Question

A charge Q is fixed at each of two opposite corners of a square. A charge q is placed at each of the other two corners. If the resultant electric force on Q is zero, then Q and q are related as

• A. $Q=\sqrt{2}q$
• B. $Q=-2\sqrt{2}q$
• C. $Q=Q=-\sqrt{2}q$
• D. $Q=2\sqrt{2}q$

Answer 1

The correct option is B $Q=-2\sqrt{2}q$


Here $\frac{Q^2}{(\sqrt{2}a{)}^2}$
$=-\frac{2Qqcos{45}^{\circ }}{a^2}$
Or $\frac{Q^2}{2a^2}=-\frac{2Qq}{\sqrt{2}{a}^2}$
$Q=-2\sqrt{2}q$