Chapter 5, Problem 42P

(a)

To determine

Find the value of polarization vector $P_1$.

Answer to Problem 42P

The value of polarization vector is $P_1=0.177a_x-0.10624a_y+0.2125a_z\text{nC}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

Consider the general formula for polarization.

$P_1={\chi }_e{\epsilon }_{\text{o}}{E}_1$

$P_1=\left({\epsilon }_{r1}-1\right){\epsilon }_{\text{o}}{E}_1\left\{\because {\chi }_e={\epsilon }_{r1}-1\right\}$        (1)

Consider the general formula for permittivity of the dielectric.

${\epsilon }_1={\epsilon }_{\text{o}}{\epsilon }_{r1}$

Substitute $3{\epsilon }_{\text{o}}$ for ${\epsilon }_1$ in above equation.

$\begin{array}{l}3{\epsilon }_{\text{o}}={\epsilon }_{\text{o}}\left({\epsilon }_{r1}\right)\\ {\epsilon }_{r1}=3\end{array}$

Substitute 3 for ${\epsilon }_{r1}$ Equation (1).

$P_1=\left(3-1\right){\epsilon }_{\text{o}}{E}_1$

Substitute $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$ and $10a_x-6a_y+12a_z$ for $E_1$ in above equation.

$\begin{array}{c}{P}_1=\left(3-1\right)\left(\frac{{10}^{-9}}{36\pi }\right)\left(10a_x-6a_y+12a_z\right)\\ =\left(1.768\times {10}^{-11}\right)\left(10a_x-6a_y+12a_z\right)\\ =0.177a_x-0.10624a_y+0.2125a_z\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the value of polarization vector is $P_1=0.177a_x-0.10624a_y+0.2125a_z\text{nC}/{\text{m}}^{\text{2}}$.

(b)

To determine

Find the value of electric field $E_2$ in the second dielectric and the angle $E_2$ makes with the y-axis.

Answer to Problem 42P

The value of electric field is $E_2=10a_x+12a_z\text{V}/\text{m}$ and the angle $E_2$ makes with the y-axis is ${\theta }_2=75.64°$.

Explanation of Solution

Calculation:

Consider the general formula for permittivity of the dielectric.

${\epsilon }_2={\epsilon }_{\text{o}}{\epsilon }_{r2}$

Substitute $4.5{\epsilon }_{\text{o}}$ for ${\epsilon }_2$ in above equation.

$\begin{array}{l}4.5{\epsilon }_{\text{o}}={\epsilon }_{\text{o}}\left({\epsilon }_{r2}\right)\\ {\epsilon }_{r2}=4.5\end{array}$

Consider the general expression for normal component of the electric flux density.

$D_{1n}=D_{2n}$

${\epsilon }_{r1}{E}_{1n}={\epsilon }_{r2}{E}_{2n}$        (2)

Find the value of normal component of the electric field.

$E_{1n}=E_1\cdot a_y\left\{\because a_n=a_y\right\}$

Substitute $10a_x-6a_y+12a_z$ for $E_1$ in above equation.

$\begin{array}{c}{E}_{1n}=\left(10a_x-6a_y+12a_z\right)\cdot a_y\\ =-6\end{array}$

Therefore,

$E_{1n}=-6a_y\text{V}/\text{m}$

Substitute $-6a_y$ for $E_{1n}$, 3 for ${\epsilon }_{r1}$, and 4.5 for ${\epsilon }_{r2}$ in Equation (2).

$\begin{array}{l}\left(3\right)\left(-6a_y\right)=\left(4.5\right)E_{2n}\\ E_{2n}=-4a_y\text{V}/\text{m}\end{array}$

Find the tangential component of electric field for the first dielectric.

$E_{1t}=E_1-E_{1n}$

Substitute $10a_x-6a_y+12a_z$ for $E_1$ and $-6a_y$ for $E_{1n}$ in above equation.

$\begin{array}{c}{E}_{1t}=\left(10a_x-6a_y+12a_z\right)-\left(-6a_y\right)\\ =10a_x+12a_z\text{V}/\text{m}\end{array}$

As the tangential component of electric field are continuous at the dielectric interface, therefore,

$\begin{array}{c}{E}_{2t}=E_{1t}\\ =10a_x+12a_z\text{V}/\text{m}\end{array}$

Find the total electric field in the second dielectric.

$E_2=E_{2t}+E_{2n}$

Substitute $10a_x+12a_z$ for $E_{2t}$ and $-4a_y$ for $E_{2n}$ in above expression.

$\begin{array}{c}{E}_2=10a_x+12a_z-4a_y\\ =10a_x-4a_y+12a_z\text{V}/\text{m}\end{array}$

Find the angle made by the electric field $E_2$ with the y-axis.

$\begin{array}{c}\tan {\theta }_2=\frac{\left|E_{2t}\right|}{\left|E_{2n}\right|}\\ =\frac{\sqrt{{10}^2+{12}^2}}{4}\\ {\theta }_2={\tan }^{-1}\left(3.905\right)\\ {\theta }_2=75.64°\end{array}$

Conclusion:

Thus, the value of electric field is $E_2=10a_x+12a_z\text{V}/\text{m}$ and the angle $E_2$ makes with the y-axis is ${\theta }_2=75.64°$.

(c)

To determine

Calculate the energy density in each region.

Answer to Problem 42P

The energy density in region 1 and region 2 is $w_{E1}=3.7136\text{nJ}/{\text{m}}^{\text{3}}$ and $w_{E2}=5.1725\text{nJ}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Calculation:

Find the energy density in region 1.

$\begin{array}{c}{w}_{E1}=\frac{1}{2}{\epsilon }_1{\left|E_1\right|}^2\\ =\frac{1}{2}{\epsilon }_{r1}{\epsilon }_{\text{o}}{\left|E_1\right|}^2\end{array}$

Substitute 3 for ${\epsilon }_{r1}$, $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$, and $10a_x-6a_y+12a_z$ for $E_1$ in above equation.

$\begin{array}{c}{w}_{E1}=\frac{1}{2}\left(3\right)\left(\frac{{10}^{-9}}{36\pi }\right)\left({10}^2+6^2+{12}^2\right)\\ =3.7136\text{nJ}/{\text{m}}^{\text{3}}\end{array}$

Find the energy density in region 2.

$w_{E2}=\frac{1}{2}{\epsilon }_{r2}{\epsilon }_{\text{o}}{\left|E_2\right|}^2$

Substitute 4.5 for ${\epsilon }_{r2}$, $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$, and $10a_x-4a_y+12a_z$ for $E_1$ in above equation.

$\begin{array}{c}{w}_{E2}=\frac{1}{2}\left(4.5\right)\left(\frac{{10}^{-9}}{36\pi }\right)\left({10}^2+4^2+{12}^2\right)\\ =5.1725\text{nJ}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the energy density in region 1 and region 2 is $w_{E1}=3.7136\text{nJ}/{\text{m}}^{\text{3}}$ and $w_{E2}=5.1725\text{nJ}/{\text{m}}^{\text{3}}$.