Question

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

by the expression

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

for some constant k, what is the charge density?

Short answer

Answer

The charge density is $\mathit{\rho }\mathbf{=}\mathbf{3}\mathit{k}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\left(1+cos2\theta sin\varphi \right)}{{\mathbf{r}}^{\mathbf{2}}}$.

Step-by-step solution

Define functions

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,

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

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

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

Determine charge density

From the equation (1), the values of ${\mathit{E}}_{\mathbf{\gamma }}$,${\mathit{E}}_{\mathbf{\theta }}$and ${\mathit{E}}_{\mathbf{\varphi }}$.

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

Substitutes the values of ${\mathit{E}}_{\mathbf{\gamma }}$,${\mathit{E}}_{\mathbf{\theta }}$and ${\mathit{E}}_{\mathbf{\varphi }}$in equation (3), then

$$\begin{gathered} \mathbf{\nabla }\mathbf{·}\mathit{E}\mathbf{=}\left\{\frac{1}{{\gamma }^2}\frac{\partial }{\partial r}\left(r^2\left[\frac{3k}{r}\right]\right)+\frac{1}{rsin\theta }\frac{\partial }{\partial \theta }\left(sin\theta \left[\frac{k\left(2sin\theta cos\theta sin\varphi \right)}{r}\right]\right)+\frac{1}{rsin\theta }\frac{\partial }{\partial \varphi }\left(\frac{k\left(sin\theta cos\varphi \right)}{r}\right)\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left\{\frac{1}{{\gamma }^2}\left(3k\right)+\frac{2k\left(sin\varphi \right)}{r^{2}sin\theta }\left(2sin\theta co{s}^2\theta +si{n}^2\theta \left(-sin\theta \right)\right)+\frac{k}{r^{2}sin\theta }\left(sin\theta \left(-sin\varphi \right)\right)\right\} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{k}\left(4co{s}^2\theta -2si{n}^2\theta \right)\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\varphi }\mathbf{+}\mathbf{k}\left(-sin\varphi \right)}{{\mathbf{r}}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{3}\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\left(\left(4co{s}^2\theta -2si{n}^2\theta \right)-1\right)\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi } \end{gathered}$$

Determine charge density using the identity

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

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

Solve further as,

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

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

$$\begin{gathered} \mathit{\rho }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\left(\nabla ·E\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathit{k}{\mathit{\epsilon }}_{\mathbf{0}}\frac{(1+\cos 2\theta \mathrm{sin\varphi })}{{\mathbf{r}}^{\mathbf{2}}} \end{gathered}$$

Thus, the charge density is $\mathit{\rho }\mathbf{=}\mathbf{3}\mathit{k}{\mathit{\epsilon }}_{\mathbf{0}}\frac{(1+\cos 2\theta \mathrm{sin\varphi })}{{\mathbf{r}}^{\mathbf{2}}}$.