Q. Four charges equal to$- \mathcal{Q}$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium the value of $q$ is
A) $- \frac{\mathcal{Q}}{4}(1 + 2\sqrt{2})$
B) $\frac{\mathcal{Q}}{4}\left( 1 + 2\sqrt{2} \right)$
C) $- \frac{\mathcal{Q}}{2}\left( 1 + 2\sqrt{2} \right)$
D) $\frac{\mathcal{Q}}{2}\left( 1 + 2\sqrt{2} \right)$
Explanation
(b) $F^{''} = \sqrt{2F} + \frac{F}{2} = F\left( \sqrt{2} + \frac{1}{2} \right)$
[image:1_image293.emf]
$\frac{q\mathcal{Q}}{\left( \frac{\sqrt{2a}}{2} \right)^{2}} = \frac{\mathcal{Q}^{2}}{a^{2}}\left( \sqrt{2} + \frac{1}{2} \right)$
$q = \frac{\mathcal{Q}}{2}\left( \frac{2\sqrt{2} + 1}{2} \right)$
q =$\frac{\mathcal{Q}}{4}(2\sqrt{2} + 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.