Chapter 4, Problem 52P

(a)

To determine

The electric field intensity and the volume charge density.

Explanation of Solution

Given:

The potential distribution $\left(V\right)$ is $\left(2x^2+4y^2\right)\text{V}$.

Calculation:

Calculate the electric field intensity $\left(E\right)$ using the relation.

  $E=-\nabla V$

  $\begin{array}{c}E=-\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\left(\left(2x^2+4y^2\right)\text{V}\right)\\ =-\left(\frac{\partial \left(2x^2+4y^2\text{V}\right)}{\partial x}{a}_x+\frac{\partial \left(2x^2+4y^2\text{V}\right)}{\partial y}{a}_y+\frac{\partial \left(2x^2+4y^2\text{V}\right)}{\partial z}{a}_z\right)\\ =-\left(\left(4x+0\right)a_x+\left(0+8y\right)a_y+\left(0\right)a_z\right)\text{V}/\text{m}\\ =-4x{a}_x-8y{a}_y\text{V}/\text{m}\end{array}$

Take the permeability of free space $\left({\epsilon }_0\right)$ as $8.854\times {10}^{-12}\text{F}/\text{m}$.

Calculate the volume charge density $\left({\rho }_v\right)$ using the relation.

  ${\rho }_v={\epsilon }_0\nabla \cdot E$

  $\begin{array}{c}{\rho }_v=\left(8.854\times {10}^{-12}\text{F}/\text{m}\right)\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(-4x{a}_x-8y{a}_y\text{V}/\text{m}\right)\\ =\left(8.854\times {10}^{-12}\text{F}/\text{m}\right)\left(\frac{\partial \left(-4x\text{V}/\text{m}\right)}{\partial x}+\frac{\partial \left(-8y\text{V}/\text{m}\right)}{\partial y}+\frac{\partial \left(0\text{V}/\text{m}\right)}{\partial z}\right)\\ =\left(8.854\times {10}^{-12}\text{F}/\text{m}\right)\left(-4\text{V}/{\text{m}}^2-8\text{V}/{\text{m}}^2+0\text{V}/{\text{m}}^2\right)\\ =\left(-12\text{V}/{\text{m}}^2\right)\left(8.854\times {10}^{-12}\text{F}/\text{m}\right)\end{array}$

  ${\rho }_v=-1.06248\times {10}^{-10}\text{C}/{\text{m}}^3$

Thus, the electric field intensity is $\underset{\_}{-4x{a}_x-8y{a}_y\text{V}/\text{m}}$ and the volume charge density is $\underset{\_}{-1.06248\times {10}^{-10}\text{C}/{\text{m}}^3}$.

(b)

To determine

The electric field intensity and the volume charge density.

Explanation of Solution

Given:

The potential distribution $\left(V\right)$ is $\left(10{\rho }^2\sin \varphi +6\rho z\right)\text{V}$.

Calculation:

Calculate the electric field intensity $\left(E\right)$ using the relation.

  $E=-\nabla V$

  $\begin{array}{c}E=-\left(\frac{\partial }{\partial \rho }{a}_{\rho }+\frac{1}{\rho }\frac{\partial }{\partial \varphi }{a}_{\varphi }+\frac{\partial }{\partial z}{a}_z\right)\left(\left(10{\rho }^2\sin \varphi +6\rho z\right)\text{V}\right)\\ =-\left(\begin{array}{l}\frac{\partial \left(10{\rho }^2\sin \varphi +6\rho z\text{V}\right)}{\partial \rho }{a}_{\rho }+\frac{1}{\rho }\frac{\partial \left(10{\rho }^2\sin \varphi +6\rho z\text{V}\right)}{\partial \varphi }{a}_{\varphi }\\ +\frac{\partial \left(10{\rho }^2\sin \varphi +6\rho z\text{V}\right)}{\partial z}{a}_z\end{array}\right)\\ =-\left(\left(20\rho \sin \varphi +6z\right)a_{\rho }+\frac{1}{\rho }\left(10{\rho }^2\cos \varphi +0\right)a_{\varphi }+\left(0+6\rho \right)a_z\right)\text{V}/\text{m}\\ =-\left(20\rho \sin \varphi +6z\right)a_{\rho }-10\rho \cos \varphi a_{\varphi }-6\rho a_z\text{V}/\text{m}\end{array}$

Calculate the volume charge density $\left({\rho }_v\right)$ using the relation.

  ${\rho }_v={\epsilon }_0\nabla \cdot E$

  $\begin{array}{c}{\rho }_v={\epsilon }_0\left[\begin{array}{l}\left(\frac{1}{\rho }\frac{\partial }{\partial \rho }\rho a_{\rho }+\frac{1}{\rho }\frac{\partial }{\partial \varphi }{a}_{\varphi }+\frac{\partial }{\partial z}{a}_z\right)\\ \cdot \left(-\left(20\rho \sin \varphi +6z\right)a_{\rho }-10\rho \cos \varphi a_{\varphi }-6\rho a_z\text{V}/\text{m}\right)\end{array}\right]\\ ={\epsilon }_0\left(\begin{array}{l}\frac{1}{\rho }\frac{\partial \left(-\rho \left(20\rho \sin \varphi +6z\right)\text{V}/\text{m}\right)}{\partial \rho }+\frac{1}{\rho }\frac{\partial \left(-10\rho \cos \varphi \text{V}/\text{m}\right)}{\partial \varphi }\\ +\frac{\partial \left(-6\rho \text{V}/\text{m}\right)}{\partial z}\end{array}\right)\\ ={\epsilon }_0\left(\frac{1}{\rho }\left(-40\rho \sin \varphi -6z\text{V}/{\text{m}}^2\right)+\frac{1}{\rho }\left(10\rho \sin \varphi \text{V}/{\text{m}}^2\right)+0\right)\\ ={\epsilon }_0\left(-40\sin \varphi -\frac{6z}{\rho }+10\sin \varphi \text{V}/{\text{m}}^2\right)\end{array}$

  ${\rho }_v=\left(-30{\epsilon }_0\sin \varphi -\frac{6z}{\rho }{\epsilon }_0\right)\text{C}/{\text{m}}^3$

Thus, the electric field intensity is $\underset{\_}{-\left(20\rho \sin \varphi +6z\right)a_{\rho }-10\rho \cos \varphi a_{\varphi }-6\rho a_z\text{V}/\text{m}}$ and the volume charge density is $\underset{\_}{\left(-30{\epsilon }_0\sin \varphi -\frac{6z}{\rho }{\epsilon }_0\right)\text{C}/{\text{m}}^3}$.

(c)

To determine

The electric field intensity and the volume charge density.

Explanation of Solution

Given:

The potential distribution $\left(V\right)$ is $\left(5r^2\cos \theta \sin \varphi \right)\text{V}$.

Calculation:

Calculate the electric field intensity $\left(E\right)$ using the relation.

  $E=-\nabla V$

  $\begin{array}{c}E=-\left(\frac{\partial }{\partial r}{a}_r+\frac{1}{r}\frac{\partial }{\partial \theta }{a}_{\theta }+\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }{a}_{\varphi }\right)\left(\left(5r^2\cos \theta \sin \varphi \right)\text{V}\right)\\ =-\left(\begin{array}{l}\frac{\partial \left(5r^2\cos \theta \sin \varphi \text{V}\right)}{\partial r}{a}_r+\frac{1}{r}\frac{\partial \left(5r^2\cos \theta \sin \varphi \text{V}\right)}{\partial \theta }{a}_{\theta }\\ +\frac{1}{r\sin \theta }\frac{\partial \left(5r^2\cos \theta \sin \varphi \text{V}\right)}{\partial \varphi }{a}_{\varphi }\end{array}\right)\\ =-\left(\begin{array}{l}\left(10r\cos \theta \sin \varphi \right)a_r+\frac{1}{r}\left(-5r^2\sin \theta \sin \varphi \right)a_{\theta }+\\ \frac{1}{r\sin \theta }\left(5r^2\cos \theta \cos \varphi \right)a_{\varphi }\end{array}\right)\text{V}/\text{m}\\ =-10r\cos \theta \sin \varphi a_r+5r\sin \theta \sin \varphi a_{\theta }-5r\cot \theta \cos \varphi a_{\varphi }\text{V}/\text{m}\end{array}$

Calculate the volume charge density $\left({\rho }_v\right)$ using the relation.

  ${\rho }_v={\epsilon }_0\nabla \cdot E$

  $\begin{array}{c}{\rho }_v={\epsilon }_0\left[\begin{array}{l}\left(\frac{1}{r^2}\frac{\partial }{\partial r}{r}^2{a}_r+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\sin \theta a_{\theta }+\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }{a}_{\varphi }\right)\\ \cdot \left(-10r\cos \theta \sin \varphi a_r+5r\sin \theta \sin \varphi a_{\theta }-5r\cot \theta \cos \varphi a_{\varphi }\text{V}/\text{m}\right)\end{array}\right]\\ ={\epsilon }_0\left(\begin{array}{l}\frac{1}{r^2}\frac{\partial \left(-10r^3\cos \theta \sin \varphi \text{V}/\text{m}\right)}{\partial r}+\frac{1}{r\sin \theta }\frac{\partial \left(5r{\sin }^2\theta \sin \varphi \text{V}/\text{m}\right)}{\partial \theta }\\ +\frac{1}{r\sin \theta }\frac{\partial \left(-5r\cot \theta \cos \varphi \text{V}/\text{m}\right)}{\partial \varphi }\end{array}\right)\\ ={\epsilon }_0\left(\begin{array}{l}\frac{1}{r^2}\left(-30r^2\cos \theta \sin \varphi \text{V}/{\text{m}}^2\right)+\frac{1}{r\sin \theta }\left(5r\left(2\sin \theta \cos \theta \right)\sin \varphi \text{V}/{\text{m}}^2\right)\\ +\frac{1}{r\sin \theta }\left(5r\cot \theta \sin \varphi \text{V}/{\text{m}}^2\right)\end{array}\right)\\ ={\epsilon }_0\left(-30\cos \theta \sin \varphi +10\cos \theta \sin \varphi +\frac{5\cos \theta \sin \varphi }{{\sin }^2\theta }\text{V}/{\text{m}}^2\right)\end{array}$

  ${\rho }_v=\left(-20\cos \theta \sin \varphi {\epsilon }_0+\frac{5\cos \theta \sin \varphi }{{\sin }^2\theta }{\epsilon }_0\right)\text{C}/{\text{m}}^3$

Thus, the electric field intensity is $\underset{\_}{-10r\cos \theta \sin \varphi a_r+5r\sin \theta \sin \varphi a_{\theta }-5r\cot \theta \cos \varphi a_{\varphi }\text{V}/\text{m}}$ and the volume charge density is $\underset{\_}{\left(-20\cos \theta \sin \varphi {\epsilon }_0+\frac{5\cos \theta \sin \varphi }{{\sin }^2\theta }{\epsilon }_0\right)\text{C}/{\text{m}}^3}$.