# Dielectric slab with a uniform field

## Homework Statement

An example I am reading has the following setup: a dielectric slab of dielectric constant εr exists between z=0 and z=d, whilst an external electric field E0=E0k is applied with k a unit vector in the z direction. This setup exists for all x and y. The rest of space is a vacuum.

The aim is to compute the electric displacement, the electric field, the polarization and the bound charges of the system.

One of the first steps is to conclude that outside of the slab, the electric field is E0. There is no explanation for this, and it seems sort of obvious - for example assuming a parallel plate capacitor produces the uniform field and then placing the slab inside this capacitor, the field in the vacuum would remain the same as what it was in the capacitor without the slab. I can prove this using Gauss' law. However how can I actually prove it for this problem without such an analogy - ideally mathematically but any simple physical reasoning will do. When I think of the situation without the capacitor analogy, I'm wondering how it is we would know that the polarization of the dielectric would not have any influence on the vacuum field.

Thanks.

Last edited:

rude man
Homework Helper
Gold Member
I'm not sure if the following is redundant with what you have already concluded, but:

you can show by Gauss that the E field outside the capacitor is E0. It has nothing to do with the dielectric constant inside the capacitor.

I'm not sure if the following is redundant with what you have already concluded, but:

you can show by Gauss that the E field outside the capacitor is E0. It has nothing to do with the dielectric constant inside the capacitor.

Yes but there isn't actually a capacitor in the problem - I was just making the analogy that the uniform field could have been produced by a capacitor and in this way the whole thing would make sense. However I was wondering if there was a better way without having to bring in imaginary capacitors...

rude man
Homework Helper
Gold Member
You don't have to think capacitor. You have an area where there is a dielectric ( 0<z<d) and you have an area where there's vacuum (z< 0 and z > d). Construct a Gaussian surface within the vacuum area.

You don't have to think capacitor. You have an area where there is a dielectric ( 0<z<d) and you have an area where there's vacuum (z< 0 and z > d). Construct a Gaussian surface within the vacuum area.

Ok, so no charge in the vacuum, so for any Gaussian surface the net flux must be zero. How does this tell me that the field is E0?

I guess the symmetry tells me any field must be in the z direction, so I can construct a cylindrical surface with an axis parallel to the z axis. Then as the flux is zero, the field must be the same at both caps of the cylinder. As the field is E0 at infinity, it must have this value anywhere else in the vacuum as I can make the cylinder as long as I want.

rude man
Homework Helper
Gold Member
Ok, so no charge in the vacuum, so for any Gaussian surface the net flux must be zero. How does this tell me that the field is E0?

I guess the symmetry tells me any field must be in the z direction, so I can construct a cylindrical surface with an axis parallel to the z axis. Then as the flux is zero, the field must be the same at both caps of the cylinder. As the field is E0 at infinity, it must have this value anywhere else in the vacuum as I can make the cylinder as long as I want.

That sounds good. I would have said that, since your Gaussian volume has no internal free charge, there is no net E flux thru its surface. So the external E field is unchanged and must therefore be E0.