91Ó°ÊÓ

Write the expression of electric filed in a certain region,

........ (1)

Here, k is constant.

Now using the Gauss Law in electrostatics, the expression the charge density in terms of electric field,

$\mathit{ÒÏ}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\left(∇×E\right)$ ….. (2)

In spherical co-ordinates, the value of $\mathbf{∇}\mathbf{·}\mathbf{E}$is,

$\mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{γ}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\mathbf{\left(}{\mathbf{r}}^{\mathbf{2}}{\mathbf{E}}_{\mathbf{θ}}\mathbf{\right)}\mathbf{+}\mathit{E}\frac{\mathbf{1}}{\mathbf{²õ¾\pm ²Ôθ}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{θ}}\mathbf{\left(}\mathbf{²õ¾\pm ²Ôθ}\mathbf{\right)}\frac{\mathbf{1}}{\mathbf{r²õ¾\pm ²Ôθ}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{Ï•}}\mathbf{(}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{)}$…… (3)

From the equation (1), the values of ${\mathit{E}}_{\mathbf{γ}}$,${\mathit{E}}_{\mathbf{θ}}$and ${\mathit{E}}_{\mathbf{ϕ}}$.

$$\begin{gathered} {\mathit{E}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{\mathbf{r}} \\ {\mathit{E}}_{\mathbf{θ}}\mathbf{=}\frac{\mathbf{k}\mathbf{(}\mathbf{2}\mathbf{²õ¾\pm ²Ôθ}\mathbf{}\mathbf{³¦´Ç²õθ}\mathbf{}\mathbf{²õ¾\pm ²ÔÏ•}\mathbf{)}}{\mathbf{r}} \\ {\mathit{E}}_{\mathbf{Ï•}}\mathbf{=}\frac{\mathbf{k}\mathbf{(}\mathbf{²õ¾\pm ²Ôθ}\mathbf{}\mathbf{³¦´Ç²õÏ•}\mathbf{)}}{\mathbf{r}} \end{gathered}$$

Substitutes the values of ${\mathit{E}}_{\mathbf{γ}}$,${\mathit{E}}_{\mathbf{θ}}$and ${\mathit{E}}_{\mathbf{ϕ}}$in equation (3), then

$$\begin{gathered} \mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\left\{\frac{1}{{\mathrm{γ}}^2}\frac{∂}{∂r}\left(r^2\left[\frac{3k}{r}\right]\right)+\frac{1}{rsin\mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(sin\mathrm{θ}\left[\frac{k\left(2sin\mathrm{θ}cos\mathrm{θ}sin\mathrm{ϕ}\right)}{r}\right]\right)+\frac{1}{rsin\mathrm{θ}}\frac{∂}{∂\mathrm{ϕ}}\left(\frac{k\left(sin\mathrm{θ}cos\mathrm{ϕ}\right)}{r}\right)\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left\{\frac{1}{{\mathrm{γ}}^2}\left(3k\right)+\frac{2k\left(sin\mathrm{ϕ}\right)}{r^{2}sin\mathrm{θ}}\left(2sin\mathrm{θ}co{s}^2\mathrm{θ}+si{n}^2\mathrm{θ}\left(-sin\mathrm{θ}\right)\right)+\frac{k}{r^{2}sin\mathrm{θ}}\left(sin\mathrm{θ}\left(-sin\mathrm{ϕ}\right)\right)\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{k}\left(4co{s}^2\mathrm{θ}-2si{n}^2\mathrm{θ}\right)\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{ϕ}\mathbf{+}\mathbf{k}\left(-sin\mathrm{ϕ}\right)}{{\mathbf{r}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\left(\left(4co{s}^2\mathrm{θ}-2si{n}^2\mathrm{θ}\right)-1\right)\mathit{s}\mathit{i}\mathit{n}\mathit{ϕ} \end{gathered}$$

Using the identity $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{θ}\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{θ}\mathbf{=}\mathbf{1}$in above simplification,

$$\begin{gathered} \mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\left(\left(4co{s}^2\mathrm{θ}-2si{n}^2\mathrm{θ}\right)-\left(si{n}^2\mathrm{θ}+co{s}^2\mathrm{θ}\right)\right)\mathit{s}\mathit{i}\mathit{n}\mathit{ϕ} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\left(3co{s}^2\mathrm{θ}-3si{n}^2\mathrm{θ}\right)\mathit{s}\mathit{i}\mathit{n}\mathit{ϕ} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\left(co{s}^2\mathrm{θ}-si{n}^2\mathrm{θ}\right)\mathit{s}\mathit{i}\mathit{n}\mathit{ϕ} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\left(cos2\mathrm{θ}\right)\mathit{s}\mathit{i}\mathit{n}\mathit{ϕ} \end{gathered}$$

Solve further as,

$\mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\mathbf{3}\mathit{k}\frac{(1+\cos 2\mathrm{θ}\mathrm{²õ¾\pm ²ÔÏ•})}{{\mathbf{r}}^{\mathbf{2}}}$

Substitute the $\mathbf{3}\mathit{k}\frac{(1+\cos 2\mathrm{θ}\mathrm{²õ¾\pm ²ÔÏ•})}{{\mathbf{r}}^{\mathbf{2}}}$for $\mathbf{∇}\mathbf{·}\mathbf{E}$in the equation (2) to solve for $\mathit{ÒÏ}$.

$$\begin{gathered} \mathit{ÒÏ}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\left(∇·E\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathit{k}{\mathit{Î\mu }}_{\mathbf{0}}\frac{(1+\cos 2\mathrm{θ}\mathrm{²õ¾\pm ²ÔÏ•})}{{\mathbf{r}}^{\mathbf{2}}} \end{gathered}$$

Thus, the charge density is $\mathit{ÒÏ}\mathbf{=}\mathbf{3}\mathit{k}{\mathit{Î\mu }}_{\mathbf{0}}\frac{(1+\cos 2\mathrm{θ}\mathrm{²õ¾\pm ²ÔÏ•})}{{\mathbf{r}}^{\mathbf{2}}}$.