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Wavefront's power losses due to perpendicular collision with dielectric material.

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant 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.


    3. 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 =(
     
    Last edited: Mar 1, 2012
  2. jcsd
  3. Mar 1, 2012 #2

    fluidistic

    User Avatar
    Gold Member

    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.
     
  4. Mar 1, 2012 #3
    OH YES!
    Thank you kind sir.
     
  5. Mar 1, 2012 #4

    rude man

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    Homework Helper
    Gold Member

    Penetrated power = incident power minus reflected power.
     
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