Chapter 5, Problem 22P

(a)

To determine

Find the value of electric field intensity E.

Answer to Problem 22P

The value of electric field intensity is $E=-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\text{V}/\text{m}$.

Explanation of Solution

Calculation:

The potential field is,

$V=10x^{2}yz-5z^2\text{V}$

Consider the general expression for electric field intensity.

$\begin{array}{c}E=-\nabla V\\ =-\frac{\partial V}{\partial x}{a}_x-\frac{\partial V}{\partial y}{a}_y-\frac{\partial V}{\partial z}{a}_z\end{array}$

Find the electric field intensity.

$\begin{array}{c}E=-\frac{\partial }{\partial x}\left(10x^{2}yz-5z^2\right)a_x-\frac{\partial }{\partial y}\left(10x^{2}yz-5z^2\right)a_y-\frac{\partial }{\partial z}\left(10x^{2}yz-5z^2\right)a_z\\ =-\left(20xyz\right)a_x-\left(10x^{2}z\right)a_y-\left(10x^{2}y-10z\right)a_z\\ =-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\text{V}/\text{m}\end{array}$

Conclusion:

Thus, the value of electric field intensity is $E=-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\text{V}/\text{m}$.

(b)

To determine

Find the value of electric flux density D.

Answer to Problem 22P

The value of electric flux density is

$D=-0.8842xyz{a}_x-0.4421x^{2}z{a}_y-0.4421\left(x^{2}y-z\right)a_z\text{nC}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

Consider the general expression for electric flux density.

$D=\epsilon E$

Here,

$\epsilon$ is permittivity.

Substitute $5{\epsilon }_{\text{o}}$ for $\epsilon$ and $-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z$ for E in above equation.

$D=\left(5{\epsilon }_{\text{o}}\right)\left[-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\right]$

Substitute $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$ in above equation.

$\begin{array}{c}D=\left(5\times \frac{{10}^{-9}}{36\pi }\right)\left[-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\right]\\ =\left(4.4209\times {10}^{-11}\right)\left[-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\right]\\ =\left[-0.8842xyz{a}_x-0.4421x^{2}z{a}_y-0.4421\left(x^{2}y-z\right)a_z\right]\times {10}^{-9}\\ =-0.8842xyz{a}_x-0.4421x^{2}z{a}_y-0.4421\left(x^{2}y-z\right)a_z\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the value of electric flux density is

$D=-0.8842xyz{a}_x-0.4421x^{2}z{a}_y-0.4421\left(x^{2}y-z\right)a_z\text{nC}/{\text{m}}^{\text{2}}$.

(c)

To determine

Find the value of polarization P.

Answer to Problem 22P

The value of polarization is $P=-0.7073xyz{a}_x-0.3537x^{2}z{a}_y-0.3537\left(x^{2}y-z\right)a_z\text{nC}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

Consider the general expression for polarization field.

$P={\chi }_e{\epsilon }_{\text{o}}E$        (1)

Here,

${\chi }_e$ is the electric susceptibly,

${\epsilon }_{\text{o}}$ is the permittivity of free space, and

E is the electric field intensity.

Consider the general relationship for electric susceptibility and relative permittivity.

$1+{\chi }_e={\epsilon }_r$

${\chi }_e={\epsilon }_r-1$        (2)

Here,

${\epsilon }_r$ is the dielectric constant.

Consider the general expression for permittivity.

$\epsilon ={\epsilon }_{\text{o}}{\epsilon }_r$

Substitute $5{\epsilon }_{\text{o}}$ for $\epsilon$ in above equation.

$\begin{array}{l}5{\epsilon }_{\text{o}}={\epsilon }_{\text{o}}{\epsilon }_r\\ {\epsilon }_r=5\end{array}$

Substitute 5 for ${\epsilon }_r$ in Equation (2).

$\begin{array}{c}{\chi }_e=5-1\\ =4\end{array}$

Substitute 4 for ${\chi }_e$, $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$, and $-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z$ for E in Equation (1).

$\begin{array}{c}P=\left(4\right)\left(\frac{{10}^{-9}}{36\pi }\right)\left[-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\right]\\ =\left(3.5367\times {10}^{-11}\right)\left[-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\right]\\ =\left[-0.7073xyz{a}_x-0.3537x^{2}z{a}_y-0.3537\left(x^{2}y-z\right)a_z\right]\times {10}^{-9}\\ =-0.7073xyz{a}_x-0.3537x^{2}z{a}_y-0.3537\left(x^{2}y-z\right)a_z\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the value of polarization is

$P=-0.7073xyz{a}_x-0.3537x^{2}z{a}_y-0.3537\left(x^{2}y-z\right)a_z\text{nC}/{\text{m}}^{\text{2}}$.

(d)

To determine

Find the value of volume charge density ${\rho }_v$.

Answer to Problem 22P

The value of volume charge density is ${\rho }_v=-0.8842yz+0.4421\text{nC}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Calculation:

Consider the general expression for volume charge density.

${\rho }_v=-\epsilon {\nabla }^{2}V$        (3)

Refer to Part (a),

$-\nabla V=-20xyz{a}_x-10x^{2}z{a}_y-10\left(x^{2}y-z\right)a_z\frac{\text{V}}{\text{m}}$

Therefore,

$\begin{array}{c}{\nabla }^{2}V=\frac{\partial }{\partial x}\left(20xyz\right)+\frac{\partial }{\partial y}\left(10x^{2}z\right)+\frac{\partial }{\partial z}10\left(x^{2}y-z\right)\\ =20yz+\left(0\right)+\left(0-10\right)\\ =20yz-10\end{array}$

Substitute $5{\epsilon }_{\text{o}}$ for $\epsilon$ and $20yz-10$ for ${\nabla }^{2}V$ in Equation (3).

${\rho }_v=-\left(5{\epsilon }_{\text{o}}\right)\left(20yz-10\right)$

Substitute $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$ in above equation.

$\begin{array}{c}{\rho }_v=-\left(5\times \frac{{10}^{-9}}{36\pi }\right)\left(20yz-10\right)\\ =-\left(4.4209\times {10}^{-11}\right)\left(20yz-10\right)\\ =\left(-0.8842yz+0.4421\right)\times {10}^{-9}\\ =-0.8842yz+0.4421\text{nC}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the value of volume charge density is ${\rho }_v=-0.8842yz+0.4421\text{nC}/{\text{m}}^{\text{3}}$.