Chapter 4, Problem 28P

(a)

To determine

The charge density.

Explanation of Solution

Given:

The electric flux density $D$ is $2\rho \left(z+1\right)\cos \varphi a_{\rho }-\rho \left(z+1\right)\sin \varphi a_{\varphi }+{\rho }^2\cos \varphi a_z\mu \text{C}/{\text{m}}^{\text{2}}$.

Calculation:

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

  $\begin{array}{c}{\rho }_V=\nabla \cdot D\\ =\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho D_{\rho }\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(D_{\varphi }\right)+\frac{\partial }{\partial z}\left(D_z\right)\end{array}$

  ${\rho }_V=\left\{\begin{array}{c}\frac{1}{\rho }\frac{\partial }{\partial \rho }\left[2{\rho }^2\left(z+1\right)\right]\\ +\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left[-\rho \left(z+1\right)\sin \varphi \right]\\ +\frac{\partial }{\partial z}\left[{\rho }^2\cos \varphi \right]\end{array}\right\}$

  ${\rho }_V=\left\{\begin{array}{c}\frac{1}{\rho }\left[4\rho \left(z+1\right)\right]\\ +\frac{1}{\rho }\left(-\rho \left(z+1\right)\cos \varphi \right)\\ +0\end{array}\right\}\mu \text{C}/{\text{m}}^3$

  $\begin{array}{c}{\rho }_V=\left[4\left(z+1\right)\cos \varphi -\left(z+1\right)\cos \varphi \right]\mu \text{C}/{\text{m}}^3\\ {\rho }_V=3\left(z+1\right)\cos \varphi \mu \text{C}/{\text{m}}^3\end{array}$

Thus, the charge density is $3\left(z+1\right)\cos \varphi \mu \text{C}/{\text{m}}^3$.

(b)

To determine

The total charge enclosed by the volume $0<\rho <2,0<\varphi <\pi /2,0<z<4$.

Explanation of Solution

Given:

The electric flux density $D$ is $2\rho \left(z+1\right)\cos \varphi a_{\rho }-\rho \left(z+1\right)\sin \varphi a_{\varphi }+{\rho }^2\cos \varphi a_z\mu \text{C}/{\text{m}}^{\text{2}}$.

Calculation:

Take the elemental volume $\left(dv\right)$ as $\rho d\rho d\varphi dz$.

Calculate the total charge $\left(Q\right)$ using relation.

  $Q={\displaystyle {\int }_V{\rho }_V\cdot dv}$

  $\begin{array}{c}Q={\displaystyle {\int }_V\left(2\rho \left(z+1\right)a_{\rho }-\rho \left(z+1\right)\sin \varphi a_{\varphi }+{\rho }^2\cos \varphi a_z\right)\rho d\rho d\varphi dz}\\ ={\displaystyle \underset{0}{\overset{4}{\int }}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}{\displaystyle \underset{0}{\overset{2}{\int }}\left(3\left(z+1\right)\cos \varphi \mu \text{C}/{\text{m}}^3\right)\rho d\rho d\varphi dz}}}\\ =\left(3\mu \text{C}\right){\left[\frac{{\rho }^2}{2}\right]}_0^2{\left[\frac{z^2}{2}+z\right]}_0^4{\left[\sin \varphi \right]}_0^{\pi /2}\\ =72\mu \text{C}\end{array}$

Thus, the total charge enclosed by the volume $0<\rho <2,0<\varphi <\pi /2,0<z<4$ is $72\mu \text{C}$.

(c)

To determine

The net flux through the surface of the volume in $\left(\text{b}\right)$.

Explanation of Solution

Given:

The electric flux density $D$ is $2\rho \left(z+1\right)\cos \varphi a_{\rho }-\rho \left(z+1\right)\sin \varphi a_{\varphi }+{\rho }^2\cos \varphi a_z\mu \text{C}/{\text{m}}^{\text{2}}$.

Calculation:

The net flux in cylinder does not vary with $z$, the flux from top to bottom is equal and opposite in nature.

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

  $\Psi ={\displaystyle {\int }_{S}D\cdot dS}$

  $\begin{array}{c}\Psi ={\displaystyle {\int }_S\left(2\rho \left(z+1\right)\cos \varphi a_{\rho }\mu \text{C}/{\text{m}}^{\text{2}}-\rho \left(z+1\right)\sin \varphi a_{\varphi }\mu \text{C}/{\text{m}}^{\text{2}}+{\rho }^2\cos \varphi a_z\mu \text{C}/{\text{m}}^{\text{2}}\right)\rho d\varphi dz{a}_{\rho }}\\ ={\displaystyle {\int }_0^{\frac{\pi }{2}}{\displaystyle {\int }_0^4\left(2\left(2\right)\left(z+1\right)\cos \varphi \right)\left(2\mu \text{C}/{\text{m}}^{\text{2}}\right)dzd\varphi }}\\ =\left(8\mu \text{C}\right){\left[\frac{z^2}{2}+z\right]}_0^4{\left[\sin \varphi \right]}_0^{\frac{\pi }{2}}\\ =\left(8\mu \text{C}\right)\left[\frac{4^2-0}{2}+4-0\right]\left[\sin \frac{\pi }{2}-\sin 0\right]\end{array}$    $\begin{array}{c}\Psi =\left(8\mu \text{C}\right)\left(12\right)\left(1-0\right)\\ =96\mu \text{C}\end{array}$

Calculate the flux density through the flat surfaces as a component of $\varphi =0°$ using the relation.

  $D_1=-\rho \left(z+1\right)\sin \varphi$

  $\begin{array}{c}{D}_1=-\rho \left(z+1\right)\sin 0°\\ =0\end{array}$

Calculate the flux density through the flat surfaces as component of $\varphi =90°$ using the relation.

  $D_2=-\rho \left(z+1\right)\sin \varphi$

  $\begin{array}{c}{D}_2=-\rho \left(z+1\right)\sin 90°\\ =-\rho \left(z+1\right)\end{array}$

Calculate the flux through the flat surface $\left(\Psi \right)$ at $\varphi =0°$ using the relation.

  ${\Psi }_2={\displaystyle {\int }_0^2{\displaystyle {\int }_0^4-}}\rho \left(z+1\right)d\rho dz$

    $\begin{array}{c}{\Psi }_2=-{\left[\frac{z^2}{2}+z\right]}_0^4{\left[\frac{{\rho }^2}{2}\right]}_0^2\\ =-\left[\frac{16-0}{2}+4-0\right]\left[\frac{4-0}{2}\right]\\ =-24\mu \text{C}\end{array}$

Calculate the net flux $\left({\Psi }_{\text{net}}\right)$ through the surface using the relation.

  ${\Psi }_{\text{net}}={\Psi }_1+{\Psi }_2$

  $\begin{array}{c}{\Psi }_{\text{net}}=96\mu \text{C}+\left(-24\mu \text{C}\right)\\ =72\mu \text{C}\end{array}$

Thus, the net flux through the surface of the volume is $72\mu \text{C}$.