Question

The plane \(z=0\) defines the boundary between two dielectrics. For \(z<0\), \(\epsilon_{r 1}=9, \epsilon_{r 1}^{\prime \prime}=0\), and \(\mu_{1}=\mu_{0} .\) For \(z>0, \epsilon_{r 2}^{\prime}=3, \epsilon_{r 2}^{\prime \prime}=0\), and \(\mu_{2}=\mu_{0}\) Let \(E_{x 1}^{+}=10 \cos (\omega t-15 z) \mathrm{V} / \mathrm{m}\) and find \((a) \omega ;(b)\left\langle\mathbf{S}_{1}^{+}\right\rangle ;(c)\left\langle\mathbf{S}_{1}^{-}\right\rangle\); (d) \(\left\langle\mathbf{S}_{2}^{+}\right\rangle\).

Short answer

Question: Calculate the average Poynting vectors for the incident, reflected, and transmitted waves. Answer: For the incident wave in the first region (z<0): \(\left\langle\mathbf{S}_{1}^{+}\right\rangle = \frac{50\sqrt{9\epsilon_0}}{\sqrt{\mu_0}}\hat{z}\) For the reflected wave in the first region (z<0): \(\left\langle\mathbf{S}_{1}^{-}\right\rangle = -\frac{50\sqrt{9\epsilon_0}(1-\sqrt{3})^2}{\sqrt{\mu_0}(2\sqrt{3}-3)^2}\hat{z}\) For the transmitted wave in the second region (z>0): \(\left\langle\mathbf{S}_{2}^{+}\right\rangle = \frac{50\sqrt{3\epsilon_0}(4\sqrt{3})^2}{\sqrt{\mu_0}(2\sqrt{3}-3)^2}\hat{z}\)

Step-by-step solution

Find the angular frequency, \(\omega\)

By simply examining the given electric field, we can determine the angular frequency, \(\omega\). The electric field is given as \(E_{x1}^+=10\cos(\omega t-15z) V/m\). In this expression, the term multiplying the time variable, t, is the angular frequency. Therefore, the angular frequency is: \(\omega = 15\)

Calculate the Wave Impedance

Next, we need to compute the wave impedance for both regions. Recall that the wave impedance is given by the formula: \(Z = \frac{\sqrt{\mu}}{\sqrt{\epsilon}}\) For the first region (z < 0), the wave impedance is: \(Z_1 = \frac{\sqrt{\mu_1}}{\sqrt{\epsilon_{r1}\epsilon_0}} = \frac{\sqrt{\mu_0}}{\sqrt{9\epsilon_0}}\) For the second region (z > 0), the wave impedance is: \(Z_2 = \frac{\sqrt{\mu_2}}{\sqrt{\epsilon_{r2}'\epsilon_0}} = \frac{\sqrt{\mu_0}}{\sqrt{3\epsilon_0}}\)

Calculate Fresnel Reflection and Transmission Coefficients

Next, we will calculate the Fresnel reflection and transmission coefficients. For normal incidence at the boundary between two dielectric regions, we have the following formulas: Reflection coefficient: \(R = \frac{Z_2 - Z_1}{Z_1 + Z_2}\) Transmission coefficient: \(T = \frac{2Z_2}{Z_1 + Z_2}\) Calculating these coefficients: \(R = \frac{\frac{\sqrt{\mu_0}}{\sqrt{3\epsilon_0}} - \frac{\sqrt{\mu_0}}{\sqrt{9\epsilon_0}}}{\frac{\sqrt{\mu_0}}{\sqrt{3\epsilon_0}} + \frac{\sqrt{\mu_0}}{\sqrt{9\epsilon_0}}} = \frac{1 - \sqrt{3}}{2\sqrt{3} - 3}\) \(T = \frac{2\frac{\sqrt{\mu_0}}{\sqrt{3\epsilon_0}}}{\frac{\sqrt{\mu_0}}{\sqrt{3\epsilon_0}} + \frac{\sqrt{\mu_0}}{\sqrt{9\epsilon_0}}} = \frac{4\sqrt{3}}{2\sqrt{3} - 3}\)

Calculate Reflected and Transmitted Electric Field Components

Now, we will use the reflection and transmission coefficients to find the reflected and transmitted electric field components in each region. Since the incident electric field is given, we can find the reflected field in the first region (z < 0) as: \(E_{x1}^- = RE_{x1}^+ = 10 (\frac{1-\sqrt{3}}{2\sqrt{3}-3})\cos(\omega t-15z) V/m\) In the second region (z > 0), the transmitted electric field component is: \(E_{x2}^+ = TE_{x1}^+ = 10 (\frac{4\sqrt{3}}{2\sqrt{3}-3})\cos(\omega t-15z) V/m\)

Calculate the Average Poynting Vectors

The last step is to compute the average Poynting vectors for the incident, reflected, and transmitted waves using the Poynting theorem: \(\left\langle\mathbf{S}\right\rangle = \frac{1}{2}\Re[\mathbf{E}\times\mathbf{H}^*]\) For the incident wave in the first region (z<0): \(\left\langle\mathbf{S}_{1}^{+}\right\rangle = \frac{1}{2Z_1}({E_{x1}^+})^2 \hat{z} = \frac{1}{2\frac{\sqrt{\mu_0}}{\sqrt{9\epsilon_0}}}(100) \hat{z} = \frac{50\sqrt{9\epsilon_0}}{\sqrt{\mu_0}}\hat{z}\) For the reflected wave in the first region (z<0): \(\left\langle\mathbf{S}_{1}^{-}\right\rangle = -\frac{1}{2Z_1}({E_{x1}^-})^2 \hat{z} = -\frac{1}{2\frac{\sqrt{\mu_0}}{\sqrt{9\epsilon_0}}}(100\frac{(1-\sqrt{3})^2}{(2\sqrt{3}-3)^2}) \hat{z} = -\frac{50\sqrt{9\epsilon_0}(1-\sqrt{3})^2}{\sqrt{\mu_0}(2\sqrt{3}-3)^2}\hat{z}\) For the transmitted wave in the second region (z>0): \(\left\langle\mathbf{S}_{2}^{+}\right\rangle = \frac{1}{2Z_2}({E_{x2}^+})^2 \hat{z} = \frac{1}{2\frac{\sqrt{\mu_0}}{\sqrt{3\epsilon_0}}}(100\frac{(4\sqrt{3})^2}{(2\sqrt{3}-3)^2}) \hat{z} = \frac{50\sqrt{3\epsilon_0}(4\sqrt{3})^2}{\sqrt{\mu_0}(2\sqrt{3}-3)^2}\hat{z}\)

Key concepts

Wave Impedance

Understanding wave impedance is crucial for analyzing electromagnetic waves traveling through different media. Wave impedance, often denoted by the symbol Z, is a characteristic of the medium that the wave is traveling through and is defined by the ratio of the electric field strength to the magnetic field strength. It can be mathematically represented as Z = \( \frac{\sqrt{\mu}}{\sqrt{\varepsilon}} \), where \( \mu \) is the magnetic permeability and \( \varepsilon \) is the electric permittivity of the medium.

In the given exercise, to find wave impedance, we first identify the permittivity and permeability for each region, and then apply the formula. Higher permittivity or lower permeability results in a lower impedance, affecting how waves are reflected or transmitted at an interface. Since the formula involves taking the square root of the product of permittivity and permeability, any change in these values will affect the wave impedance non-linearly. For students, it's essential to grasp wave impedance as it directly influences wave behavior at boundaries.

Fresnel Reflection and Transmission Coefficients

The Fresnel equations offer a descriptive way to calculate how much of an electromagnetic wave is reflected or transmitted at the boundary between two media, which is characterized by the Fresnel reflection and transmission coefficients. These coefficients, represented as R and T, depend on the wave impedance of the involved media.

For the reflection coefficient, we use R = \( \frac{Z_2 - Z_1}{Z_1 + Z_2} \), and for the transmission coefficient, T = \( \frac{2Z_2}{Z_1 + Z_2} \). These coefficients relate to the conservation of energy at the boundary. The reflection coefficient indicates how much of the incident energy is sent back into the first medium, while the transmission coefficient tells us how much energy enters the second medium.

It's important for students to understand that these coefficients are crucial in designing systems like antennas and sensors, where controlling the reflection and transmission of waves is key for optimal functionality. The calculated Fresnel coefficients not only help in predicting the behavior of waves at an interface but also provide insight into the efficiency of energy transfer across different media.

Average Poynting Vector

The average Poynting vector holds significant relevance in electromagnetics as it represents the average power per unit area carried by an electromagnetic wave. This vector, typically denoted by \( \left\langle\mathbf{S}\right\rangle \), quantifies the directional energy flux (or the rate of energy transfer) of the wave. For sinusoidal steady-state fields, it's given by \( \left\langle\mathbf{S}\right\rangle = \frac{1}{2}\Re[\mathbf{E}\times\mathbf{H}^*] \), where \( \mathbf{E} \) and \( \mathbf{H}^* \) denote the electric field and the complex conjugate of the magnetic field, respectively.

In the provided exercise solution, we consider the normal incident wave and apply the formula to find the average Poynting vectors for the incident, reflected, and transmitted waves. A positive average Poynting vector indicates power flow in the direction of propagation, while a negative value denotes power flow in the opposite direction, which is the case for the reflected wave.

Students should understand that the average Poynting vector provides more than just the power; it also tells about the directionality of power distribution in space, which is vital for applications such as wireless power transfer, microwave heating, and understanding energy absorption in materials.