Field due to an infinite sheet of charge with dielectric on one side

Question

Suppose you have a surface charge density $\sigma$ on a conducting plane $z=0$. The region $z<0$ is filled with a dielectric of permittivity $\varepsilon$. What is the field everywhere?

I tried taking a limit of a setup where we have a 'slab' conductor of thickness $t$ (so the region $0\leq z\le t$ is conducting) and then taking a Gaussian pillbox (a cylinder of cross sectional area $A$ with length along $z$ axis, one face on $z=t/2$ and the other at $z=z_0$ where $z_0<0$ is fixed. Then, we can find $\mathbf D$ using this pillbox, using the formula $\int_{\partial V}\mathbf D\cdot\mathrm d\mathbf S=Q_{V}$, and the fact that we can assume $\mathbf D(\mathbf r)=D(z)\mathbf e_z$. Since $\mathbf D\cdot\mathrm d\mathbf S$ vanishes for the curved surface for the cylinder, and $\mathbf D=\varepsilon\mathbf E=\mathbf0$ inside the conductor, all we get is $$\mathbf D(z_0)\mathbf e_z(-\mathbf e_z)A=\sigma_{z=0}A$$ Thus, $\mathbf D(z_0)=\sigma_{z=0}$, and so get the field for $z<0$, $$\mathbf E=\frac{\sigma_{z=0}}{\varepsilon}(-\mathbf e_z)$$ And similarly, we can get the field for $z>t$ as $$\mathbf E=\frac{\sigma_{z=t}}{\varepsilon}(+\mathbf e_z)$$ We can take the limit, but how do we know what the densities on the two surfaces are going to be?

This question popped up when I was trying to see if the result for charge distribution on parallel plates still holds if we have dielectrics in the middle. So, for that we need to know the field due to a surface charge density in presence of a dielectric on one side. I'm guessing it will still hold as the dielectric will only affect the field for the charge densities on the inner side, and it will be the same factor for both plates, which should cancel since the charge is equal and opposite. But I'm not sure how to prove it. Any help regarding proving this, or the original problem will be appreciated.

Answer 1

Your making things unnecessarily hard for yourself. First, do it without the dielectric, such that - by symmetry - in your approach $\sigma_{z=0} = \sigma_{z=t} = \frac{\sigma}{2}$. Or rather, you can use the interface conditions or your pillbox without the finite slab width to find that $$ \left(\vec{D}_{z<0} - \vec{D}_{z>0}\right)\cdot \vec{n} = \sigma $$ with the normal vector $\vec{n}$. By symmetry the values are the same apart from their sign and the absolute value is $$ D = \frac{\sigma}{2}. $$ Divide by $\varepsilon_0$ and you get the electric field near the plate in vacuum.

Now, add the dielectric, the only thing that changes is the permittivity on one side of the plate...

Edit: Two steps of my chain of thought that I didn't mention explicitly above:

1. Use an analogy to a parallel-plate condensator where you can do something similar in experiment: charge the capacitor to a given charge, the field inside is $E_0$, insert a dielectric with relative permittivity $\varepsilon$ (keeping the charge constant by keeping the plates isolated) and the field inside will drop to $E_0 / \varepsilon$.

2. Calculating the field of a charged plate in vacuum is super-easy, a one-dimensional Poisson problem $$ \partial_x \partial_x \phi = \delta $$ (modulo some physical constants/factors and with $\phi$ the potential such that $E = -\partial_x \phi$ and $\delta$ the Dirac delta). From the fundamental solution you find that the electric field is the Heaviside function plus an arbitrary additive constant! So the solution is non-unique. That is until you choose the only one that is physically reasonable: the one with a mirror symmetry (the same mirror symmetry the charge distribution has...). In the problem you describe you face the same kind of non-uniqueness potentially (unless you give some boundary conditions at fixed $z$?!) and the two-step approach seems reasonable: first look at the situation without the dielectric such that there is a symmetry and a reasonable choice for the additive constant the field might potentially have, and only then add the dielectric...