Suppose we have a system of two parallel conducting plates charges $Q$ and $-Q$ and charge densities $\sigma$ and $-\sigma$. In this case, to find the electric field, Gauss' Law is used with a Gaussian Surface being a cylinder going from the interior of one plate to somewhere between the plates. Why is this result valid if the other plate isn't being included in the Gaussian surface?

  • $\begingroup$ Gauss' law does not require the entire system to be inside the Gaussian surface, however see the answers here for more about this problem. $\endgroup$
    – jacob1729
    Commented Oct 18, 2019 at 15:47

1 Answer 1


It is simply because we have no need of the other surface to find the electric field. Our basic need is to find the electric field between the plates, for which the inclusion of the second surface is not necessary.

You may have read in your reference material a list of things/rules/points that has to be taken care of before Gauss law is applied, they are listed below;

  • Gaussian surface should not pass through a discrete point charge.
  • Gauss theorem is mostly used to find the electric field of symmetric and extended charge distribution.
  • While applying Gauss law directions of the electric field is to be known beforehand.
  • The electric field term 'E' appearing in Gauss law is the electric contributed by all charges.

The above are listed to make the process of simplification much easier and to also showcase the fact wherein it is viable to apply Gauss law.

Reading the third point it is clear that as to why we do not need to include the second surface.

Also if we were to include the second surface in the cylinder the electric field lines would originate and terminate inside the cylinder itself and we would not get the necessary flux term in the Gaussian equation!

  • $\begingroup$ I think I got it, but suppose the negatively charged plate has a charge -2Q. Personally, I would find this with superposition, but wouldn't the result be the same as with a charge -Q using just one surface? $\endgroup$
    – Johnn.27
    Commented Oct 18, 2019 at 16:10
  • $\begingroup$ Remember I stated the rules to be followed, having -2Q disrupts symmetry of the system and using Gauss law here would lead to complications. $\endgroup$ Commented Oct 18, 2019 at 16:14
  • $\begingroup$ Alright, I think I got it, thanks! $\endgroup$
    – Johnn.27
    Commented Oct 18, 2019 at 18:13

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