Question

Figure 24-24 shows eight particles that form a square, with distance dbetween adjacent particles. What is the net electric potential at point Pat the center of the square if we take the electric potential to be zero at infinity?

Short answer

Thenet electric potential at point P at the center of the square is $-\frac{q}{\pi {\epsilon }_0{d}^{-}}$.

Step-by-step solution

Listing the given quantities:

Number of particles forming a square n = 8

Distance between adjacent particles is d.

Understanding the concept of electric potential at a point:

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:

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

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

Calculation of the net electric potential at point P:

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

$\begin{array}{rcl}\mathrm{V}& =& \frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}\\ & =& \frac{\mathrm{kq}}{\mathrm{r}}\end{array}$

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

Here,

$\begin{array}{rcl}k& =& \frac{1}{4\pi {\epsilon }_0}\\ & =& 9\times {10}^{9}N{m}^2/C^2\end{array}$

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;

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

Hence, the electric potential at point P at the center of the square is $-\frac{q}{\pi {\epsilon }_{0}d}$.