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Homework Help: Attenuation constant (low-loss dielectrics)

  1. Nov 10, 2006 #1
    [tex]\alpha =\frac {\omega \epsilon \prime \prime} {2} \sqrt {\mu \epsilon \prime} [/tex]

    My question is what exactly does the [tex]\epsilon \prime \prime[/tex] term signify??

    I see sections of my notes that say [tex]\epsilon \prime \prime << \epsilon [/tex], but what is it?

    I mean when I am trying to find the attenuation constant and the material happens to be a low-loss dielectric and you are given the actual permittivity [tex]( \epsilon )[/tex] I don't know what to do with the [tex] \epsilon \prime \prime[/tex] term. It's the only value I don't know in the problem?

    I would REALLY appreciate any help on this because I really need to understand this before my exam on Monday.
     
    Last edited: Nov 10, 2006
  2. jcsd
  3. Nov 10, 2006 #2
    Sorry I was looking up how to correctly write up the primes using Latex. It's fixed now.
     
  4. Nov 10, 2006 #3

    OlderDan

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    Last edited: Nov 10, 2006
  5. Nov 10, 2006 #4

    quasar987

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    Dan is probably right.

    The basic equation for the absorption [itex]\alpha[/itex] is

    [tex]\alpha = 2\Im(k) = 2\omega \frac{\mu}{2}\sqrt{\sqrt{\Re(\epsilon)^2+\Im(\epsilon)^2}-\Re(\epsilon)}[/tex]

    And my guess is that your expression is some approximation of that.

    But are there really 3 epsilons?! a non prime, a primed and a double primed?!?
     
  6. Nov 10, 2006 #5

    OlderDan

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    [tex] \epsilon = \epsilon \prime + i \epsilon \prime \prime[/tex]
     
  7. Nov 10, 2006 #6

    quasar987

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    That is the most sensible guess, but then the expression e''<<e doesn't make sense since e is not real.
     
  8. Nov 10, 2006 #7
    The [tex]\epsilon[/tex] in my notes is the permittivity given by this expression:

    [tex]\epsilon = \epsilon_r \epsilon_0 [/tex]

    Thanks for the quick replies guys and I understand this a little better, but could you just tell me if I have this right.

    OK let's say that you are wanting to find the attenuation constant for a low-loss dielectric with the following properties:

    [tex]\sigma=5.80*10^-^2 (S/m)[/tex]
    [tex]\omega = 100 GHz[/tex]
    [tex]\epsilon_r = 1[/tex]
    [tex]\mu = \mu_0[/tex]

    You first test to see if it is a low-loss dielectric or a good conductor.
    Then finding that at that the material is a low-loss dielectric you can proceed to use the previsously stated equation for the attenuation constant (in first post). Do you say that it doesn't behave like a perfect dielectric, which means that the conduction current is not negligible and this means that you can reexpress the magnitude of the complex part as follows:

    [tex]\epsilon \prime \prime = \frac {\sigma} {\omega}[/tex]

    Am I right in doing this??
    ,
     
    Last edited: Nov 10, 2006
  9. Nov 11, 2006 #8

    OlderDan

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    That apperas to be consitent with the approach taken in that link I posted earlier
     
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