\mathrm{D}$. The change in the potential energy of the system is \(\frac{q_3 k}{4 \pi \in_0}\), where $k$ is:

• A

  $8 q_1$
• B

  $6 q_1$
• Correct answer
  C

  $8 q_2$
• D

  $6 q_2$

Solution

$$ \begin{aligned} & V_i=\frac{1}{4 \pi \epsilon_0}\left[\frac{q_1 q_3}{(0.4)}+\frac{q_1 q_2}{(0.3)}+\frac{q_2 q_3}{(0.5)}\right] \\ & v_f=\frac{1}{4 \pi \epsilon_0}\left[\frac{q_1 q_3}{(0.4)}+\frac{q_1 q_2}{(0.3)}+\frac{q_2 q_3}{(0.1)}\right] \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \Delta V=V f-V_1 \\ & =\frac{1}{4 \pi \epsilon_0} q_2 q_3\left(\frac{1}{0.1}-\frac{1}{0.5}\right) \\ & =\frac{q_2 q_3}{\pi \epsilon_0}\left(10^{-2}\right)=\frac{q_3}{4 \pi \epsilon_0}\left(8 q_2\right) \\ & \Rightarrow k=8 q_2 \end{aligned} $$

Hence, the correct answer is option: 'C'.