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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?

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  • $\begingroup$ 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. $\endgroup$
    – Steeven
    Mar 8, 2015 at 16:33
  • $\begingroup$ @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.) $\endgroup$
    – Sofia
    Mar 8, 2015 at 16:36
  • $\begingroup$ 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. $\endgroup$
    – user2694307
    Mar 8, 2015 at 16:38
  • $\begingroup$ 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. $\endgroup$
    – MonkeysUncle
    Mar 8, 2015 at 16:45
  • $\begingroup$ @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. $\endgroup$
    – Sofia
    Mar 8, 2015 at 16:57

1 Answer 1

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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.

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  • $\begingroup$ The charge distribution on a very big flat metallic plate is bound to be uniform, and all of it on the surface of the plates. Also, between two infinite plates, the field is uniform and constant, it doesn't depend on the distance from one plate or the other, -dV/dz = const. $\endgroup$
    – Sofia
    Mar 8, 2015 at 17:17
  • $\begingroup$ The fact that the interior of a conductor is free from charges is a consequence of only Coulomb's law. Old experiments on the limit of the photon mass made used of that by measuring the electric field inside a cavity of a hollow conductor that was put in a strong electric field. The slightest deviation from Coulomb's law would cause a charge buildup in the interior of the conductor which then would cause an electric field inside the cavity. See e.g. here $\endgroup$
    – Count Iblis
    Mar 8, 2015 at 17:23
  • $\begingroup$ The problem of the user is not with the Coulomb law. $\endgroup$
    – Sofia
    Mar 8, 2015 at 17:26
  • $\begingroup$ "uncharged due to attraction between two plates" suggests otherwise. $\endgroup$
    – Count Iblis
    Mar 8, 2015 at 17:42

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