Two point charges $4q$ and $-q$ are fixed on the $x$-axis at $x = -\dfrac d2$ and $x = \dfrac d2$, respectively. If a third point charge $q$ is taken from the origin to $x = d$ along a semicircle (centred at $x = \dfrac d2$), the energy of the charge will:
Answer: (D) decrease by $\dfrac{4q^{2}}{3\pi\varepsilon_0d}$
The change in potential energy depends only on the end points. With $k = \dfrac1{4\pi\varepsilon_0}$:
At the origin: $V_i = k\left(\dfrac{4q}{d/2} - \dfrac{q}{d/2}\right) = \dfrac{6kq}{d}$.
At $x = d$: $V_f = k\left(\dfrac{4q}{3d/2} - \dfrac{q}{d/2}\right) = \dfrac{2kq}{3d}$.
$$\Delta U = q(V_f - V_i) = -\frac{16kq^{2}}{3d} = -\frac{4q^{2}}{3\pi\varepsilon_0d}$$
The energy decreases by $\dfrac{4q^{2}}{3\pi\varepsilon_0d}$.
Solution by Sreeraj P, M.Sc Physics