91Ó°ÊÓ

Write the expression of electric filed in a certain region,

$\mathit{E}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{k}}{\mathbf{r}}\mathbf{[}\mathbf{3}\stackrel{\mathbf{Áåœ}}{\mathbf{r}}\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{θ}\mathbf{s}\mathbf{i}\mathbf{n}\stackrel{\mathbf{Áåœ}}{\mathbf{θ}}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{f}\stackrel{\mathbf{Áåœ}}{\mathbf{f}}\mathbf{]}$ ........ (1)

Here,$k$ is constant.

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

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

In spherical co-ordinates, the value of $∇-E$ is,

$\mathbf{∇}\mathbf{.}\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{γ}}^{\mathbf{2}}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{r}}\left(r^2{E}_r\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{°ù²õ¾\pm ²Ôθ}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{θ}}\left({\mathrm{²õ¾\pm ²Ôθ·¡}}_{\mathrm{θ}}\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{°ù²õ¾\pm ²Ôθ}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{Ï•}}\left(E_f\right)$ …… (3)

From the equation (1), the values of$E_{\mathrm{γ}}$,$E_{\mathrm{θ}}$and $E_{\mathrm{ϕ}}$.

$$\begin{gathered} E_r=\frac{3k}{r} \\ {E}_{\mathrm{θ}}=\frac{k\left(2\sin \mathrm{θ}\cos \mathrm{θ}\sin \mathrm{θ}\right)}{r} \\ {E}_{\mathrm{ϕ}}=\frac{k\left(\sin \mathrm{θ}\cos \mathrm{ϕ}\right)}{r} \end{gathered}$$

Substitutes the values of $E_{\mathrm{γ}}$,$E_{\mathrm{θ}}$and $E_{\mathrm{ϕ}}$in equation (3), then

$$\begin{gathered} ∇.E=\left\{\frac{1}{{\mathrm{γ}}^2}\frac{∂}{∂r}\left(r^2\left[\frac{3k}{r}\right]\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{θ}}\left(\sin \mathrm{θ}\left[\frac{k\left(2\sin \mathrm{θ}\cos \mathrm{θ}\sin \mathrm{ϕ}\right)}{r}\right]\right)+\frac{1}{r\sin \mathrm{θ}}\frac{∂}{∂\mathrm{ϕ}}\left(\frac{k\left(\sin \mathrm{θ}\cos \mathrm{ϕ}\right)}{r}\right)\right\} \\ =\left\{\frac{1}{r^2}\left(3k\right)+\frac{2k\left(\sin \mathrm{ϕ}\right)}{r^2\sin \mathrm{θ}}\left(2\sin \mathrm{θ}{\cos }^2\mathrm{θ}+{\sin }^2\mathrm{θ}\left(-\sin \mathrm{θ}\right)\right)+\frac{k}{r^2\sin \mathrm{θ}}\left(\sin \mathrm{θ}\left(-\sin \mathrm{ϕ}\right)\right)\right\} \\ =\frac{3k}{r^2}+\frac{k\left(4{\cos }^2\mathrm{θ}-2{\sin }^2\mathrm{θ}\right)\sin \mathrm{ϕ}+k\left(-\sin \mathrm{ϕ}\right)}{r^2} \\ =\frac{3k}{r^2}+\frac{k}{r^2}\left(\left(4{\cos }^2\mathrm{θ}-2{\sin }^2\mathrm{θ}\right)-1\right)\sin \mathrm{ϕ} \end{gathered}$$

Using the identity${\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}=1$in above simplification,

$$\begin{gathered} ∇.E=\frac{3k}{r^2}+\frac{k}{r^2}\left(\left(4{\cos }^2\mathrm{θ}-2{\sin }^2\mathrm{θ}\right)-\left({\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}\right)\right)\sin \mathrm{ϕ} \\ =\frac{3k}{r^2}+\frac{k}{r^2}\left(3{\cos }^2\mathrm{θ}-3{\sin }^2\mathrm{θ}\right)\sin \mathrm{ϕ} \\ =\frac{3k}{r^2}+\frac{3k}{r^2}\left({\cos }^2\mathrm{θ}-{\sin }^2\mathrm{θ}\right)\sin \mathrm{ϕ} \\ =\frac{3k}{r^2}+\frac{3k}{r^2}\left(\cos 2\mathrm{θ}\right)\sin \mathrm{ϕ} \end{gathered}$$

Solve further as,

$∇.E=3k\frac{\left(1+\cos 2\mathrm{θ}\sin \mathrm{ϕ}\right)}{r^2}$

Substitute the$3k\frac{\left(1+\cos 2\mathrm{θ}\sin \mathrm{ϕ}\right)}{r^2}$for$∇.E$in the equation (2) to solve for .

role="math" localid="1657360938595" $$\begin{gathered} \mathrm{ÒÏ}={\mathrm{Î\mu }}_0\left(∇.E\right) \\ =3k{\mathrm{Î\mu }}_0\frac{\left(1+\cos 2\mathrm{θ}\sin \mathrm{Ï•}\right)}{r^2} \end{gathered}$$

Thus, the charge density is$\mathrm{ÒÏ}=3k{\mathrm{Î\mu }}_0\frac{\left(1+\cos 2\mathrm{θ}\sin \mathrm{Ï•}\right)}{r^2}$ .