91Ó°ÊÓ

Number of particles forming a square n = 8

Distance between adjacent particles is d.

The problem is based on the concept of electric potential. It is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field.

Formula:

$V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{q}{r}$

Where, r is the distance between the charge particles, q is the charge particle, ${\mathrm{Î\mu }}_0$is the permittivity of free space.

The electrostatic potential at a point due to a charge q is given by,

$\begin{array}{rcl}\mathrm{V}& =& \frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_0\mathrm{r}}\\ & =& \frac{\mathrm{kq}}{\mathrm{r}}\end{array}$

Where, r is the distance of the point from the charge q, and k is the Coulomb’s constant.

Here,

$\begin{array}{rcl}k& =& \frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\\ & =& 9×{10}^{9}N{m}^2/C^2\end{array}$

Draw the figure as given below.

Consider the following charges.

The charge on point A, q_A = -4q

The charge on point B,q_B = -2q

The charge on point C,q_C = q

The charge on point D,q_D = 5q

The charge on point E,q_E = -5q

The charge on point F,q_F = -q

The charge on point G,q_G = -2q

The charge on point H, q_H = 4q

From the above diagram,

The distance, $AP=CP=FP=HP=r=\sqrt{2}d$

The distance,$BP=DP=EP=GP=d$

Since the electric potential is a scalar quantity, so the electrostatic potential at point P due to all the charges is given by;

$\begin{array}{rcl}{V}_P& =& \frac{k{q}_A}{r}+\frac{k{q}_B}{d}+\frac{k{q}_C}{r}+\frac{k{q}_D}{d}+\frac{k{q}_E}{d}+\frac{k{q}_F}{r}+\frac{k{q}_G}{d}+\frac{k{q}_H}{r}\\ & =& \frac{k\left(-4q\right)}{r}+\frac{k\left(-2q\right)}{d}+\frac{k\left(q\right)}{r}+\frac{k\left(5q\right)}{d}+\frac{k\left(-5q\right)}{d}+\frac{k\left(-q\right)}{r}+\frac{k\left(-2q\right)}{d}+\frac{k\left(4q\right)}{r}\\ V_P& =& \frac{k\left(-2q\right)}{d}+\frac{k\left(-2q\right)}{d}\\ & =& -\frac{4kq}{d}\\ & =& -\frac{4q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}d}\\ & =& -\frac{q}{\mathrm{Ï€}{\mathrm{Î\mu }}_{0}d}\end{array}$

Hence, the electric potential at point P at the center of the square is $-\frac{q}{\mathrm{Ï€}{\mathrm{Î\mu }}_{0}d}$.