Chapter 4, Problem 57P

To determine

Whether the electric flux density is a genuine electric flux density and the magnitude of flux crossing the given plane.

Explanation of Solution

Given:

The electric flux density $\left(D\right)$ is $\left(2\rho \sin \varphi a_{\rho }-\frac{\cos \varphi }{2\rho }{a}_{\varphi }\right)\text{C}/{\text{m}}^2$.

The limit of the radial axis $\left(\rho \right)$ is $\rho =1$.

The limit of the angular axis $\left(\varphi \right)$ is $0\le \varphi \le \pi /4$.

The limit of axial axis $\left(z\right)$ is $0<z<1$.

Calculation:

Calculate the curl of field $\left(\nabla \times D\right)$ using the relation.

  $\nabla \times D=\frac{1}{\rho }\left|\begin{array}{ccc}{a}_{\rho }& \rho a_{\varphi }& a_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ D_{\rho }& \rho D_{\varphi }& D_z\end{array}\right|$

  $\begin{array}{c}\nabla \times D=\frac{1}{\rho }\left|\begin{array}{ccc}{a}_{\rho }& \rho a_{\varphi }& a_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ 2\rho \sin \varphi & \rho \left(-\frac{\cos \varphi }{2\rho }\right)& 0\end{array}\right|\\ =\frac{1}{\rho }\left[\begin{array}{l}{a}_{\rho }\left(\frac{\partial }{\partial \varphi }\left(0\right)-\frac{\partial }{\partial z}\left(-\frac{\cos \varphi }{2}\right)\right)-\rho a_{\varphi }\left(\frac{\partial }{\partial \rho }\left(0\right)-\frac{\partial }{\partial z}\left(2\rho \sin \varphi \right)\right)\\ +a_z\left(\frac{\partial }{\partial \rho }\left(-\frac{\cos \varphi }{2}\right)-\frac{\partial }{\partial \varphi }\left(2\rho \sin \varphi \right)\right)\end{array}\right]\\ =\frac{1}{\rho }\left[a_{\rho }\left(0-0\right)-\rho a_{\varphi }\left(0-0\right)+a_z\left(0-2\rho \cos \varphi \right)\right]\\ =-\frac{2\rho \cos \varphi }{\rho }{a}_z\end{array}$

  $\begin{array}{c}\nabla \times D=-2\cos \varphi a_z\\ \ne 0\end{array}$

The curl of the field is not equal to zero. Thus, the field is not genuine.

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

  ${\rho }_v=\nabla \cdot D$

  $\begin{array}{c}{\rho }_v=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho D_{\rho }\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(D_{\varphi }\right)+\frac{\partial }{\partial z}\left(D_z\right)\\ =\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(2{\rho }^2\sin \varphi \text{C}/{\text{m}}^2\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(-\frac{\cos \varphi }{2\rho }\text{C}/{\text{m}}^2\right)+\frac{\partial }{\partial z}\left(0\text{C}/{\text{m}}^2\right)\\ =\frac{1}{\rho }\left(4\rho \sin \varphi \text{C}/{\text{m}}^3\right)+\frac{1}{\rho }\left(-\frac{\left(-\sin \varphi \right)}{2\rho }\text{C}/{\text{m}}^3\right)+0\\ =4\sin \varphi +\frac{\sin \varphi }{2{\rho }^2}\text{C}/{\text{m}}^3\end{array}$

Calculate the flux $\left(\Psi \right)$ using the relation.

  $\Psi ={\displaystyle {\int }_v{\rho }_{v}dv}$

  $\begin{array}{c}\Psi ={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{\frac{\pi }{4}}{\displaystyle {\int }_0^1\left(4\sin \varphi +\frac{\sin \varphi }{2{\rho }^2}\text{C}/{\text{m}}^3\right)\rho d\rho d\varphi dz}}}\\ ={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{\frac{\pi }{4}}\left(4\rho \sin \varphi +\frac{\sin \varphi }{2\rho }\text{C}/{\text{m}}^2\right)d\rho d\varphi {\left[z\right]}_0^1}}\\ =\left[1-0\right]{\displaystyle {\int }_0^1{\displaystyle {\int }_0^{\frac{\pi }{4}}\left(4\rho \sin \varphi +\frac{\sin \varphi }{2\rho }\text{C}/{\text{m}}^2\right)d\rho d\varphi }}\\ ={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{\frac{\pi }{4}}\left(4\rho \sin \varphi +\frac{\sin \varphi }{2\rho }\text{C}/{\text{m}}^2\right)d\rho d\varphi }}\end{array}$

  $\begin{array}{c}\Psi ={\displaystyle {\int }_0^1{\left[4\rho \left(-\cos \varphi \right)+\frac{-\cos \varphi }{2\rho }\text{C}/\text{m}\right]}_0^{\frac{\pi }{4}}d\rho }\\ ={\displaystyle {\int }_0^1\left[4\rho \left(-\cos \left(\frac{\pi }{4}\right)\right)+\frac{-\cos \left(\frac{\pi }{4}\right)}{2\rho }-\left(4\rho \left(-\cos \left(0\right)\right)+\frac{-\cos \left(0\right)}{2\rho }\right)\text{C}/\text{m}\right]d\rho }\\ ={\displaystyle {\int }_0^1\left[4\rho \left(-\frac{\sqrt{2}}{2}\right)+\frac{-\frac{\sqrt{2}}{2}}{2\rho }-\left(4\rho \left(-1\right)+\frac{-1}{2\rho }\right)\text{C}/\text{m}\right]d\rho }\\ ={\displaystyle {\int }_0^1\left[-2\sqrt{2}\rho -\frac{\sqrt{2}}{4\rho }+4\rho +\frac{1}{2\rho }\text{C}/\text{m}\right]d\rho }\end{array}$

  $\begin{array}{c}\Psi ={\left[-2\sqrt{2}\left(\frac{{\rho }^2}{2}\right)-\frac{\sqrt{2}}{4}\left(\ln \rho \right)+4\left(\frac{{\rho }^2}{2}\right)+\frac{1}{2}\left(\ln \rho \right)\text{C}\right]}_0^1\\ =\{\begin{array}{c}\left[-2\sqrt{2}\left(\frac{1^2}{2}\right)-\frac{\sqrt{2}}{4}\left(\ln 1\right)+4\left(\frac{1^2}{2}\right)+\frac{1}{2}\left(\ln 1\right)\text{C}\right]\\ -\left[-2\sqrt{2}\left(\frac{0^2}{2}\right)-\frac{\sqrt{2}}{4}\left(\ln 0\right)+4\left(\frac{0^2}{2}\right)+\frac{1}{2}\left(\ln 0\right)\text{C}\right]\end{array}\\ =\left[\left(-\sqrt{2}-\frac{\sqrt{2}}{4}\left(0\right)+2+\frac{1}{2}\left(0\right)\right)-\left(0-0+0+0\right)\right]\text{C}\\ =2-\sqrt{2}\text{C}\end{array}$

  $\begin{array}{c}\Psi =2-1.4142\text{C}\\ =0.58578\text{C}\end{array}$

Thus, the given electric flux density $\left(D\right)$ is $\underset{\_}{\text{not genuine}}$ and the magnitude of the flux crossing the given surface is $\underset{\_}{0.58578\text{C}}$.