# 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.

## 1 Answer

In the first case, the case of charge sheet, the charge density on the sheet is able to generate electric field on both sides of the sheet. But in the second case, the case of conductor, whole charge density is generating electric field only on one side. That is where the factor 2 disappeared. This is all intuition. But if you want to know real reason then it is all about the gaussian surface you draw. In the first case, the case of charge sheet, you don't have thickness and the gaussian surface you draw must cross the sheet. But in the second case you have thickness and some of your gaussian surface ends inside the conductor. Understanding things mathematically is the best way to interpret. Neglect the words 'forces cancel out' in the picture.