Why is dielectric constant for metals is infinity?

Answer 1

It is not infinite. At some extreme values electron-positron pairs will start forming rapidly what will result current flow.

However from practical standpoint vacuum dielectric strength is limited by field emission from electrodes. Best results are about 40MV/m , what is far worse than some good dielectrics can provide.

Answer 2

The dielectric constant is related to the electronic susceptance in an isotropic material.
The susceptance is basically the ratio of polarization to applied electric field.
You can think about a conductor as having "bound" electrons in that they cannot leave the entire material, but are free to polarize across the entire length of a conductor.
When you apply an external electric field to a conductor, you polarize the entire conductor, such that the polarization causes the electric field inside the conductor to be zero (electrostatic equilibrium). In a normal dielectric, the bound electrons cannot move as far as in a conductor, they have a much smaller polarization.
The definition of a dipole moment is charge separation times separation distance.
In a conductor, the induced dipoles have distances of the magnitude of the size of the macroscopic object, which is much much larger than the dipole distances of a dielectric.
Hence, the polarization vectors in a conductor are nearly infinite compared to the polarization vectors of a dielectric (whose order of magnitude we are accustomed to dealing with). The susceptance is therefore near infinite, and so is the relative permittivity.

I believe I can reconcile the two opposing "proofs" for this problem.
The capacitance proof takes advantage of the fact that capacitance is proportional to permittivity. You can construct a capacitor with a specific dielectric, and compare its capacitance to a capacitor with identical geometry, but no dielectric.
Let C_0 be the capacitance with no dielectric and the C be the capacitance with a dielectric.
You therefore have
C / C_0 = e / e_0, I use e instead of epsilon.
C / C_0 = e_r = relative permittivity
For the purpose of this discussion, let us assume we are working with parallel plate capacitors.
The capacitance argument goes: If you replace the dielectric material with a conductor, you will measure no capacitance. Therefore C = 0 and e_r = 0/C_0 = 0.

Here is my rebuttal.
This sort of test is invalid when determining the permittivity properties of a conductor.
The reason is that the wires and parallel plates are made of the same material as the "dielectric" which in this case is a conductor.
You would not trust the results of this capacitance test if the wires and the plates were made of the same material as an insulating dielectric. You no longer have a capacitor, but merely a piece of bulk material with a strange shape.
The key thing when dealing with permittivity is the effects of polarization in the presence of an external electric field.
Since we are interested in measuring the corresponding property for a conductor, we must construct a test in which the only differences in the result are due to the polarization differences between materials.
We can find a sufficient test by assuming that no charge transfer can occur between the dielectric conductor and the wires/parallel plates.
Assume you have constructed some ingenious way to prevent charge transfer from occurring between the dielectric conductor and the wire/parallel plates.
Now, apply an ideal voltage source with voltage V_0 to our "capacitor".
This potential difference will create an electric field inside of our conductor.
The typical response of a conductor to an electric field is the near instantaneous movement of charge in order to cancel out the applied field and reach static equilibrium (electric field reaching zero inside the conductor). This typically happens in less then a femto-second.
That is the typical case. In our case, the conductor will experience charges migrating due to the electric field that try to cancel out the field. If we truly have an ideal voltage source however, the electric field inside the conductor will persist no matter how many charges move within the conductor. These charges cannot escape from the dielectric conductor either because of our earlier assumption. In this ideal model, the charges will build up on each side of the dielectric conductor indefinitely. In other words, Q approaches infinity. The capacitance by definition is Q / V_0 and since Q is infinite, so is the capacitance.
Hence by this modified capacitance test, we have C/C_0 = infinity = e_r.
The relative permittivity is infinite as well.
I think this should settle the debate.