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 $\frac{Q}{4}(1+2\sqrt{2})$

If all charges are in equilibrium, system is also in equilibrium.
Charge at centre : charge q is in equilibrium because no net force acting on it corner charge :If we consider the charge at corner B. This charge will experience following forces
$F_A=k\frac{Q^2}{d^2},F_C=\frac{k{Q}^2}{d^2},F_D=\frac{k{Q}^2}{(a\sqrt{2}{)}^2}{F}_O=\frac{KQq}{(a\sqrt{2}{)}^2}$

Force at B away from the centre $=F_{AC}+F_D$
$=\sqrt{F_A^2+F_C^2}+F_D=\sqrt{2}\frac{k{Q}^2}{a^2}+\frac{k{Q}^2}{2a^2}=\frac{k{Q}^2}{a^2}(\sqrt{2}+\frac{1}{2})$
Force at B towards the centre $=F_0=\frac{2kQq}{a^2}$
For equilibrium of charge at B, $F_{AC}+F_D=F_O$
$\Rightarrow \frac{K{Q}^2}{a^2}(\sqrt{2}+\frac{1}{2})=\frac{2KQq}{a^2}\Rightarrow q=\frac{Q}{4}(1+2\sqrt{2})$