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## Main Question or Discussion Point

What are the dielectric constants of metals and i have heard that some materials have dielectric constants that are complex numbers. Please tell me which materials have complex Dielectric constants?

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What are the dielectric constants of metals and i have heard that some materials have dielectric constants that are complex numbers. Please tell me which materials have complex Dielectric constants?

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Born2bwire

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If i am not wrong, the dielectric constant is different for AC, DC and stationary charges. Sir what i have concluded is that the dielectric constant is dependent on the frequency of AC. What is the dielectric constant of metals in case of DC and stationay charges? What is the physical sgnificance of imaginary part of dielectric constant of metals?

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Born2bwire

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The imaginary part of the permittivity is the loss factor. You can also relate it to the conductivity of the material. The very high conductivities of metals means that most electromagnetic waves, with the exception of those in very very low frequencies, will not penetrate any appreciable distance into a metal before being completely attenuated. The real part of the permittivity, that regulates wave phenomenon like refratction, will not be much of a factor since the waves cannot penetrate far into the material. Of course, this is still dependent on the frequency, the type of material that you are working with and the physical dimensions of the material.

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The imaginary part of the permittivity is related to the ratio between the optical conductivity and the frequency omega. So it would appear that in the limit omega --> 0

The imaginary part of the permittivity is the loss factor. You can also relate it to the conductivity of the material. The very high conductivities of metals means that most electromagnetic waves, with the exception of those in very very low frequencies, will not penetrate any appreciable distance into a metal before being completely attenuated. The real part of the permittivity, that regulates wave phenomenon like refratction, will not be much of a factor since the waves cannot penetrate far into the material. Of course, this is still dependent on the frequency, the type of material that you are working with and the physical dimensions of the material.

the imaginary part of the permittivity should diverge, for any metal with finite conductivity, right ?

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DrDu

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well, my point is that:

Im eps = 4 pi sigma / omega

with eps(omega) = eps(omega,k-->0).

This means that for omega --> 0 then Im eps must diverge if sigma is finite, as it is in

metals. Or am I missing something ?

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Born2bwire

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That is correct, although there is a limit to the range of frequencies over which the conductivity is valid. At DC, any good conductor will basically behave like a perfect electrical conductor for any static problems. Even for slowly varying problems, like with quasi-statics I feel this should probably hold true as well. This is of course reflected in what you stated in that the imaginary part of the permittivity will diverge. A perfect electrical conductor has an infinite imaginary part.The imaginary part of the permittivity is related to the ratio between the optical conductivity and the frequency omega. So it would appear that in the limit omega --> 0

the imaginary part of the permittivity should diverge, for any metal with finite conductivity, right ?

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I assume by "perfect conductor" you mean a material with zero resistance, or, which is theThat is correct, although there is a limit to the range of frequencies over which the conductivity is valid. At DC, any good conductor will basically behave like a perfect electrical conductor for any static problems. Even for slowly varying problems, like with quasi-statics I feel this should probably hold true as well. This is of course reflected in what you stated in that the imaginary part of the permittivity will diverge. A perfect electrical conductor has an infinite imaginary part.

same, infinite conductivity sigma.

I believe the imaginary part of the permittivity should diverge at zero frequency even for a

"non perfect" electrical conductor, i.e. a conductor with a finite value of

sigma (finite DC electrical conductivity), as it is the omega in the denominator to make the

sigma/omega ratio diverge for omega --> 0, isn't it ?

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Born2bwire

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No, the imaginary part for good conductors does not actually diverge in real life. Copper, alumin(i)um, silver, gold, etc. are all good conductors but they have a finite conductivity at DC. This can easily be seen by the fact that a copper wire will have a very small, but non-zero, resistance even in DC circuits. But currents do not fall under the heading of statics that I mentioned earlier. If we have an electrostatic situation, then a good conductor behaves like a perfect conductor because we assume that the system is allowed to settle into a steady state. In which case the charges have had time to migrate in response to any fields so that in the end result, after sufficient time over which the applied fields and sources are held constant, the system arranges itself like that using a PEC. The better the conductor, the less time we need to wait for these transients to sort themselves out. Hence, a very good conductor can also behave like a perfect conductor over long length scales, like in the quasi-static regime.I assume by "perfect conductor" you mean a material with zero resistance, or, which is the

same, infinite conductivity sigma.

I believe the imaginary part of the permittivity should diverge at zero frequency even for a

"non perfect" electrical conductor, i.e. a conductor with a finite value of

sigma (finite DC electrical conductivity), as it is the omega in the denominator to make the

sigma/omega ratio diverge for omega --> 0, isn't it ?

The conductivity model is a fairly simple model so we should not be surprised to see that it diverges from empirical results when we go to the extremely low frequency or infrared frequency range. It would be an interesting exercise to see what the real part of the permittivity should be assuming the traditional conductivity model for the imaginary part. The Kramers-Kronig relation dictates that the real and imaginary parts of the permittivity/permeability are related to each other by Hilbert transforms. So if we have a lossy material, we must have a dispersive media.

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Born2bwire

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While we are discussing the actual effects of the permittivity with the behavior of the electric fields, I should note, to prevent confusion, that generally the behavior of a good conductor is more closer to that of a perfect conductor at high frequencies, say microwave and above. While the imaginary part in the model does diverge as we go down to DC, there are two factors. One, as I mentioned above, real world good conductors have a finite conductivity. Two, the electrical length becomes infinite as we approach DC. This becomes significant because when we talk about electromagnetic waves, we generally deal with the product kr, where k is the wave number (\omega\sqrt{\epsilon\mu}) and r is distance. If we think of this in terms of wavelength, where r = a\lambda, then the product kr ~ a*2\pi*\sqrt{\epsilon_r\mu_r}.

So if the permittivity and permeability are frequency independent, then the amount of attenuation (and phase change) over a single wavelength is constant. This creates a problem where even if the loss factor is slightly divergent near DC, the fields still penetrate through the entirety of the material since the most materials are going to be much smaller than the wavelength (easily on the order of 10's-100's of km even in the low KHZ). Where as in the microwave regime, the loss factor will be smaller and the material will be much larger than a wavelength in many situations. This means that the waves will penetrate only shallowly into the material. For a perfect conductor, the waves would not penetrate at all. So we get another odd behavior where at DC and maybe a few very extremely low frequencies we can estimate a good conductor as being a PEC and at very high frequencies we can do the same. But in between we cannot do so, even though we may have a very large loss factor in the low frequencies.

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