Question

It is required to hold equal charges, $q$ in equilibrium at the corners of a square. What charge when placed at the centre of the square will do this?

Answer 1

Let $ABCD$ be a square of side $a$, and $Q$ be the charge placed at the centre.
$r=\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}$
${\overrightarrow{F}}_{BA}=\frac{k{q}^2}{a^2}\hat{i}$, ${\overrightarrow{F}}_{BD}=\frac{-k{q}^2}{a^2}\hat{j}$
${\overrightarrow{F}}_{BA}=\frac{k{q}^2}{(a{\sqrt{2}}^{)}2}(\cos {45}^o\hat{i}-\sin {45}^o\hat{j})$
${\overrightarrow{F}}_{BA}=\frac{kQq}{(a/{\sqrt{2}}^{)}2}(\cos {45}^o\hat{i}-\sin {45}^o\hat{j})$
Here $\hat{i}$ and $\hat{j}$ have usual meaning.
Net force on the charge at $B$ is
${\overrightarrow{F}}_R=(\frac{k{q}^2}{a^2}+\frac{k{q}^2}{(a\sqrt{2}{)}^2}\cos {45}^o+\frac{kQq}{(a/\sqrt{2}{)}^2}\cos {45}^o)\hat{i}-(\frac{k{q}^2}{a^2}+\frac{k{q}^2}{(a\sqrt{2}{)}^2}\sin {45}^o+\frac{kQq}{(a/\sqrt{2}{)}^2}\sin {45}^o)\hat{j}=F_x\hat{i}+F_y\hat{j}$
For change, $q$ to be in equilibrium at $B$, the net force on it must be zero.
$\therefore F_x=0$ & $F_y=0$
$⟹k[\frac{q^2}{a^2}+\frac{q^2}{(a/\sqrt{2}{)}^2}.\frac{1}{\sqrt{2}}+\frac{Qq}{(a/\sqrt{2}{)}^2}.\frac{1}{\sqrt{2}}]=0$
$\therefore Q=-\frac{q}{4}(1+2\sqrt{2})$
Similarly, $F_y=0,$ if $Q=-\frac{q}{4}(1+2\sqrt{2})$.