91Ó°ÊÓ

1. Edge length of the square, a =0.04 m
2. Charge of the particle, q =+6e

Using the concept of the electric potential of all the points charges present on the square, we can get the difference between the electric potential at the center of the square. Thus, this value multiplied by the charge on which external work is done determines the work done.

Formulae:

Expression for the electric potential at a point due to a system of point charges is given by the relation as follows: $V=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\underset{}{∑}\frac{q}{r}$ (i)

Here, q is the charge and r is the distance between the charges.

Expression for the work done to move the given charge from the infinity to the given point is given by,$W=q\mathrm{Δ}V$(ii)

The work done to bring the charge to the center from the infinity is given as:

$$\begin{gathered} W=q\left(V_f-V_i\right) \\ =q\left(V_f-0\right) \\ =q{V}_f.......................\left(a\right) \end{gathered}$$

Here, $V_f-V_i$ is the change in electric potential, but the particle is initially at rest.

Now, the electric potential at the center of square due to diagonally opposite charges is given using equation (i) as:

$$\begin{gathered} V_{f1}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}m}\left[\left(-e\right)+\left(e\right)+\left(-3e\right)+\left(3e\right)\right] \\ \text{(where, m is the distance of the corner to the center)} \\ =0 \end{gathered}$$

Length of the diagonal of the square is,

$$\begin{gathered} d=\sqrt{a^2+a^2} \\ =\sqrt{2}a \end{gathered}$$

Distance of charge at any corner to the center of square is,

Distance of each of the remaining charges to the center of square,

$$\begin{gathered} n=\frac{a}{2} \\ =\frac{0.04\text{m}}{2} \\ =0.02\text{m} \end{gathered}$$

The value of the electric potential at the center of square due to other charges is given using equation (i) as:

$$\begin{gathered} V_{f2}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}n}\left[\left(3e\right)+\left(2e\right)+\left(-2e\right)\right] \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_{0}n}\left(3e\right) \\ =\frac{\left(9×{10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(3\right)\left(1.6×{10}^{-19}\text{C}\right)}{0.02\text{m}} \\ =21.6×{10}^{-8}\text{V} \end{gathered}$$

Thus, the final potential at the center of square as follows:

$$\begin{gathered} V_f=V_{f1}+V_{f2} \\ =21.6×{10}^{-8}\text{V}+0 \\ =21.6×{10}^{-8}\text{V} \end{gathered}$$

Now, the work done is given using the above value in equation (a) as:

$$\begin{gathered} W=6\left(1.6×{10}^{-19}\text{C}\right)\left(21.6×{10}^{-8}\text{V}\right) \\ =2.07×{10}^{-25}\text{J} \end{gathered}$$

Therefore, the required work done is$2.07×{10}^{-25}\text{J}$ .