Chapter 4, Problem 4.24P

A certain spherically symmetric charge configuration in free space produces an electric field given in spherical coordinates by $E\left(r\right)=\{{}_{(100p_0)/\left({\epsilon }_0{r}^2\right)a_{r}V/m\left(r\ge 10\right)}^{\left(p_0{r}^2\right)/{\left({100}_{c0}\right)}_{a}V/m\left(r\le 10\right)}$

where p₀ is a constant. (a) Find the charge density as a function of position, (b) Find the absolute potential as a function of position in the two regions, r < 10 and r > 10. (c) Check your result for part b by using the gradient, (d) Find the stored energy in the charge by an integral of the form of Eq. (42). (e) Find the stored energy in the field by an integral of the form of Eq. (44).

To determine

(a)

Charge density as a function of position.

Answer to Problem 4.24P

   ${\rho }_v\left(r\le 10\right)$ $=\frac{{\rho }_0{r}^2}{25}{\text{C/m}}^{\text{3}}$

   ${\rho }_v\left(r\ge 10\right)=$ $=0{\text{C/m}}^{\text{3}}$

Explanation of Solution

Given:

   $E\left(r\right)=\{\begin{array}{c}\left(\frac{{\rho }_0{r}^2}{100{\epsilon }_0}\right)a_r\text{V/m}\left(r\le 10\right)\\ \left(\frac{100{\rho }_0}{{\epsilon }_0{r}^2}\right)a_r\text{V/m}\left(r\ge 10\right)\end{array}$

   ${\rho }_0$ is constant.

Concept used:

   $\begin{array}{l}{\rho }_v=\nabla \cdot D\\ {\rho }_v=\nabla \cdot {\epsilon }_{0}E\end{array}$

Calculation:

Formula for charge density is:

   $\begin{array}{l}{\rho }_v=\nabla \cdot D\\ {\rho }_v=\nabla \cdot {\epsilon }_{0}E\end{array}$

Plugging value of E in the formula shown above:

   $\begin{array}{c}{\rho }_v\left(r\le 10\right)=\nabla \cdot \left(\frac{{\epsilon }_0{\rho }_0{r}^2}{100{\epsilon }_0}\right)a_r\\ =\frac{1}{r^2}\cdot \frac{d}{dr}\left(\frac{{\rho }_0{r}^2}{100}\right)\\ =\frac{{\rho }_0{r}^2}{25}{\text{C/m}}^{\text{3}}\\ {\rho }_v\left(r\ge 10\right)=\nabla \cdot \left(\frac{{\epsilon }_{0}100{\rho }_0}{{\epsilon }_0{r}^2}\right)a_r\\ =\frac{1}{r^2}\cdot \frac{d}{dr}\left(\frac{r^{2}100{\rho }_0}{r^2}\right)\\ =0{\text{C/m}}^{\text{3}}\end{array}$

To determine

(b)

Absolute potential as a function of position in the given regions.

Answer to Problem 4.24P

   $V\left(r\le 10\right)=\frac{40{\rho }_0}{3{\epsilon }_0}\text{V}$

   $V\left(r\ge 10\right)=\frac{100{\rho }_0}{{\epsilon }_{0}r}\text{V}$

Explanation of Solution

Given:

   $r\le 10$ and $r\ge 10$

   $E\left(r\right)=\{\begin{array}{c}\left(\frac{{\rho }_0{r}^2}{100{\epsilon }_0}\right)a_r\text{V/m}\left(r\le 10\right)\\ \left(\frac{100{\rho }_0}{{\epsilon }_0{r}^2}\right)a_r\text{V/m}\left(r\ge 10\right)\end{array}$

   ${\rho }_0$ is constant.

Concept used:

   $V=-{\displaystyle \underset{r_1}{\overset{r_2}{\int }}E{a}_r.a_{r}dr}$

Calculation:

Plugging value of E in the formula shown above:

   $\begin{array}{c}V\left(r\le 10\right)=-{\displaystyle \underset{\infty }{\overset{10}{\int }}\frac{100{\rho }_0}{{\epsilon }_0{r}^2}}{a}_r\cdot a_{r}dr-{\displaystyle \underset{10}{\overset{r}{\int }}\frac{{\rho }_0{r}^{\text{'}}{}^2}{100{\epsilon }_0}}{a}_r\cdot a_{r}d{r}^{\text{'}}\\ ={\left(\frac{100{\rho }_0}{{\epsilon }_{0}r}\right)}_{\infty }^{10}-{\left(\frac{{\rho }_0{r}^{\text{'}}{}^3}{300{\epsilon }_0}\right)}_{10}^r\\ =\frac{10{\rho }_0}{3{\epsilon }_0}\left[4-\left(0.001r^3\right)\right]\\ =\frac{40{\rho }_0}{3{\epsilon }_0}V\\ V\left(r\ge 10\right)=-{\displaystyle \underset{r}{\overset{\infty }{\int }}\frac{100{\rho }_0}{{\epsilon }_0{r}^{\text{'}}{}^2}}{a}_r\cdot a_{r}d{r}^{\text{'}}\\ ={\left(\frac{100{\rho }_0}{{\epsilon }_0{r}^{\text{'}}}\right)}_{\infty }^r\\ =\frac{100{\rho }_0}{{\epsilon }_{0}r}V\end{array}$

To determine

(c)

To verify:

The result obtained in part (b) by method of the gradient.

Answer to Problem 4.24P

   $E\left(r\right)=\{\begin{array}{c}\left(\frac{{\rho }_0{r}^2}{100{\epsilon }_0}\right)a_r\text{V/m}\left(r\le 10\right)\\ \left(\frac{100{\rho }_0}{{\epsilon }_0{r}^2}\right)a_r\text{V/m}\left(r\ge 10\right)\end{array}$

Explanation of Solution

Given:

   $r\le 10$ and $r\ge 10$

   $V\left(r\le 10\right)=\frac{40{\rho }_0}{3{\epsilon }_0}\text{V}$

   $V\left(r\ge 10\right)=\frac{100{\rho }_0}{{\epsilon }_{0}r}\text{V}$

   ${\rho }_0$ is constant.

Concept used:

   $E=-\nabla V$

Calculation:

Formula for electric field is shown above.

Plugging value of V in the formula shown above.

   $\begin{array}{l}E=-\nabla V\\ E_1=-\nabla V\left(r\le 10\right)\\ E_1=-\frac{d}{dr}\left(\frac{10{\rho }_0}{3{\epsilon }_0}\left[4-\left(0.001r^3\right)\right]\right)a_r\\ E_1=\frac{10{\rho }_0}{3{\epsilon }_0}\left(3r^2\right)0.001a_r\\ E_1=\frac{{\rho }_0}{100{\epsilon }_0}\left(r^2\right)a_r\\ E=-\nabla V\\ E_2=-\nabla V\left(r\ge 10\right)\\ E_2=-\frac{d}{dr}\frac{100{\rho }_0}{{\epsilon }_{0}r}{a}_r\\ E_2=\frac{100{\rho }_0}{r^2{\epsilon }_0}{a}_r\end{array}$

The result is same using both the methods.

To determine

(d)

The stored energy in the charge by an integral.

Answer to Problem 4.24P

   $W=7.18\times {10}^3\frac{{\rho }_0{}^2}{{\epsilon }_0}\text{J}$

Explanation of Solution

Given:

   $\begin{array}{l}V\left(r\le 10\right)=\frac{40{\rho }_0}{3{\epsilon }_0}\text{V}\\ {\rho }_v=\frac{{\rho }_0{r}^2}{25}{\text{C/m}}^{\text{3}}\end{array}$

   $\begin{array}{l}V\left(r\ge 10\right)=\frac{100{\rho }_0}{{\epsilon }_{0}r}\text{V}\\ {\rho }_v=0{\text{C/m}}^{\text{3}}\end{array}$

   ${\rho }_0$ is constant.

Concept used:

   $W=\frac{1}{2}{\displaystyle \underset{v}{\int }{\rho }_{v}Vdv}$

Calculation:

Formula for stored energy is shown above.

Plugging value of V and ${\rho }_v$ in the formula shown above.

   $\begin{array}{l}W=\frac{1}{2}{\displaystyle \underset{v}{\int }{\rho }_{v}Vdv}\\ W=\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \underset{0}{\overset{10}{\int }}\frac{{\rho }_0{r}^2}{25}\frac{10{\rho }_0}{3{\epsilon }_0}\left[4-\left(0.001r^3\right)\right]r^2}}}\sin \theta drd\theta d\varphi \\ W=\frac{4\pi {\rho }_0{}^2}{150{\epsilon }_0}{\displaystyle \underset{0}{\overset{10}{\int }}\left[40r^3-\frac{r^6}{100}\right]}dr\\ W=\frac{4\pi {\rho }_0{}^2}{150{\epsilon }_0}{\left[10r^4-\frac{r^7}{700}\right]}_0^{10}\\ W=7.18\times {10}^3\frac{{\rho }_0{}^2}{{\epsilon }_0}\text{J}\end{array}$

To determine

(e)

The stored energy in the electric field by an integral.

Answer to Problem 4.24P

   $W=7.18\times {10}^3\frac{{\rho }_0{}^2}{{\epsilon }_0}\text{J}$

Explanation of Solution

Given:

   $E\left(r\right)=\{\begin{array}{c}\left(\frac{{\rho }_0{r}^2}{100{\epsilon }_0}\right)a_r\text{V/m}\left(r\le 10\right)\\ \left(\frac{100{\rho }_0}{{\epsilon }_0{r}^2}\right)a_r\text{V/m}\left(r\ge 10\right)\end{array}$

   ${\rho }_0$ is constant.

Concept used:

   $W=\frac{1}{2}{\displaystyle \underset{v}{\int }{\epsilon }_0{E}^{2}dv}$

Calculation:

Formula for stored energy is shown above.

Plugging value of E in the formula shown above.

   $\begin{array}{l}W=\frac{1}{2}{\displaystyle \underset{r\le 10}{\int }{\epsilon }_0{E}^{2}dv}+\frac{1}{2}{\displaystyle \underset{r\le 10}{\int }{\epsilon }_0{E}^{2}dv}\\ W=\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \underset{0}{\overset{10}{\int }}{\epsilon }_0{\left(\frac{{\rho }_0{r}^2}{100{\epsilon }_0}\right)}^2{r}^2}}}\sin \theta drd\theta d\varphi +\frac{1}{2}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \underset{10}{\overset{\infty }{\int }}{\epsilon }_0{\left(\frac{100{\rho }_0}{{\epsilon }_0{r}^2}\right)}^2{r}^2}}}\sin \theta drd\theta d\varphi \\ W=\frac{2\pi {\rho }_0{}^2}{{\epsilon }_0}\left[{10}^{-4}{\displaystyle {\int }_0^{10}{r}^{6}dr+{10}^{-4}{\displaystyle {\int }_{10}^{\infty }{r}^{-2}dr}}\right]\\ W=\frac{2\pi {\rho }_0{}^2}{{\epsilon }_0}\left(\frac{1000}{7}+1000\right)\end{array}$

   $W=7.18\times {10}^3\frac{{\rho }_0{}^2}{{\epsilon }_0}\text{J}$