Chapter 3, Problem 3.20P

A radial electric field distribution in free space is given in spherical coordinates as:

$\begin{array}{l}E=\frac{r{p}_0:}{3\theta }ar\left(r\le a\right)\\ E_2=\frac{\left(2a^3-r^2\right)p_0}{3_{\theta }{p}^2}ar\left(a\le r\le b\right)\\ E_3=\frac{\left(2a^3-b^2\right)p_0}{3\theta r^2}ar\left(r\ge b\right)\end{array}$

where p₀, a, and b are constants. (a) Determine the volume charge density in the entire region (0 < r < âˆ�) by the appropriate use of V, D = p_v. (b) In terms of given parameters, find the total charge, Q, within a sphere of radius r where r>b.

To determine

(a)

The volume charge density in the entire region ( $0\le r\le \infty$ ) by the appropriate use of $\nabla \cdot D={\rho }_v$.

Answer to Problem 3.20P

The volume charge density is given by

:

   $\begin{array}{l}{\rho }_{v1}={\rho }_0\text{ (}r\le a)\\ {\rho }_{v2}=-{\rho }_0\text{ (}a\le r\le \text{b) }\\ {\rho }_{v3}=0\text{ (}r\ge \text{b)}\end{array}$

Explanation of Solution

Given information:

The radial electric field distribution in free space is given as:

   $\begin{array}{l}{E}_1=\frac{r{\rho }_0}{3{\epsilon }_0}{a}_r\text{ (}r\le a)\\ E_2=\frac{(2a^3-r^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{ (}a\le r\le \text{b)}\\ E_3=\frac{(2a^3-b^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{ (}r\ge \text{b)}\end{array}$

Concept used:

The volume charge density ${\rho }_v$ can be evaluated given the electric flux density D using the Gauss's law.

   $\nabla \cdot D={\rho }_v$

We know that $D={\epsilon }_{0}E$

So, for each radius values we can obtain the corresponding flux density values and hence charge density from Gauss's law.

Calculation:

We are given electric field distribution in free space as:

   $\begin{array}{l}{E}_1=\frac{r{\rho }_0}{3{\epsilon }_0}{a}_r\text{ (}r\le a)\\ E_2=\frac{(2a^3-r^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{ (}a\le r\le \text{b)}\\ E_3=\frac{(2a^3-b^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{ (}r\ge \text{b)}\end{array}$

We have spherical coordinate coordinate system here. The divergence of flux density $D$ is given by

   $\nabla \cdot D=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2{D}_r)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }(D_{\theta }\sin \theta )+\frac{1}{r\sin \theta }\frac{\partial D_{\varphi }}{\partial \varphi }$

Here only radial component, $D_r$ exists and the remaining components are zero. Hence divergence is given by:

   $\nabla \cdot D=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2{D}_r)$

By Gauss's law:

   $\nabla \cdot D={\rho }_v$

Hence,

   ${\rho }_v=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2{D}_r)$

Substituting

   $\begin{array}{l}{E}_1=\frac{r{\rho }_0}{3{\epsilon }_0}{a}_r\text{ (}r\le a)\\ E_2=\frac{(2a^3-r^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{ (}a\le r\le \text{b)}\\ E_3=\frac{(2a^3-b^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{ (}r\ge \text{b)}\end{array}$

So, electric flux densities are given by:

   $\begin{array}{l}{D}_1={\epsilon }_0\frac{r{\rho }_0}{3{\epsilon }_0}{a}_r\text{ = }\frac{r{\rho }_0}{3}{a}_r\text{ (}r\le a)\\ D_2={\epsilon }_0\frac{(2a^3-r^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{=}\frac{(2a^3-r^3){\rho }_0}{3r^2}{a}_r\text{ (}a\le r\le \text{b)}\\ D_3={\epsilon }_0\frac{(2a^3-b^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{=}\frac{(2a^3-b^3){\rho }_0}{3r^2}{a}_r\text{ (}r\ge \text{b)}\end{array}$

   $\begin{array}{l}{\rho }_{v1}=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{r{\rho }_0}{3})={\rho }_0\text{ (}r\le a)\\ {\rho }_{v2}=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{(2a^3-r^3){\rho }_0}{3r^2})=-{\rho }_0\text{ (}a\le r\le \text{b) }\\ {\rho }_{v3}=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2\frac{(2a^3-b^3){\rho }_0}{3r^2})=0\text{ (}r\ge \text{b)}\end{array}$

Hence, we can summarise the charge density values as:

   $\begin{array}{l}{\rho }_{v1}={\rho }_0\text{ (}r\le a)\\ {\rho }_{v2}=-{\rho }_0\text{ (}a\le r\le \text{b) }\\ {\rho }_{v3}=0\text{ (}r\ge \text{b)}\end{array}$

To determine

(b)

The total charge, Q, within a sphere of radius r where r >b.

Answer to Problem 3.20P

The total charge $Q_{enc}$ within a sphere of radius r >b $\text{=4}\pi \left(\frac{2a^3}{3}-\frac{b^3}{3}\right){\rho }_0\text{ C}$ .

Explanation of Solution

Given information:

The electric field distribution in free space is given in spherical coordinates as:

   $\begin{array}{l}{E}_1=\frac{r{\rho }_0}{3{\epsilon }_0}{a}_r\text{ (}r\le a)\\ E_2=\frac{(2a^3-r^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{ (}a\le r\le \text{b)}\\ E_3=\frac{(2a^3-b^3){\rho }_0}{3{\epsilon }_0{r}^2}{a}_r\text{ (}r\ge \text{b)}\end{array}$

From the previous part a), we have volume charge densities as:

   $\begin{array}{l}{\rho }_{v1}={\rho }_0\text{ (}r\le a)\\ {\rho }_{v2}=-{\rho }_0\text{ (}a\le r\le \text{b) }\\ {\rho }_{v3}=0\text{ (}r\ge \text{b)}\end{array}$

Concept used:

We have to integrate the volume charge density over the volume of sphere of radius b using the piece-wise distribution given.

Calculation:

We have the volume charge densities as:

   $\begin{array}{l}{\rho }_{v1}={\rho }_0\text{ (}r\le a)\\ {\rho }_{v2}=-{\rho }_0\text{ (}a\le r\le \text{b) }\\ {\rho }_{v3}=0\text{ (}r\ge \text{b)}\end{array}$

So, the enclosed charge is given by:

   $\begin{array}{l}{Q}_{enc}={\displaystyle \underset{V}{\iiint }{\rho }_{v}dV}\text{=}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \underset{0}{\overset{a}{\int }}{\rho }_0}{r}^2\sin \theta drd\theta d\varphi }}-{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \underset{a}{\overset{b}{\int }}{\rho }_0}{r}^2\sin \theta drd\theta d\varphi }}\\ \text{ = 4}\pi {\rho }_0·{\left[\frac{r^3}{3}\right]}_0^a-\text{4}\pi {\rho }_0·{\left[\frac{r^3}{3}\right]}_a^b\\ \text{ =4}\pi \left(\frac{2a^3}{3}-\frac{b^3}{3}\right){\rho }_0\text{ C}\end{array}$