Complex permitivity of good conductors

[iVenky]

We can define complex permitivity of any medium as

[tex]\epsilon=\epsilon'-j\epsilon''[/tex]

And the loss tangent as

[tex]tan \delta = \frac{\omega \epsilon'' + \sigma}{\omega \epsilon'} [/tex]

The question that I have is for good conductors. I read that for good conductors, we are dominated by σ rather than displacement current, which makes sense. What I don't get it, for good conductors, ε''>>ε'. Why is that? I understand ε'' is dominant but ε' (=ε₀ε_r -> ∞ since ε_r -> ∞ for good conductors). What's the fallacy in this logic?

 

[Meir Achuz]

\epsilon' does not approach infinity for good conductors. You have been misled by the fact that letting epsilon' become infinite in a relation for a dielectric in a static electric field gives the corresponding relation for a conductor, but that is just a convenient mathematical trick. The physical epsilon' does not get particularly large in a conductor.

 

[iVenky]

\epsilon' does not approach infinity for good conductors. You have been misled by the fact that letting epsilon' become infinite in a relation for a dielectric in a static electric field gives the corresponding relation for a conductor, but that is just a convenient mathematical trick. The physical epsilon' does not get particularly large in a conductor.

Thanks, how much is the epsilon for a conductor?

 

[Meir Achuz]

epsilon is a complex function of wave number (k=1/lambda) for a conductor. For copper, the real part of epsilon varies from about 10^4 for k=10^3 cm^-1
to 10 for k=10^4 cm^-1. A summary of permittivity for conductors is given in:
muri.lci.kent.edu/References/NIM_Papers/Permittivity/1983_Ordal_optical.pdf
which I found on google.

 

[Meir Achuz]

The imaginary part of epsilon varies from 10^4 for k=10^3 to 1 for k=10^4, so it is less than the real part.