Question

A two point charges $4q$ and $-q$ are fixed on the $x-axis$ at $x=-\frac{d}{2}$ and $x=\frac{d}{2},$ respectively. If a third point charge ${}^{\prime }{q}^{\prime }$ 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 $\frac{2q^2}{3\pi {\epsilon }_{0}d}$
• B. increase by $\frac{3q^2}{4\pi {\epsilon }_{0}d}$
• C. decrease by $\frac{4q^2}{3\pi {\epsilon }_{0}d}$
• D. decrease by $\frac{q^2}{4\pi {\epsilon }_{0}d}$

Answer 1

Answer: (C)

Potential of $-q$ is same as initial and final point of the path therefore potential due to $4q$ 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
$=q\left[\frac{k4q}{d/2}-\frac{k4q}{3d/2}\right]$
$=\frac{4q^2}{3\pi {\epsilon }_{0}d}$