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

$\text{Step 1 : Calculate all the required distances between the charges :}$

Let the side length of the square be $a$.

Refer Figure, In $△ACD$,

$\Rightarrow A{D}^2+C{D}^2=A{C}^2$

$\Rightarrow a^2+a^2=A{C}^2$ $\Rightarrow AC=\sqrt{2}a$

However, $OC=\frac{AC}{2}$ $=\frac{a}{\sqrt{2}}$

$\text{Step 2 : Equilibrium of central charge}$

By symmetry, Force on central charge will be equal and opposite due to the diagonally opposite charges, which will cancel each other.

Hence, Net force on central charge will always be zero, irrespective of value of charge $q$.

Therefore, to find value of $q$ we have to check equilibrium of any one charge at the corner.

$\text{Step 3 : Force due to all the charges at point C}$

Force on charge at $C$ due to $B$, $F_1=\frac{K{Q}^2}{a^2}$

Force on charge at $C$ due to $D$, $F_2=\frac{K{Q}^2}{a^2}$

Force on charge at $C$ due to $A$, $F_4=\frac{K{Q}^2}{A{C}^2}=\frac{K{Q}^2}{2a^2}$

Force on charge at $C$ due to $q$ at centre, $F_3=\frac{KqQ}{O{C}^2}=\frac{2KqQ}{a^2}$

$\text{Step 4 : Apply the equilibrium condition at C :}$

For the system to be in equilibrium, net force acting on charge at C must be zero.

So, $\overrightarrow{F_3}+\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_4}=0$

Resultant of $F_1$ and $F_2$ (Along OC, by symmetry) = $\sqrt{F_1^2+F_2^2}=\sqrt{2}{F}_1$ Since $(|F_1|=|F_2|)$

Also, $F_3$ & $F_4$ are along OC, Therefore, magnitudes of sum of these forces should be zero.

$\Rightarrow$ $|F_3|+\sqrt{2}|F_1|+|F_4|=0$

$\Rightarrow \frac{2KqQ}{a^2}$ $+\sqrt{2}\frac{K{Q}^2}{a^2}$ $+\frac{K{Q}^2}{2a^2}=0$

$\Rightarrow 2q=-(\sqrt{2}Q+\frac{Q}{2})$

$\Rightarrow q=-\frac{Q}{4}(1+2\sqrt{2})$