Q. A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then $Q/q$ equals
A) $- 2\sqrt{2}$
B) $- 1$
C) 1
D) $- \frac{1}{\sqrt{2}}$
Explanation
(a) It is given that $\overrightarrow{F_{A}} + \ \overrightarrow{F_{B}} + \ \overrightarrow{F_{D}} = 0$ where$\ \overrightarrow{F_{A}},\overrightarrow{F_{B}}$ and $\overrightarrow{F_{D}}$ are the forces applied by charges placed at $A,B$ and $D$ on the charge placed at $C$
[image:1_image320.emf]
$\Rightarrow \ \overrightarrow{F_{B}} + \overrightarrow{F_{D}} = - \overrightarrow{F_{A}}$
$\left| \overrightarrow{F_{B}} + \overrightarrow{F_{D}} \right| = \ \sqrt{2}\frac{kqQ}{a^{2}}\&$
$\left| \overrightarrow{F_{A}} \right| = \frac{kQ^{2}}{2a^{2}}$
$\therefore\sqrt{2}\ \frac{kqQ}{a^{2}} = \frac{kQ^{2}}{2a^{2}} \Rightarrow \frac{Q}{q} = - 2\sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.