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. $\frac{Q}{2}(1+2\sqrt{2})$

Answer 1

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

$\begin{array}{c}\frac{\sqrt{2}K{Q}^2}{R^2}+\frac{K{Q}^2}{2R^2}=\frac{2KQq}{R^2}\\ \Rightarrow \sqrt{2}Q+\frac{Q}{2}=2\left|q\right|\\ \Rightarrow \left|q\right|=\frac{Q}{2}\left(\sqrt{2}+\frac{1}{2}\right)\\ =\frac{Q}{4}\left(1+2\sqrt{2}\right)\\ \Rightarrow q=\frac{Q}{4}\left(1+2\sqrt{2}\right)\end{array}$

Hence,

option $\left(B\right)$ is correct answer.