Chapter 4, Problem 4.23P

To determine

(a)

The electric field intensity of an electric potential given in cylindrical co-ordinates.

Answer to Problem 4.23P

   $E=\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}{a}_{\rho }\text{V}/\text{m}$.

Explanation of Solution

Given:

   $V\left(\rho \right)=\frac{a^2{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}$

Concept used:

   $E=-\nabla V$

Calculation:

Formula for electric field is formula shown below

   $\begin{array}{l}E=-\nabla V\\ E=-\frac{dV}{d\rho }\\ \text{Plugging}\text{value}\text{of}\text{potential}\text{in}\text{above}\text{formula}\\ E=-\frac{d\frac{a^2{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}}{d\rho }\\ E=\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}\end{array}$

Conclusion:

Hence,electric field is $E=\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}$.

To determine

(b)

The volume charge density for given electric potential.

Answer to Problem 4.23P

   ${\rho }_v=e^{-\frac{\rho }{a}}\frac{a{\rho }_0}{1}\left[\frac{1}{\rho }-1\right]$

Explanation of Solution

Given:

   $V\left(\rho \right)=\frac{a^2{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}$

   $E=\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}$

Concept used:

   ${\rho }_v=\left|\nabla \cdot D\right|$

   $D={\epsilon }_{0}E$

Calculation:

First calculate D from above formula

   $\begin{array}{l}D={\epsilon }_{0}E\\ D={\epsilon }_0\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}\\ D=a{\rho }_0{e}^{-\frac{\rho }{a}}\end{array}$

Therefore

   $\begin{array}{l}{\rho }_v=\left|\nabla \cdot D\right|\\ {\rho }_v=\frac{1}{\rho }\frac{d}{d\rho }\left(\rho D_{\rho }\right)\\ {\rho }_v=\frac{1}{\rho }\left[\rho \frac{d}{d\rho }\left(D_{\rho }\right)+D_{\rho }\frac{d}{d\rho }\left(\rho \right)\right]\\ {\rho }_v=\frac{1}{\rho }\left[\frac{\rho d}{d\rho }\left(a{\rho }_0{e}^{-\frac{\rho }{a}}\right)+a{\rho }_0{e}^{-\frac{\rho }{a}}\frac{d\rho }{d\rho }\right]\\ {\rho }_v=\frac{1}{\rho }\left[\frac{a{\rho }_0\rho }{1}\left(-\frac{1}{a}\right)\left(e^{-\frac{\rho }{a}}\right)+a{\rho }_0{e}^{-\frac{\rho }{a}}\right]\\ {\rho }_v=a{\rho }_0{e}^{-\frac{\rho }{a}}\left[-1+\frac{1}{\rho }\right]\\ {\rho }_v=a{\rho }_0{e}^{-\frac{\rho }{a}}\left[\frac{1}{\rho }-1\right]\end{array}$

Conclusion:

Hence, thevolume charge density is ${\rho }_v=a{\rho }_0{e}^{-\frac{\rho }{a}}\left[\frac{1}{\rho }-1\right]c/m^3$

To determine

(c)

Stored energy in the region $0\le \rho \le \infty ,0\le \varphi \le 2\pi \text{and}0\le z\le 1$.

Answer to Problem 4.23P

   $W_E=\frac{\pi a^4{\rho }_0{}^2}{4{\epsilon }_0}\text{J}$

Explanation of Solution

Given:

   ${\rho }_v=e^{-\frac{\rho }{a}}\frac{a{\rho }_0}{1}\left[\frac{1}{\rho }-1\right]$

   $V\left(\rho \right)=\frac{a^2{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}$

   $E=\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}$

Concept used:

   $W_E=\frac{1}{2}{\displaystyle \underset{Vol}{\int }{\rho }_v}Vdv$

Calculation:

Although,

   $W_E=\frac{1}{2}{\displaystyle \underset{Vol}{\int }{\rho }_v}Vdv$

Plugging value of ${\rho }_v$ and $V\left(\rho \right)$ in formula of the energy shown above:

   $\begin{array}{l}{W}_E=\frac{1}{2}{\displaystyle \underset{Vol}{\int }{e}^{-\frac{\rho }{a}}\frac{a{\rho }_0}{1}\left[\frac{1}{\rho }-1\right]}\frac{a^2{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}dv\\ W_E=\frac{1}{2}{\displaystyle \underset{Vol}{\int }{e}^{-\frac{\rho }{a}}\frac{a{\rho }_0}{1}\left[\frac{1}{\rho }-1\right]}\frac{a^2{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}dv\\ W_E=\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-\frac{\rho }{a}}\frac{a{\rho }_0}{1}\left[\frac{1}{\rho }-1\right]\frac{a^2{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}\rho d\rho d\varphi dz}}}\end{array}$

On solving the above integral:

   $W_E=\frac{\pi a^4{\rho }_0{}^2}{4{\epsilon }_0}\text{J}$

Conclusion:

Hence stored energy in the region $0\le \rho \le \infty ,0\le \varphi \le 2\pi \text{and}0\le z\le 1$ is $W_E=\frac{\pi a^4{\rho }_0{}^2}{4{\epsilon }_0}\text{J}$.

To determine

(d)

Stored energy in the region $0\le \rho \le \infty ,0\le \varphi \le 2\pi \text{and}0\le z\le 1$.

Answer to Problem 4.23P

   $W_E=\frac{\pi a^4{\rho }_0{}^2}{4{\epsilon }_0}\text{J}$

Explanation of Solution

Given:

   $E=\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}$

Concept used:

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

Calculation:

Although

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

Plugging value of $E$ in formula of the energy shown above

   $$

   $\begin{array}{l}{W}_E=\frac{1}{2}{\displaystyle \underset{Vol}{\int }{\epsilon }_0}{E}^{2}dv\\ W_E=\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\epsilon }_0{\left(\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{{\epsilon }_0}\right)}^2\rho d\rho d\varphi dz}}}\\ W_E=\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\infty }{\int }}{\left(\frac{a{\rho }_0{e}^{-\frac{\rho }{a}}}{1}\right)}^2\rho d\rho d\varphi dz}}}\end{array}$

On Solving above integral

   $W_E=\frac{\pi a^4{\rho }_0{}^2}{4{\epsilon }_0}\text{J}$

Conclusion:

Hence stored energy in the region $0\le \rho \le \infty ,0\le \varphi \le 2\pi \text{and}0\le z\le 1$ is $W_E=\frac{\pi a^4{\rho }_0{}^2}{4{\epsilon }_0}\text{J}$.