Q. The charges$\mathcal{Q}$, +q and +q are placed at the vertices of an equilateral triangle of side$\ l$. If the net electrostatic potential energy of the system is zero, then $\mathcal{Q}$ is equal to
A) $- \frac{q}{2}$
B) $- q$
C) $\frac{+ q}{2}$
D) Zero
Explanation
(a) Potential energy of the system
$U = \frac{K\mathcal{Q}q}{l} + \frac{Kq^{2}}{l} + \frac{Kq\mathcal{Q}}{l} = 0$
$\Longrightarrow \ \frac{Kq}{l}\left( \mathcal{Q +}q + \mathcal{Q} \right) = 0\ $
$\Longrightarrow \ \ \ \ \ \ \ \ \mathcal{Q = -}\frac{q}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.