# Electric field for a capacitor and for a flat conductor

From Gauss' law, one can easily derive that the electric field at some distance from an infinite sheet of charge density $\sigma$ is $E=\frac{\sigma}{2\epsilon _0}$. Now when one considers a conductor instead, because there is no electric field within the body of a conductor, the electric field somewhere above the surface of this infinite, flat conductor is $E=\frac{\sigma}{\epsilon _0}$. Now I am struggling to see which formula you can use in a physical situation.

In particular, I was considering a capacitor where the charge on each plate was Q. The well-known results say that the electric field within the conductor is $E=\sigma / \epsilon _0$, and since the electric fields from each plate add this must mean that each plate seperately is being considered as in the first case: a sheet of charge density $\sigma$ and not a conductor itself. I do not see why we doo not consider each plate as a conductor.

I am trying to reason this out by considering the fact that, since the plates are connected by a wire, technically to think of this as a single conductor I would have to think about the whole system. Then both of the plates would belong to the system/conductor, so above the surface of this whole system I could say that the electric field is $E=\sigma / \epsilon _0$, but to take both sheets into account I have to look between the plates.

But this argument is really unconvincing to me. I was wondering if anyone has any better explanation as to the lack of a factor of two in the expression for the electric field.