Attenuation constant (low-loss dielectrics)

[big man]

\alpha =\frac {\omega \epsilon \prime \prime} {2} \sqrt {\mu \epsilon \prime}

My question is what exactly does the \epsilon \prime \prime term signify??

I see sections of my notes that say \epsilon \prime \prime << \epsilon, 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 ( \epsilon ) I don't know what to do with the \epsilon \prime \prime 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.

 

[big man]

Sorry I was looking up how to correctly write up the primes using Latex. It's fixed now.

 

[OlderDan]

My guess is it's the real and imaginary parts of the "actual permittivity". This seems to support that interpretation

http://en.wikipedia.org/wiki/Permittivity#Complex_permittivity

 

[quasar987]

Dan is probably right.

The basic equation for the absorption \alpha is

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

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?!?

 

[OlderDan]

Dan is probably right.

The basic equation for the absorption \alpha is

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

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?!?

\epsilon = \epsilon \prime + i \epsilon \prime \prime

 

[quasar987]

That is the most sensible guess, but then the expression e''<<e doesn't make sense since e is not real.

 

[big man]

The \epsilon in my notes is the permittivity given by this expression:

\epsilon = \epsilon_r \epsilon_0

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:

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

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:

\epsilon \prime \prime = \frac {\sigma} {\omega}

Am I right in doing this??
,

 

[OlderDan]

\epsilon \prime \prime = \frac {\sigma} {\omega}

Am I right in doing this??

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