Question

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. lf 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

Answer: (B)

The system is in equilibrium means the force experienced by each charge is zero. It is clear that charge placed at centre would be in equilibrium for any value of $q$, so we are considering the equilibrium of charge placed at at any corner.
$F_{CD}+F_{CA}cos45^{\circ }+F_{CO}cos45^{\circ }=0$
$\Rightarrow \frac{1}{4\pi {\epsilon }_0}.\frac{(-Q)(-Q)}{a^2}+\frac{1}{4\pi {\epsilon }_0}\frac{(-Q)(-Q)}{\sqrt{2}{a}^2}\times \frac{1}{\sqrt{2}}$
$+\frac{1}{4\pi {\epsilon }_0}\frac{(-Q)q}{(\sqrt{2}a/2{)}^2}\times \frac{1}{2}=0$
$\Rightarrow \frac{1}{4\pi {\epsilon }_0}.\frac{(Q^2}{a^2}+\frac{1}{4\pi {\epsilon }_0}.\frac{Q^2}{2a^2}.\frac{1}{\sqrt{2}}$

$-\frac{1}{4\pi {\epsilon }_0}.\frac{2Qq}{a^2}\times \frac{1}{\sqrt{2}}=0$
$\Rightarrow Q+\frac{Q}{2\sqrt{2}}-\sqrt{2}q=0$
$\Rightarrow 2\sqrt{2}Q+Q-4q=0$
4q = $(2\sqrt{2}+1)Q$
q = $\frac{Q}{4}(2+2\sqrt{2})$