Chapter 5, Problem 43P

(a)

To determine

Find the value of $E_1$ and $D_1$.

Answer to Problem 43P

The value of $D_1$ is $12a_{\rho }-14a_{\varphi }+21a_z\text{nC}/{\text{m}}^{\text{2}}$ and $E_1$ is $387.8a_{\rho }-452.4a_{\varphi }+678.6a_z\text{V}/\text{m}$.

Explanation of Solution

Calculation:

Consider the general expression for dielectric interface.

$D_{1n}=D_{2n}$        (1)

Consider the general expression for $D_{2n}$.

$D_{2n}=D_2\cdot a_p\left\{\because a_n=a_p\right\}$

Substitute $\left(12a_{\rho }-6a_{\varphi }+9a_z\right)\times {10}^{-9}$ for $D_2$ in above equation.

$\begin{array}{c}{D}_{2n}=\left(12a_{\rho }-6a_{\varphi }+9a_z\right)\times {10}^{-9}\cdot a_p\\ =\left(12a_{\rho }\times {10}^{-9}\right)-0+0\\ =12a_{\rho }\text{nC}/{\text{m}}^{\text{2}}\end{array}$

From Equation (1),

$D_{1n}=D_{2n}=12a_{\rho }\text{nC}/{\text{m}}^{\text{2}}$

Find the value of $D_{2t}$.

$D_{2t}=D_2-D_{2n}$

Substitute $12a_{\rho }\times {10}^{-9}$ for $D_{2n}$ and $\left(12a_{\rho }-6a_{\varphi }+9a_z\right)\times {10}^{-9}$ for $D_2$ in above expression.

$\begin{array}{c}{D}_{2t}=\left[\left(12a_{\rho }-6a_{\varphi }+9a_z\right)\times {10}^{-9}\right]-\left(12a_{\rho }\times {10}^{-9}\right)\\ =\left(12a_{\rho }-6a_{\varphi }+9a_z-12a_{\rho }\right)\times {10}^{-9}\\ =-6a_{\varphi }+9a_z\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Consider general expression for dielectric interface.

$E_{2t}=E_{1t}$

The above equation becomes,

$\begin{array}{l}\frac{D_{2t}}{{\epsilon }_{r2}}=\frac{D_{1t}}{{\epsilon }_{r1}}\\ D_{1t}={\epsilon }_{r1}\frac{D_{2t}}{{\epsilon }_{r2}}\end{array}$

Substitute 3.5 for ${\epsilon }_{r1}$, 1.5 for ${\epsilon }_{r2}$, and $\left(-6a_{\varphi }+9a_z\right)\times {10}^{-9}$ for $D_{2t}$ in above equation.

$\begin{array}{c}{D}_{1t}=\left(3.5\right)\frac{\left(-6a_{\varphi }+9a_z\right)\times {10}^{-9}}{\left(1.5\right)}\\ =\left(-14a_{\varphi }+21a_z\right)\times {10}^{-9}\\ =-14a_{\varphi }+21a_z\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Find the value of $D_1$.

$D_1=D_{1t}+D_{1n}$

Substitute $\left(-14a_{\varphi }+21a_z\right)\times {10}^{-9}$ for $D_{1t}$ and $12a_{\rho }\times {10}^{-9}$ for $D_{1n}$ in above equation.

$\begin{array}{c}{D}_1=\left[\left(-14a_{\varphi }+21a_z\right)\times {10}^{-9}\right]+\left(12a_{\rho }\times {10}^{-9}\right)\\ =\left(12a_{\rho }-14a_{\varphi }+21a_z\right)\times {10}^{-9}\\ =12a_{\rho }-14a_{\varphi }+21a_z\text{nC}/{\text{m}}^{\text{2}}\end{array}$

Consider the general expression to find the value of electric field $E_1$.

$E_1=\frac{D_1}{{\epsilon }_{r1}{\epsilon }_{\text{o}}}$

Substitute $\left(12a_{\rho }-14a_{\varphi }+21a_z\right)\times {10}^{-9}$ for $D_1$, 3.5 for ${\epsilon }_{r1}$, and $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$ in above equation.

$\begin{array}{c}{E}_1=\frac{\left(12a_{\rho }-14a_{\varphi }+21a_z\right)\times {10}^{-9}}{\left(3.5\right)\left(\frac{{10}^{-9}}{36\pi }\right)}\\ =\frac{\left(12a_{\rho }-14a_{\varphi }+21a_z\right)\times {10}^{-9}}{\left(3.09467\times {10}^{-11}\right)}\\ =\left(387.8a_{\rho }-452.4a_{\varphi }+678.6a_z\right)\\ =387.8a_{\rho }-452.4a_{\varphi }+678.6a_z\text{V}/\text{m}\end{array}$

Conclusion:

Thus, the value of $D_1$ is $12a_{\rho }-14a_{\varphi }+21a_z\text{nC}/{\text{m}}^{\text{2}}$ and $E_1$ is

$387.8a_{\rho }-452.4a_{\varphi }+678.6a_z\text{V}/\text{m}$.

(b)

To determine

Find the value of $P_2$ and ${\rho }_{pv}{}_2$.

Answer to Problem 43P

The value of ${\rho }_{pv}{}_2$ is $-\frac{4}{\rho }\text{C}/{\text{m}}^{\text{2}}$ and $P_2$ is $4a_{\rho }-2a_{\varphi }+3a_z\text{nC}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Calculation:

Consider the general expression for polarization.

$P_2={\chi }_e{\epsilon }_{\text{o}}{E}_2$        (2)

Consider the general expression for susceptibility.

${\chi }_e={\epsilon }_{r2}-1$        (3)

Substitute Equation (3) in Equation (2).

$P_2=\left({\epsilon }_{r2}-1\right){\epsilon }_{\text{o}}{E}_2$

Substitute $\frac{D_2}{{\epsilon }_{r2}{\epsilon }_{\text{o}}}$ for $E_2$ in above equation.

$\begin{array}{c}{P}_2=\left({\epsilon }_{r2}-1\right){\epsilon }_{\text{o}}\left(\frac{D_2}{{\epsilon }_{r2}{\epsilon }_{\text{o}}}\right)\\ =\left(\frac{{\epsilon }_{r2}-1}{{\epsilon }_{r2}}\right)\left(D_2\right)\end{array}$

Substitute 1.5 for ${\epsilon }_{r2}$ and $\left(12a_{\rho }-6a_{\varphi }+9a_z\right)\times {10}^{-9}$ for $D_2$ in above expression.

$\begin{array}{c}{P}_2=\left(\frac{1.5-1}{1.5}\right)\left(12a_{\rho }-6a_{\varphi }+9a_z\right)\times {10}^{-9}\\ =\left(0.333\right)\left(12a_{\rho }-6a_{\varphi }+9a_z\right)\times {10}^{-9}\\ =\left(4a_{\rho }-2a_{\varphi }+3a_z\right)\times {10}^{-9}\end{array}$

$P_2=4a_{\rho }-2a_{\varphi }+3a_z\text{nC}/{\text{m}}^{\text{2}}$        (4)

Consider the general expression to find the volume charge density.

$\begin{array}{c}{\rho }_{pv}{}_2=-\nabla \cdot P_2\\ =-\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho P_{2_{\rho }}\right)-\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(P_{2_{\varphi }}\right)-\frac{\partial }{\partial z}\left(P_{2_z}\right)\end{array}$

Apply Equation (4) in above equation.

${\rho }_{pv}{}_2=-\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(4\rho \right)-\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(-2\right)-\frac{\partial }{\partial z}\left(3\right)$

On differentiating,

$\begin{array}{c}{\rho }_{pv}{}_2=-\frac{1}{\rho }\left(4\right)+0+0\\ =-\frac{4}{\rho }\text{C}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Thus, the value of ${\rho }_{pv}{}_2$ is $-\frac{4}{\rho }\text{C}/{\text{m}}^{\text{2}}$ and $P_2$ is $4a_{\rho }-2a_{\varphi }+3a_z\text{nC}/{\text{m}}^{\text{2}}$.

(c)

To determine

Calculate the energy density in each region.

Answer to Problem 43P

The energy density in region 1 and region 2 is $w_{E1}=12.62\mu \text{J}/{\text{m}}^{\text{3}}$ and $w_{E2}=9.839\mu \text{J}/{\text{m}}^{\text{3}}$.

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 3.5 for ${\epsilon }_{r1}$, $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$, and $387.8a_{\rho }-452.4a_{\varphi }+678.6a_z$ for $E_1$ in above equation.

$\begin{array}{c}{w}_{E1}=\frac{1}{2}\left(3.5\right)\left(\frac{{10}^{-9}}{36\pi }\right)\left(\sqrt{{387.8}^2+{452.4}^2+{678.6}^2}\right)\\ =12.62\mu \text{J}/{\text{m}}^{\text{3}}\end{array}$

Find the energy density in region 2.

$\begin{array}{c}{w}_{E2}=\frac{1}{2}{\epsilon }_{r2}{\epsilon }_{\text{o}}{\left|E_2\right|}^2\\ =\frac{1}{2}{\epsilon }_{r2}{\epsilon }_{\text{o}}{\left|\frac{D_2}{{\epsilon }_{r2}{\epsilon }_{\text{o}}}\right|}^2\left\{\because E_2=\frac{D_2}{{\epsilon }_{r2}{\epsilon }_{\text{o}}}\right\}\\ =\frac{1}{2{\epsilon }_{r2}{\epsilon }_{\text{o}}}{\left|D_2\right|}^2\end{array}$

Substitute 1.5 for ${\epsilon }_{r2}$, $\frac{{10}^{-9}}{36\pi }$ for ${\epsilon }_{\text{o}}$, and $\left(12a_{\rho }-6a_{\varphi }+9a_z\right)\times {10}^{-9}$ for $D_2$ in above equation.

$\begin{array}{c}{w}_{E2}=\frac{1}{2\left(1.5\right)\left(\frac{{10}^{-9}}{36\pi }\right)}\left[\left(\sqrt{{12}^2+6^2+9^2}\right)\times {10}^{-9}\right]\\ =9.839\mu \text{J}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the energy density in region 1 and region 2 is $w_{E1}=12.62\mu \text{J}/{\text{m}}^{\text{3}}$ and $w_{E2}=9.839\mu \text{J}/{\text{m}}^{\text{3}}$.