Two charges $q_1$ and $q_2$ are separated by a distance of $30\ \text{cm}$. A third charge $q_3$ initially at C, as shown in the figure, is moved along the circular path of radius $40\ \text{cm}$ (centred at $q_1$) from C to D. If the difference in potential energy due to the movement of $q_3$ from C to D is given by $\dfrac{q_3K}{4\pi\epsilon_0}$, the value of $K$ is:
Answer: (A) $8q_2$
$q_3$ stays $0.4\ \text{m}$ from $q_1$, so the $q_1$ term does not change. Only the distance from $q_2$ changes:
- at C: $\sqrt{0.4^2 + 0.3^2} = 0.5\ \text{m}$;
- at D (on line AB, $0.4\ \text{m}$ from A): $0.4 - 0.3 = 0.1\ \text{m}$.
$$\Delta U = \frac{q_2q_3}{4\pi\epsilon_0}\left(\frac1{0.1} - \frac1{0.5}\right) = \frac{q_3(8q_2)}{4\pi\epsilon_0}$$
So $K = 8q_2$.
Solution by Sreeraj P, M.Sc Physics