Q46P - Introduction to Electrodynamics

Question

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

by the expression

$\mathit{E}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{k}}{\mathbf{r}}\mathbf{[}\mathbf{3}\stackrel{\mathbf{⏜}}{\mathbf{r}}\mathbf{+}\mathbf{2}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\varphi }\stackrel{\mathbf{⏜}}{\mathbf{\theta }}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\varphi }\stackrel{\mathbf{⏜}}{\mathbf{\varphi }}\mathbf{]}$

for some constant $\mathit{k}$, what is the charge density?

Step-by-Step Solution

Answer

The charge density is $p=3k{\epsilon }_0\frac{\left(1+\cos 2\theta \sin \varphi \right)}{r^2}$.

Step 1: Define functions

Write the expression of electric filed in a certain region,

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

Here,$k$ is constant.

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

$\mathit{p}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\left(\nabla \times E\right)$                                                                   ….. (2)

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

$\mathbf{\nabla }\mathbf{.}\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\gamma }}^{\mathbf{2}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left(r^2{E}_r\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{rsin\theta }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\theta }}\left({\mathrm{sin\theta E}}_{\theta }\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{rsin\theta }}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\varphi }}\left(E_f\right)$                                    …… (3)

Step 2: Determine charge density

From the equation (1), the values of$E_{\gamma }$,$E_{\theta }$ and $E_{\varphi }$.

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

Substitutes the values of $E_{\gamma }$,$E_{\theta }$ and $E_{\varphi }$in equation (3), then

$$\begin{gathered} \nabla .E=\left\{\frac{1}{{\gamma }^2}\frac{\partial }{\partial r}\left(r^2\left[\frac{3k}{r}\right]\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \left[\frac{k\left(2\sin \theta \cos \theta \sin \varphi \right)}{r}\right]\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }\left(\frac{k\left(\sin \theta \cos \varphi \right)}{r}\right)\right\} \\ =\left\{\frac{1}{r^2}\left(3k\right)+\frac{2k\left(\sin \varphi \right)}{r^2\sin \theta }\left(2\sin \theta {\cos }^2\theta +{\sin }^2\theta \left(-\sin \theta \right)\right)+\frac{k}{r^2\sin \theta }\left(\sin \theta \left(-\sin \varphi \right)\right)\right\} \\ =\frac{3k}{r^2}+\frac{k\left(4{\cos }^2\theta -2{\sin }^2\theta \right)\sin \varphi +k\left(-\sin \varphi \right)}{r^2} \\ =\frac{3k}{r^2}+\frac{k}{r^2}\left(\left(4{\cos }^2\theta -2{\sin }^2\theta \right)-1\right)\sin \varphi \end{gathered}$$

Step 3: Determine charge density using the identity

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

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

Solve further as,

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

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

$$\begin{gathered} \rho ={\epsilon }_0\left(\nabla .E\right) \\ =3k{\epsilon }_0\frac{\left(1+\cos 2\theta \sin \varphi \right)}{r^2} \end{gathered}$$

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