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A two point charges $4 q$ and $-q$ 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
increase by $\frac{2 q^{2}}{3 \pi \varepsilon_{0} d }$
increase by $\frac{3 q^{2}}{4 \pi \varepsilon_{0} d }$
decrease by $\frac{4 q^{2}}{3 \pi \varepsilon_{0} d }$
decrease by $\frac{q^{2}}{4 \pi \varepsilon_{0} d }$
Solution
Potential of $- q$ is same as initial and final point of the path therefore potential due to $4 q$ will only change and as potential is decreasing the energy will decrease Decrease in potential energy $=q\left( V _{ i }- V _{ f }\right)$
Decrease in potential energy
$=q\left[\frac{ k 4 q }{ d / 2}-\frac{ k 4 q }{3 d / 2}\right]=\frac{4 q ^{2}}{3 \pi \varepsilon_{0} d }$
