Question

Four equal point charges, each of magnitude +Q, are to be placed in equilibrium at the corners of a square. What should be the magnitude and sign of the point charge that should be placed at the center of the square to do this job?

Answer 1

Four equal point charges +Q at the corners of square.
Now to maintain equilibrium, net force on any charged body in the system should be zero (which also means that sign of center point charge should be negative).
Let the net force on charge at A = 0.
$F_1=\frac{Q^2}{k{r}^2}\hat{i}$
$F_2=\frac{Q^2}{k{r}^2}-\hat{i}$
The resultant force of two charges will be

$F_{12}=\sqrt{F_1^2+F_2^2}$

But $F_1=-F_2$

$\therefore F_{12}=F_1\times \sqrt{2}$

$=\sqrt{2}\frac{Q^2}{k{r}^2}\hat{i}$

$F_3=\frac{Q^2}{2k{r}^2}$
$F_4=\frac{2qQ}{k{r}^2}$ (q is point charge at center)

Now, $\Sigma F=0$
$F_{12}+F_3+F_4=0$

$(\sqrt{2}+\frac{1}{2})(\frac{Q^2}{k{r}^2}+\frac{2qQ}{k{r}^2})=0$

After solving,

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