# Prove a dielectric with infinite dielectric constant behaves as a conductor for static fields

I read the following problem: Prove that a dielectric medium for which $\varepsilon \to \infty$ behaves as a perfect conductor in the presence of static electric fields.

So, the easy part is that the normal component of an electric field is cancelled when entering the medium, as in a conductor. We also know that the tangent component of the electric field is continuous through the media, but how can we prove it vanishes for this case? This is usually done for a conductor by saying that any tangent field on the surface would produce a flowing motion of charges, hence leading with a non-static situation, impliying that this component (and the normal also) vanish for a conductor. But I cannot see how to prove this for an infinite constant dielectric without using the previous argument (assuming a priori that it is a conductor with free movable charges), this would be circular reasoning.

I could sum up the problem by asking: how do we prove that such infinite dielectric has an equipotential surface when exposed to static electric fields? or equivalently: how do we prove that the total field is actually normal to the surface of such dielectric?

Thanks, any help is appreciated.

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Did you get any result on the continuity of the normal component of D (electric flux)?? –  Gotaquestion Sep 26 '13 at 20:44
All I got from the continuity is that the normal component of the field must vanish inside the dielectric. The tangent component is another story. –  Rogelio Molina Sep 28 '13 at 5:47

Before proving anything, one should remember that the most general boundary conditions are:

For the perfect conductor case, we know intuitively that the electric field is zero. So by setting the electric field and flux equal to zero in medium 2, the boundary conditions become:

As I said, we know that intuitively. The physical argument is that if electric field existed it would induce charge that will cancel its existence in the material. Now for your problem, you need to prove intuitively that as permittivity goes to infinity, the electric field becomes zero. Unlike perfect conductor it is a little bit tricky and requires some mathematics. Starting from this equation:

The previous equation says that the total electric flux in material is the sum of external electric flux (first term in right hand side) and internal flux generated by material (second term in right hand side) which is called Polarization. The polarization can be written as:

You can think of polarization intuitively as being the flux of a locally generated charge density which is generated by the external electric field. The relation between polarization and induced charge density is given by (see Sadiku chapter 5):

Now speaking of your case, when permittivity goes to infinity it means the relative permittivity goes to infinity, X goes to infinity and polarization goes to infinity. Infinite polarization means infinite charge density. Since we know that charge is conserved and the material has non-zero volume, that implies the charge density has to be limited, which means the polarization can’t be infinite. The only possible mathematical case in which the product of two numbers is finite with one of them being infinite is having the other to be zero. In this case it is the electric field that has to be zero.

In shorter description, infinite relative permittivity implies infinite polarization which implies infinite charge density which violates the conservation of charge principle, UNLESS the electric field is zero.

Having the electric field in the material equal to zero, that makes it practically a perfect conductor.

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thanks, it makes sense now. –  Rogelio Molina Sep 27 '13 at 22:29
Simply consider Gauss law : $\nabla E=\frac{\rho}{\epsilon}$ so with $\epsilon\rightarrow\infty$ there is no $\nabla E$ : the electric field, if it exists, must be divergence-free.
However, if you don't have any variable magnetic field, $\nabla \times E=0$ so any electric field in your material has to be uniform.
At this point you only have to prove that it must be zero ; you can say that the energy density $\epsilon E^2/2 \rightarrow \infty$ for any $E\neq 0$ ; but there is probably a more elegant proof at this point.
I am not sure that settles the question, because in any part of space where there are not charges, as e.g. oustide a point charge say, both equations $\nabla \cdot E = 0$ and $\nabla \times E =0$ hold. This is just Laplace equation for the potential, and it does not imply that the field be uniform, so the question stands. The energy argument on the other hand may be helpful, but I was looking for a proof without involving mechanics. –  Rogelio Molina Sep 26 '13 at 20:41