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, $q$ 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 $\frac{3q^2}{4\pi {\epsilon }_{0}d}$
• B. Increase by $\frac{2q^2}{3\pi {\epsilon }_{0}d}$
• C. Decrease by $\frac{q^2}{4\pi {\epsilon }_{0}d}$
• D. Decrease by $\frac{4q^2}{3\pi {\epsilon }_{0}d}$

Answer 1

The correct option is D Decrease by $\frac{4q^2}{3\pi {\epsilon }_{0}d}$

Change in potential energy, $\Delta u=q(V_f-V_i)$

Potential of $-q$ is the same at initial and final points of the path, as it lies at the same distance from $d$.


$\Delta u=q(\frac{k4q}{3d/2}-\frac{k4q}{d/2})=-\frac{4q^2}{3\pi {\epsilon }_{0}d}$

$-$ve sign shows the energy of the charge is decreasing.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.