A two point charges $4 \mathrm{q}$ and $-\mathrm{q}$ are fixed on the
$x$-axis at $x=-\frac{d}{2}$ and $x=\frac{d}{2}$, respectively. If
a third point charge ' $q$ ' is taken from the origin
to $x=d$ along the semicircle as shown in the
figure, the energy of the charge will:
Correct Option: , 3
Potential of $-\mathrm{q}$ is same as initial and final point of the path therefore potential due to $4 \mathrm{q}$ will only change and as potential is decreasing the energy will decrease
Decrease in potential energy $=\mathrm{q}\left(\mathrm{V}_{\mathrm{i}}-\mathrm{V}_{\mathrm{f}}\right)$
Decrease in potential energy
$=\mathrm{q}\left[\frac{\mathrm{k} 4 \mathrm{q}}{\mathrm{d} / 2}-\frac{\mathrm{k} 4 \mathrm{q}}{3 \mathrm{~d} / 2}\right]=\frac{4 \mathrm{q}^{2}}{3 \pi \varepsilon_{0} \mathrm{~d}}$
Therefore correct answer is 3 .