Three charges are situated at the comers of a square ,as shown in Fig. 2.41.

1. How much work does it take to bring in another charge, +q,from far away and place it in the fourth comer?
2. How much work does it take to assemble the whole configuration of four charges?

Write the expression for the potential is,

$\mathbf{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\pi }{\mathbf{\in }}_{\mathbf{0}}\mathbf{r}}$

Here, qis the charge, r is the distance and Vis the volume.

Consider the figure for the charges.

The figure consist the distribution of charged as shown above.

Consider the distance of the charge at point Aform the charge +qis the diagonal of the square and the value is,

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

(a)

Now write the expression for potential at point A,

$$\begin{gathered} \mathrm{V}\left(\mathrm{A}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{-\mathrm{q}}{\mathrm{a}}+\frac{\mathrm{q}}{\mathrm{a}\sqrt{2}}-\frac{\mathrm{q}}{\mathrm{a}}\right\} \\ \mathrm{Solving}\mathrm{for}\mathrm{V}\left(\mathrm{A}\right), \\ \mathrm{V}\left(\mathrm{A}\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left\{\frac{1}{\sqrt{2}}-2\right\} \\ \mathrm{Thus},\mathrm{work}\mathrm{done}\mathrm{in}\mathrm{bringing}\mathrm{the}\mathrm{fourth}\mathrm{charge}+\mathrm{q}\mathrm{at}\mathrm{from}\mathrm{infinity}. \\ \mathrm{Let}\text{'}\mathrm{s}\mathrm{assume},\mathrm{V}(\infty )=0\mathrm{is}, \\ \mathrm{w}=\mathrm{q}\left\{\mathrm{V}\left(\mathrm{r}\right)-\mathrm{V}\left(\infty \right)\right\} \\ \mathrm{Now}\mathrm{solving}\mathrm{for}\mathrm{V}\left(\mathrm{r}\right)\mathrm{by}\mathrm{substituting}\sqrt{2}\mathrm{a}\mathrm{for}\mathrm{r}\mathrm{in}\mathrm{equation}\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}. \\ \mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}\sqrt{2}} \\ \mathrm{Substitute}\mathrm{the}\infty \mathrm{for}\mathrm{the}\mathrm{r}\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}. \\ \mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0\infty } \\ =0 \\ \mathrm{Substitute}0\mathrm{for}\mathrm{V}\left(\infty \right)\mathrm{and}\mathrm{V}\left(\mathrm{r}\right)\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{w}=\mathrm{q}\left\{\mathrm{v}\left(\mathrm{r}\right)-\mathrm{v}\left(\infty \right)\right\}. \\ \mathrm{w}=\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left\{\frac{1}{\sqrt{2}}-2\right\} \\ \mathrm{Therefore},\mathrm{the}\mathrm{required}\mathrm{work}\mathrm{is}\mathrm{w}=\frac{{\mathrm{q}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}\left\{\frac{1}{\sqrt{2}}-2\right\}. \end{gathered}$$

(b)

At point ,

Write the expression for amount of work done in bringing the charge,

$\mathrm{W}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left[-\mathrm{q}+\left(-\mathrm{q}+\frac{1}{\mathrm{a}\sqrt{2}}\right)+\left(-\mathrm{q}+\frac{1}{\mathrm{a}\sqrt{2}}-\mathrm{q}\right)\right]$

Thus, the work done in assembling the whole configuration is,

$w=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{a}\left\{-1+\left(-1+\frac{1}{\sqrt{2}}\right)+\left(-q+\frac{1}{a\sqrt{2}}-1\right)\right\}$

Solve as further,

$$\begin{gathered} \mathrm{w}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{\mathrm{a}}\left\{\sqrt{2}-4\right\} \\ \mathrm{Therefore},\mathrm{the}\mathrm{work}\mathrm{need}\mathrm{to}\mathrm{be}\mathrm{done}\mathrm{is}\mathrm{w}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}^2}{\mathrm{a}}\left\{\sqrt{2}-4\right\}. \end{gathered}$$