# Gauss's law and superposition for parallel plates

Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors are given equal but opposite uniform surface charge densities +- $\sigma$. Ignore edge effects and use Gauss's law to show that for points far from the edges, the electric field between the plates is $E = > \frac{\sigma }{\epsilon_0}$.

I've searched a lot to find a solution to this problem.

In both of these links, the approach to find electrical field between the plates is

1- create a cylindrical gaussian surface

2- put one end of the cylinder to one of the plates where the area is uncharged (uncharged due to attraction between two plates)

3- put other end to be between the plates.

Since the flux will pass through only one end of this cylinder

$$EA = \frac{\sigma A}{\epsilon_0}$$ $$E = \frac{\sigma }{\epsilon_0}$$

My question is, why didn't we do the same thing for the other plate, and then use superposition principle? Or simply, why didn't we multiply what we found by 2 because of superposition?

• Because this is enough. Doing the same thing for the other plate would give you the same answer. Why would you double it? - I'm not sure what you mean by because of superposition. Mar 8, 2015 at 16:33
• @Steeven "this is enough" doesn't seem to be helpful, because this is exactly what the user does not understand - i.e. why is it enough. (I apologize for criticizing.) Mar 8, 2015 at 16:36
• I thought this calculation lacked the contribution of the other plate. Imagine two point charges along the same axis, and you are asked to find electrical field between them. When you enclose one with a gaussian sphere, you find an electrical field. But shouldn't you consider the other point charge? In this case, we just dealt with one plate, and I questioned that what the other plate is for. Mar 8, 2015 at 16:38
• Notice that they say the electric field outside of the two plates is zero. In other words, they are already considering the effects of two plates. You can draw the same surface for a single plate in which case the electric field on both faces of the cylinder will be equal and opposite, and then use superposition to determine the field everywhere. Mar 8, 2015 at 16:45
• @user2694307 I suggest you to draw the figure with the plates. On this figure, I suggest you to draw the wide cylinder. One of the plane surfaces of the cylinder is inside one of the plates, right? Well, there, inside, there is no field. If it is not clear to you why, then ask it. The other plane surface of the cylinder is between the plates. Well, in the region between the plates you have the total field, from both plates, not only from the plate in which penetrates the cylinder. Mar 8, 2015 at 16:57