Q. The charges $Q,\ + q\ and + q$ are placed at the vertices of a right-angled isosceles triangle as shown. The net electrostatic energy of the configuration is zero if $Q$ is equal to
A) $\frac{- q}{1 + \sqrt{2}}$
B) $\frac{- 2q}{2 + \sqrt{2}}$
C) $- 2q$
D) $+ q$
Explanation
(b) Net electrostatic energy $U = \frac{kQq}{a} + \frac{kq^{2}}{a} + \frac{kQq}{a\sqrt{2}} = 0$
$\Rightarrow \frac{kq}{a}\left( Q + q + \frac{Q}{\sqrt{2}} \right) = 0 \Rightarrow Q = - \frac{2q}{2 + \sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.