Question: If the electric field in some region is given (in spherical coordinates)

by the expression

$E\left(r\right)=\frac{k}{r}[3\hat{r}+2sin\theta cos\theta sin\varphi \hat{\theta }+sin\theta cos\varphi \hat{\varphi }]$

for some constant , what is the charge density?

Write the expression of electric filed in a certain region,

$E\left(r\right)=\frac{k}{r}[3\hat{r}+2sin\theta cos\theta sinf\hat{\theta }+sin\theta cos\varphi f\hat{f}]$

Here,$k$is constant.

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

role="math" localid="1657473342414" $\rho ={\epsilon }_0(\mathrm{\nabla }\times E)$

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

$\mathrm{\nabla }\cdot \mathbf{E}=\frac{1}{{\gamma }^2}\frac{\mathrm{\partial }}{\mathrm{\partial }r}\left({\mathbf{r}}^2{\mathbf{E}}_{\mathrm{r}}\right)+\frac{1}{rsin\theta }\frac{\mathrm{\partial }}{\mathrm{\partial }\theta }\left(sin\theta {\mathbf{E}}_{\theta }\right)+\frac{1}{rsin\theta }\frac{\mathrm{\partial }}{\mathrm{\partial }\varphi }\left({\mathbf{E}}_{\mathrm{f}}\right)$

From the equation $\left(1\right)$, the values of $E_{\gamma }$, $E_{\theta }$and role="math" localid="1657473869425" $E_{\varphi }$.

$E_r=\frac{3k}{r}$

$E_{\theta }=\frac{k(2\sin \theta \cos \theta \sin \varphi )}{r}$

$E_{\varphi }=\frac{k(\sin \theta \cos \varphi )}{r}$

Substitutes the values of $E_{\gamma }$, $E_{\theta }$and$E_{\varphi }$ in equation $\left(3\right)$, then
$\mathrm{\nabla }\cdot E=\left\{\frac{1}{{\gamma }^2}\frac{\mathrm{\partial }}{\mathrm{\partial }r}\left(r^2\left[\frac{3k}{r}\right]\right)+\frac{1}{r\sin \theta }\frac{\mathrm{\partial }}{\mathrm{\partial }\theta }\left(\sin \theta \left[\frac{k(2\sin \theta \cos \theta \sin \varphi )}{r}\right]\right)+\frac{1}{r\sin \theta }\frac{\mathrm{\partial }}{\mathrm{\partial }\varphi }\left(\frac{k(\sin \theta \cos \varphi )}{r}\right)\right\}$

$=\left\{\frac{1}{r^2}\left(3k\right)+\frac{2k(\sin \varphi )}{r^2\sin \theta }\left(2\sin \theta {\cos }^2\theta +{\sin }^2\theta (-\sin \theta )\right)+\frac{k}{r^2\sin \theta }(\sin \theta (-\sin \varphi \left)\right)\right\}$

$=\frac{3k}{r^2}+\frac{k\left(4{\cos }^2\theta -2{\sin }^2\theta \right)\sin \varphi +k(-\sin \varphi )}{r^2}$

$=\frac{3k}{r^2}+\frac{k}{r^2}\left(\left(4{\cos }^2\theta -2{\sin }^2\theta \right)-1\right)\sin \varphi$

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

$\mathrm{\nabla }\cdot E=\frac{3k}{r^2}+\frac{k}{r^2}\left(\left(4{\cos }^2\theta -2{\sin }^2\theta \right)-\left({\sin }^2\theta +{\cos }^2\theta \right)\right)\sin \varphi$

$=\frac{3k}{r^2}+\frac{k}{r^2}\left(3{\cos }^2\theta -3{\sin }^2\theta \right)\sin \varphi$

$=\frac{3k}{r^2}+\frac{3k}{r^2}\left({\cos }^2\theta -{\sin }^2\theta \right)\sin \varphi$

$\begin{array}{r}=\frac{3k}{r^2}+\frac{3k}{r^2}(\cos 2\theta )\sin \varphi \end{array}$

Solve further as,

$\mathrm{\nabla }\cdot E=3k\frac{(1+\cos 2\theta \sin \varphi )}{r^2}$

Substitute the$3k{\epsilon }_0\frac{(1+\cos 2\theta \sin \varphi )}{r^2}$ for$\nabla ·E$in the equation $\left(2\right)$to solve for p.

$\rho ={\epsilon }_0(\mathrm{\nabla }\cdot E)$

$=3k{\epsilon }_0\frac{(1+\cos 2\theta \sin \varphi )}{r^2}$

Thus, the charge density is$p=3k{\epsilon }_0\frac{(1+\cos 2\theta \sin \varphi )}{r^2}$