Question

Four charges each equal to $Q$ are placed at the four corners of a square and a charge $q$ is placed at the centre of the square. If the system is in equilibrium then the value of $q$ is

• A. $\frac{Q}{2}(1+2\sqrt{2})$
• B. $\frac{-Q}{4}(1+2\sqrt{2})$
• C. $\frac{Q}{4}(1+2\sqrt{2}$
• D. $\frac{-Q}{2}(1+2\sqrt{2})$

Answer 1

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

For system of charges to remain in equilibrium, net force on each charge should be balanced.



$F_1$ and $F_2$ are the repulsive forces exerted by charges at the adjacent corner on charge $Q$.
$F_1=F_2=\frac{k\left(Q\right)\left(Q\right)}{a^2}$
$F_3$ is the repulsive force exerted by charge $Q$ at the diagonally opposite corner.
Length of diagonal $=\sqrt{a^2+a^2}=a\sqrt{2}$
$\Rightarrow F_3=\frac{k\left(Q\right)\left(Q\right)}{(\sqrt{2}a{)}^2}$

The component of $F_1$ and $F_2$ along the $F_3$ is
$F_1^{\prime }=F_3^{\prime }=F_1\cos {45}^{\circ }=\frac{F_1}{\sqrt{2}}$

Thus, for equilibrium along diagonal,
$F_1^{\prime }+F_2^{\prime }+F_3+F_4=0$
$\Rightarrow \frac{2F_1}{\sqrt{2}}+\frac{k{Q}^2}{2a^2}+\frac{kQq}{{\left(\frac{\sqrt{2}}{2}a\right)}^2}=0$ $(\because F_1=F_2)$
$\Rightarrow \sqrt{2}{F}_1+\frac{k{Q}^2}{2a^2}+\frac{2kQq}{a^2}=0$

$\Rightarrow \sqrt{2}\frac{k{Q}^2}{a^2}+\frac{k{Q}^2}{2a^2}+\frac{2kQq}{a^2}=0$
$\Rightarrow -\frac{(2\sqrt{2}+1)k{Q}^2}{2a^2}=\frac{2kQq}{a^2}$
$\Rightarrow q=-\frac{(2\sqrt{2}+1)k{a}^2{Q}^2}{2\left(2k{a}^{2}Q\right)}$
$\Rightarrow q=-\frac{-(2\sqrt{2}+1)Q}{4}$
$\therefore$ The charge $\left(q\right)$ placed at centre of square must be negative

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Here force}{F}_1\text{and}{F}_2\text{are equal and angle b/w}\\ \text{them is}{90}^{\circ },\text{hence resultant force will be along angle}\\ \text{bisector i.e. at angle}{45}^{\circ }\text{along the line of force}{F}_3.\end{array}$