Q. If charge $q$ is placed at the centre of the line joining two equal charges $Q$, the system of these charges will be the same distance would be
A) $- 4Q$
B) $- Q/4$
C) $- Q/2$
D) $+ Q/2$
Explanation
(b) For equilibrium, net force on $Q = 0$
$\therefore\ \frac{kQQ}{\left( 2x^{\ } \right)^{2}} + \frac{kqQ}{x^{2}} = 0$
$\therefore q = - Q/4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.