Question

Four charges equal to -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. $-\frac{Q}{4}(1+2\sqrt{2})$
• B. $\frac{Q}{4}(1+2\sqrt{2})$
• C. $\frac{Q}{2}(1+2\sqrt{2})$
• D. $\frac{Q}{8}(1+2\sqrt{2})$

Answer 1

The correct option is B $\frac{Q}{4}(1+2\sqrt{2})$

Let the square is of side a.

Force $|F_1|=\frac{K{Q}^2}{a^2}$ where $K=\frac{1}{4\pi {ϵ}_o}$

$|F_2|=\frac{K{Q}^2}{(\sqrt{2}a{)}^2}$

$|F_3|=\frac{KQq}{(\frac{a}{\sqrt{2}}{)}^2}$

As the system is in equilibrium, thus net force is equal to zero.

$\sqrt{2}\frac{K{Q}^2}{a^2}+\frac{K{Q}^2}{(\sqrt{2}a{)}^2}=\frac{KQq}{(\frac{a}{\sqrt{2}}{)}^2}$

$⟹q=\frac{Q}{4}(1+2\sqrt{2})$