Question

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 :

• A. increase by $\frac{2 q^{2}}{3 \pi \varepsilon_{0} d}$
• B. increase by $\frac{3 q^{2}}{4 \pi \varepsilon_{0} d}$
• C. decrease by $\frac{4 q ^{2}}{3 \pi \varepsilon_{0} d }$
• D. decrease by $\frac{ q ^{2}}{4 \pi \varepsilon_{0} d }$

Answer 1

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