Question

Charges $Q$ each are placed at each of two opposite corners of a square. Charges $q$ each are placed at each of the other two corners. Find the relation between $q$ and $Q$, when the net force on Q is zero.

• A. $q=-\frac{3Q}{2\sqrt{2}}$
• B. $q=+\frac{Q}{2\sqrt{2}}$
• C. $q=-\frac{Q}{\sqrt{2}}$
• D. $q=-\frac{Q}{2\sqrt{2}}$

Answer 1

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

For C to be in equilibrium, it is necessary that Q and q should have opposite signs. And $F_D$ and $F_C$ should be equal in magnitude to $F_A$
Let the length of square be a

force on Q due to q is resultant force due to both particle and are perpendicular =$\sqrt{2}\frac{kqQ}{a^2}$

force due to Q =$\frac{KQQ}{2a^2}$

now sum of the both force should be zero.

this give $q=\frac{-Q}{2\sqrt{2}}$