c
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
\(=q\left[\frac{ k 4 q }{ d / 2}-\frac{ k 4 q }{3 d / 2}\right]=\frac{4 q ^{2}}{3 \pi \varepsilon_{0} d }\)