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.
http://aerostudents.com/files/physics/solutionsManualPhysics/PSE4_ISM_Ch22.pdf (solution number 24)
http://www.phys.utk.edu/courses/Spring 2007/Physics231/chapter22.pdf (page 21)
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?
Best Answer
You can use the superposition principle, in general this is more difficult as that requires you to know the charge distribution on all the objects. Note that this is not always as simple as it may seem. E.g. you say "uncharged due to attraction between two plates", but this is not the correct reason, attraction alone would not yield this, it's the inverse square law which is equivalent to Gauss's law that is responsible for this.
In these sorts of problems, you need both the superposition principle and Gauss' law to draw conclusions about the surface charges. So, while you could have used the superposition principle to correctly calculate the electric field between the plates, you could not have concluded that the surface charges reside on the interior sides of both plates without invoking Gauss's law.