91影视

Write the expression of electric filed in a certain region,

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

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" $蚁={蔚}_0(\mathrm{鈭�}脳E)$

In spherical co-ordinates, the value of $\mathrm{鈭�}路E$is,

$\mathrm{鈭�}鈰�\mathbf{E}=\frac{1}{{纬}^2}\frac{\mathrm{鈭�}}{\mathrm{鈭�}r}\left({\mathbf{r}}^2{\mathbf{E}}_{\mathrm{r}}\right)+\frac{1}{rsin胃}\frac{\mathrm{鈭�}}{\mathrm{鈭�}胃}\left(sin鈦�胃{\mathbf{E}}_{胃}\right)+\frac{1}{rsin胃}\frac{\mathrm{鈭�}}{\mathrm{鈭�}蠒}\left({\mathbf{E}}_{\mathrm{f}}\right)$

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

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

$E_{胃}=\frac{k(2\sin 鈦�胃\cos 鈦�胃\sin 鈦�蠒)}{r}$

$E_{蠒}=\frac{k(\sin 鈦�胃\cos 鈦�蠒)}{r}$

Substitutes the values of $E_{纬}$, $E_{胃}$and$E_{蠒}$ in equation $\left(3\right)$, then
$\mathrm{鈭�}鈰�E=\left\{\frac{1}{{纬}^2}\frac{\mathrm{鈭�}}{\mathrm{鈭�}r}\left(r^2\left[\frac{3k}{r}\right]\right)+\frac{1}{r\sin 鈦�胃}\frac{\mathrm{鈭�}}{\mathrm{鈭�}胃}\left(\sin 鈦�胃\left[\frac{k(2\sin 鈦�胃\cos 鈦�胃\sin 鈦�蠒)}{r}\right]\right)+\frac{1}{r\sin 鈦�胃}\frac{\mathrm{鈭�}}{\mathrm{鈭�}蠒}\left(\frac{k(\sin 鈦�胃\cos 鈦�蠒)}{r}\right)\right\}$

$=\left\{\frac{1}{r^2}\left(3k\right)+\frac{2k(\sin 鈦�蠒)}{r^2\sin 鈦�胃}\left(2\sin 鈦�胃{\cos }^2鈦�胃+{\sin }^2鈦�胃(鈭�\sin 鈦�胃)\right)+\frac{k}{r^2\sin 鈦�胃}(\sin 鈦�胃(鈭�\sin 鈦�蠒\left)\right)\right\}$

$=\frac{3k}{r^2}+\frac{k\left(4{\cos }^2鈦�胃鈭�2{\sin }^2鈦�胃\right)\sin 鈦�蠒+k(鈭�\sin 鈦�蠒)}{r^2}$

$=\frac{3k}{r^2}+\frac{k}{r^2}\left(\left(4{\cos }^2鈦�胃鈭�2{\sin }^2鈦�胃\right)鈭�1\right)\sin 鈦�蠒$

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

$\mathrm{鈭�}鈰�E=\frac{3k}{r^2}+\frac{k}{r^2}\left(\left(4{\cos }^2鈦�胃鈭�2{\sin }^2鈦�胃\right)鈭�\left({\sin }^2鈦�胃+{\cos }^2鈦�胃\right)\right)\sin 鈦�蠒$

$=\frac{3k}{r^2}+\frac{k}{r^2}\left(3{\cos }^2鈦�胃鈭�3{\sin }^2鈦�胃\right)\sin 鈦�蠒$

$=\frac{3k}{r^2}+\frac{3k}{r^2}\left({\cos }^2鈦�胃鈭�{\sin }^2鈦�胃\right)\sin 鈦�蠒$

$\begin{array}{r}=\frac{3k}{r^2}+\frac{3k}{r^2}(\cos 鈦�2胃)\sin 鈦�蠒\end{array}$

Solve further as,

$\mathrm{鈭�}鈰�E=3k\frac{(1+\cos 鈦�2胃\sin 鈦�蠒)}{r^2}$

Substitute the$3k{蔚}_0\frac{(1+\cos 鈦�2胃\sin 鈦�蠒)}{r^2}$ for$鈭�路E$in the equation $\left(2\right)$to solve for p.

$蚁={蔚}_0(\mathrm{鈭�}鈰�E)$

$=3k{蔚}_0\frac{(1+\cos 鈦�2胃\sin 鈦�蠒)}{r^2}$

Thus, the charge density is$p=3k{蔚}_0\frac{(1+\cos 鈦�2胃\sin 鈦�蠒)}{r^2}$