JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Electrostatics

Because electrostatic potential energy depends only on the end points and not on the semicircular path, we only need the energy of the moving charge $q$ with the two fixed charges at the start and at the finish. At the origin the charge is a distance $d/2$ from both $4q$ and $-q$, so $U_i=\frac{k(4q)(q)}{d/2}+\frac{k(-q)(q)}{d/2}=\frac{8k{q}^2}{d}-\frac{2k{q}^2}{d}=\frac{6k{q}^2}{d}$. At the final point $x=d$ it is a distance $3d/2$ from $4q$ and $d/2$ from $-q$, giving $U_f=\frac{k(4q)(q)}{3d/2}+\frac{k(-q)(q)}{d/2}=\frac{8k{q}^2}{3d}-\frac{2k{q}^2}{d}=\frac{2k{q}^2}{3d}$. The change is $\Delta U=U_f-U_i=\frac{2k{q}^2}{3d}-\frac{6k{q}^2}{d}=-\frac{16k{q}^2}{3d}$, and with $k=\frac{1}{4\pi {ϵ}_0}$ this is $-\frac{4q^2}{3\pi {ϵ}_{0}d}$. The negative sign means the energy of the charge decreases by $\frac{4q^2}{3\pi {ϵ}_{0}d}$.