Chapter 10, Problem 58P

(a)

To determine

Find the total electric field component of the wave in region 1.

Answer to Problem 58P

The total electric field component of the wave in region 1 is $\underset{\_}{5\cos \left({10}^{8}t+\frac{2}{3}y\right)a_z+\frac{5}{3}\cos \left({10}^{8}t-\frac{2}{3}y\right)a_z\text{V}/\text{m}}$.

Explanation of Solution

Calculation:

Write the expression for given incident electric field in region 1.

$E_i=5\cos \left({10}^{8}t+\beta y\right)a_z\text{V}/\text{m}$        (1)

Consider the expression to find the total electric field in region 1.

$E_1=E_i+E_r$        (2)

Here,

$E_i$ is the incident electric field and

$E_r$ is the reflected electric field.

Consider the expression for phase constant in lossless medium (region).

$\beta =\omega \sqrt{\mu \epsilon }$

Rewrite the expression for the given data.

$\beta =\omega \sqrt{{\mu }_{\text{o}}\left(4{\epsilon }_{\text{o}}\right)}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon =4{\epsilon }_{\text{o}}\end{array}\right\}$

Substitute ${10}^8\text{rad}/\text{s}$ for $\omega$, $4\pi \times {10}^{-7}\text{H}/\text{m}$ for ${\mu }_{\text{o}}$, and $\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}$ for ${\epsilon }_{\text{o}}$.

$\begin{array}{c}\beta =\left({10}^8\text{rad}/\text{s}\right)\sqrt{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\left(4\right)\left(\frac{{10}^{-9}}{36\pi }\text{F}/\text{m}\right)}\\ =\frac{2}{3}\text{rad}/\text{m}\end{array}$

Consider the expression for reflected electric field in region 1.

$E_r=E_{ro}\cos \left({10}^{8}t-\beta y\right)a_z\text{V}/\text{m}$        (3)

Here,

$E_{ro}$ is the magnitude of reflected electric field at $y=0$.

Consider the expression for reflected electric field at $y=0$.

$E_{ro}=\Gamma E_{io}$        (4)

Here,

$E_{io}$ is the magnitude of incident electric field at $y=0$, which is $5\text{V}/\text{m}$, and

$\Gamma$ is the reflection coefficient.

Write the expression to find the reflection coefficient.

$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}$        (5)

Here,

${\eta }_1$ is the intrinsic impedance for region 1 and

${\eta }_2$ is the intrinsic impedance for region 2.

As region 2 is free space, the intrinsic impedance is $120\pi \Omega$.

${\eta }_2=120\pi \Omega$

Find the intrinsic impedance for region 1.

$\begin{array}{c}{\eta }_1=\sqrt{\frac{\mu }{\epsilon }}\\ =\sqrt{\frac{{\mu }_{\text{o}}}{4{\epsilon }_{\text{o}}}}\left\{\begin{array}{l}\because \mu ={\mu }_{\text{o}}\\ \because \epsilon =4{\epsilon }_{\text{o}}\end{array}\right\}\\ =\frac{1}{2}\sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}}}\\ =\frac{1}{2}\left(120\pi \Omega \right)\left\{\because \sqrt{\frac{{\mu }_{\text{o}}}{{\epsilon }_{\text{o}}}}\cong 120\pi \Omega \right\}\end{array}$

${\eta }_1=60\pi \Omega$

Substitute $60\pi \Omega$ for ${\eta }_1$ and $120\pi \Omega$ for ${\eta }_2$ in Equation (5).

$\begin{array}{c}\Gamma =\frac{120\pi \Omega -60\pi \Omega }{120\pi \Omega +60\pi \Omega }\\ =\frac{1}{3}\end{array}$

Substitute $\frac{1}{3}$ for $\Gamma$ and $5\text{V}/\text{m}$ for $E_{io}$ in Equation (4).

$\begin{array}{c}{E}_{ro}=\left(\frac{1}{3}\right)\left(5\text{V}/\text{m}\right)\\ =\frac{5}{3}\text{V}/\text{m}\end{array}$

Substitute $\frac{5}{3}\text{V}/\text{m}$ for $E_{ro}$ and $\frac{2}{3}\text{rad}/\text{m}$ for $\beta$ in Equation (3).

$\begin{array}{c}{E}_r=\left(\frac{5}{3}\text{V}/\text{m}\right)\cos \left[{10}^{8}t-\left(\frac{2}{3}\text{rad}/\text{m}\right)y\right]a_z\\ =\frac{5}{3}\cos \left({10}^{8}t-\frac{2}{3}y\right)a_z\text{V}/\text{m}\end{array}$

Substitute $\frac{2}{3}\text{rad}/\text{m}$ for $\beta$ in Equation (1).

$E_i=5\cos \left({10}^{8}t+\frac{2}{3}y\right)a_z\text{V}/\text{m}$

Substitute $\left[5\cos \left({10}^{8}t+\frac{2}{3}y\right)a_z\text{V}/\text{m}\right]$ for $E_i$ and $\left[\frac{5}{3}\cos \left({10}^{8}t-\frac{2}{3}y\right)a_z\text{V}/\text{m}\right]$ for $E_r$ in Equation (2).

$\begin{array}{c}{E}_1=\left[5\cos \left({10}^{8}t+\frac{2}{3}y\right)a_z\text{V}/\text{m}\right]+\left[\frac{5}{3}\cos \left({10}^{8}t-\frac{2}{3}y\right)a_z\text{V}/\text{m}\right]\\ =5\cos \left({10}^{8}t+\frac{2}{3}y\right)a_z+\frac{5}{3}\cos \left({10}^{8}t-\frac{2}{3}y\right)a_z\text{V}/\text{m}\end{array}$

Conclusion:

Thus, the total electric field component of the wave in region 1 is $\underset{\_}{5\cos \left({10}^{8}t+\frac{2}{3}y\right)a_z+\frac{5}{3}\cos \left({10}^{8}t-\frac{2}{3}y\right)a_z\text{V}/\text{m}}$.

(b)

To determine

Find the time-average Poynting vector in region 1.

Answer to Problem 58P

The time-average Poynting vector in region 1 is $\underset{\_}{-0.0589a_y\text{W}/{\text{m}}^2}$.

Explanation of Solution

Calculation:

Consider the expression to find the time-average Poynting vector in region 1.

$P_{\text{ave1}}=\frac{E_{io}^2}{2{\eta }_1}\left(a_k\right)+\frac{E_{ro}^2}{2{\eta }_1}\left(-a_k\right)$        (6)

From Equation (1), the vector $a_k$ is $-a_y$.

$a_k=-a_y$

Substitute $5\text{V}/\text{m}$ for $E_{io}$, $\frac{5}{3}\text{V}/\text{m}$ for $E_{ro}$, $60\pi \Omega$ for ${\eta }_1$, and $-a_y$ for $a_k$ in Equation (6).

$\begin{array}{c}{P}_{\text{ave1}}=\frac{{\left(5\text{V}/\text{m}\right)}^2}{\left(2\right)\left(60\pi \Omega \right)}\left(-a_y\right)+\frac{{\left(\frac{5}{3}\text{V}/\text{m}\right)}^2}{\left(2\right)\left(60\pi \Omega \right)}\left[-\left(-a_y\right)\right]\\ \cong -0.0589a_y\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the time-average Poynting vector in region 1 is $\underset{\_}{-0.0589a_y\text{W}/{\text{m}}^2}$.

(c)

To determine

Find the time-average Poynting vector in region 2.

Answer to Problem 58P

The time-average Poynting vector in region 2 is $\underset{\_}{-0.0589a_y\text{W}/{\text{m}}^2}$.

Explanation of Solution

Calculation:

Consider the expression to find the time-average Poynting vector in region 2.

$P_{\text{ave2}}=\frac{E_{to}^2}{2{\eta }_2}{a}_k$        (7)

Here,

$E_{to}$ is the magnitude of transmitted electric field at $y=0$.

Consider the expression to find the magnitude of transmitted electric field at $y=0$.

$E_{to}=\tau E_{io}$        (8)

Here,

$\tau$ is the transmission coefficient.

Consider for the expression for transmission coefficient.

$\tau =\frac{2{\eta }_2}{{\eta }_2+{\eta }_1}$

Substitute $60\pi \Omega$ for ${\eta }_1$ and $120\pi \Omega$ for ${\eta }_2$.

$\begin{array}{c}\tau =\frac{\left(2\right)\left(120\pi \Omega \right)}{120\pi \Omega +60\pi \Omega }\\ =\frac{4}{3}\end{array}$

Substitute $\frac{4}{3}$ for $\tau$ and $5\text{V}/\text{m}$ for $E_{io}$ in Equation (8).

$\begin{array}{c}{E}_{to}=\left(\frac{4}{3}\right)\left(5\text{V}/\text{m}\right)\\ =\frac{20}{3}\text{V}/\text{m}\end{array}$

Substitute $\frac{20}{3}\text{V}/\text{m}$ for $E_{to}$, $120\pi \Omega$ for ${\eta }_2$, and $-a_y$ for $a_k$ in Equation (7).

$\begin{array}{c}{P}_{\text{ave2}}=\frac{{\left(\frac{20}{3}\text{V}/\text{m}\right)}^2}{\left(2\right)\left(120\pi \Omega \right)}\left(-a_y\right)\\ \cong -0.0589\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Thus, the time-average Poynting vector in region 2 is $\underset{\_}{-0.0589a_y\text{W}/{\text{m}}^2}$.