Power of Wavefront Penetrating Dielectric Material

In summary, the proportion of wavefront's power that penetrates a dielectric material's surface in a perpendicular collision from air can be calculated using the relative permittivity, wave impedance, and complex poynting vector equations. The resulting passing fraction is four times higher than the input power.
  • #1
DemoniWaari
15
0

Homework Statement


How large proportion of wavefront's power penetrates dielectric material's surface in a perpendicular collision from air. The only parameter that I have is [itex] \varepsilon_r = 16 [/itex] where [itex] \varepsilon_r [/itex] is the relative permittivity.

Homework Equations


[itex] \varepsilon = \varepsilon_r \varepsilon_0 [/itex]
[itex] \eta = \sqrt{\frac{\mu}{\varepsilon}}[/itex] Where [itex]\mu = \mu_0 [/itex] Because of the dielectric material.
[itex] \vec E(\vec r) = \vec E_0 e^{-j \vec k \cdot \vec r}[/itex]
[itex] \vec H(\vec r) = \frac{1}{\eta} \vec E_{0p} e^{-j \vec k \cdot \vec r}[/itex] Where [itex] \vec E_{0p} [/itex] is perpendicular to [itex] \vec E_0 [/itex] and has the same magnitude.
[itex] \vec S(\vec r) = \frac{1}{2} \vec E(\vec r) \times \vec H(\vec r)^{*}[/itex] The complex poyinting vector.

The Attempt at a Solution



[itex] \vec E_+(\vec r) = \vec E_0 e^{-j \vec k_1 \cdot \vec r}[/itex] wavefront in the air.
And
[itex] \vec E_-(\vec r) = \vec E_0 e^{-j \vec k_2 \cdot \vec r}[/itex] wavefront in the material
And [itex] \eta_1 = \sqrt{\varepsilon_0 \mu_0}[/itex] is the wave impedance.
Thus we get
[itex] \vec H_+(\vec r) = \frac{1}{\eta_1} \vec E_{0p} e^{-j \vec k_1 \cdot \vec r}[/itex]
And
[itex] \vec H_-(\vec r) = \frac{1}{\eta_2} \vec E_{0p} e^{-j \vec k_2 \cdot \vec r}[/itex]
Now the complex poyinting vectors are
[itex] \vec S_+(\vec r) = \frac{1}{2} \vec E_+(\vec r) \times \vec H_+(\vec r)^{*}[/itex]
We're not interested in the directions so we can just check the magnitudes thus we get.
[itex] \vec S_+(\vec r) = \frac{1}{2\eta_1} |E_0|^2[/itex]
And same thing for the other wavefront...
[itex] \vec S_-(\vec r) = \frac{1}{2\eta_2} |E_0|^2[/itex]

So the passing fraction is
[itex]\frac{ \vec S_-(\vec r)}{\vec S_+(\vec r)} =
\frac{\frac{1}{\eta_2}}{\frac{1}{\eta_1}} = \frac{\eta_1}{\eta_2} =
\frac{\sqrt{\frac{\mu_0}{\varepsilon_0}}}
{\sqrt{ \frac{\mu_0}{\varepsilon_r \varepsilon_0} }} = \sqrt{\epsilon_r}=4[/itex]

So interestingly output is four times higher than input... So I have a problem here =(
 
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  • #2
I suggest to use [itex.] rather than [tex.] or the latex will be displayed on a new line for each expression.
When you close the [itex.], use [/itex.] rather than [\itex.]. This is the reason why it doesn't work.
Without the . inside the []'s.
 
  • #3
OH YES!
Thank you kind sir.
 
  • #4
Penetrated power = incident power minus reflected power.
 
  • #5
I would first like to clarify that the power of a wavefront is not a fixed quantity that can be measured or calculated. The power of a wavefront can vary depending on the specific characteristics of the wave, such as its amplitude, frequency, and direction of propagation. Therefore, it is not accurate to talk about a specific proportion of a wavefront's power penetrating a dielectric material's surface.

That being said, the relative permittivity (\varepsilon_r) of a material does affect the propagation of electromagnetic waves through that material. The higher the \varepsilon_r, the more the material can polarize in the presence of an electric field, which can cause the wave to slow down and potentially reflect or refract at the interface between the air and the material.

Based on the given information, it is not possible to accurately calculate the proportion of a wavefront's power that would penetrate a dielectric material's surface in a perpendicular collision from air. This would require knowing the specific characteristics of the wave and the material, as well as the angle of incidence and the thickness of the material. Therefore, it is important to have more information and context in order to accurately answer this question.
 

1. What is the "Power of Wavefront Penetrating Dielectric Material"?

The "Power of Wavefront Penetrating Dielectric Material" refers to the ability of a dielectric material to allow an electromagnetic wavefront to pass through it with minimal distortion. This is important in various applications such as telecommunications, optics, and radar systems.

2. How does a dielectric material affect the power of a wavefront?

A dielectric material affects the power of a wavefront by altering the electric and magnetic fields of the wave as it passes through. This is due to the polarization of the material's molecules, which can either amplify or attenuate the wave's power depending on the dielectric constant and other properties of the material.

3. What are some common examples of dielectric materials?

Some common examples of dielectric materials include glass, plastic, rubber, ceramics, and certain types of fluids and gases. These materials can be found in everyday items such as lenses, cables, capacitors, and insulators.

4. How is the power of wavefront penetrating dielectric material measured?

The power of wavefront penetrating dielectric material is measured by its dielectric constant, which is a measure of how much the material affects the electric field of a wave passing through it. This can be determined through experiments or simulations using equipment such as a network analyzer or a vector network analyzer.

5. What are the practical applications of the power of wavefront penetrating dielectric material?

The power of wavefront penetrating dielectric material has numerous practical applications. In telecommunications, it is used to transmit and receive signals without distortion. In optics, it is used in lenses and mirrors to focus and reflect light. In radar systems, it is used to detect and track objects. It is also used in energy storage devices such as capacitors and batteries.

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