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