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Complex Impedance in metals

  • Thread starter bayners123
  • Start date
  • #1
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Homework Statement



A plane wave is passing through a metal. Show that the impedance Z can be given by
[tex] Z = \sqrt{ \frac{2 \omega \epsilon _0} {\sigma} } \frac{Z_0}{1-i} [/tex] where Zo is the impedance of free space and sigma is the conductivity.

You may assume that E is polarised in the x direction.

Homework Equations



[tex] Z_0 = \sqrt{ \frac{\epsilon_r \epsilon_0}{\mu_r \mu_0}} [/tex]

[tex] E_x = E_0 e^{i(\omega t - \tilde{k} x)} [/tex]
where [tex]\tilde{k} = k - iK[/tex]

The Attempt at a Solution



I've managed to get to the impedance in the form:
[tex] Z = \frac{ \mu_r \mu_0 \omega }{ k - iK } [/tex]
but this doesn't have any reference to the conductivity in it and I can't see how to get to the required equation from it. I thought to use [tex] \frac{\omega}{k} = \frac{c}{n} = \frac{c}{\sqrt{\epsilon_r \mu_r}} [/tex] but it didn't seem to help.
 

Answers and Replies

  • #2
31
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Actually, by subbing back into the wave equation I've got to
[tex] Z = \sqrt{ \frac{2 \omega}{\sigma \epsilon_r \epsilon_0}} \frac{\mu_r \mu_0}{1-i} Z_0 [/tex]
which is nearly there but I can't see the last bit..

---

EDIT: Solved. I was just being silly as usual: I had the wrong equation for the speed of light in terms of mu and epsilon. Please delete
 
Last edited:
  • #3
1,860
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Reasons like this are why I'm glad that I was taught EM in SI units over CGS, everyone knows c from curlB. :p
 

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