We have a charged sphere with charge Q and the charge is uniformly distributed with a charge density ρ. The sphere has a radius R.

If we construct a Gaussian Surface with radius r, with r < R. If we apply Gauss's law here, we would find the electric field at radius r only depends on how much charge is enclosed in this Gaussian Surface. But how can this be conceptually? How can the charge outside of this Gaussian Surface not contribute any electric field to a point on the Gaussian Surface?

I understand that Gauss's law determines the answer, but any conceptual explanation would be much appreciated.


2 Answers 2


Gauss’ Law says that the electric flux through the sphere due to charges outside is zero. That doesn’t mean that the field on the sphere due to charges outside is zero. It means that that the field enters the sphere in some places and exits in others, in such a way that the total flux over the whole sphere is zero.

Addendum: I think I misunderstood your question. I thought you were asking about charges outside the sphere. You were actually asking why the charges on the sphere don’t create a field inside. This is because all the charged area elements do cause field contributions, but those field contributions add up to zero. This is obvious at the center, but true anywhere inside. The field contribution due to an element of charge on one side of the sphere gets balanced out by the field contribution due to an element of charge on the other side. This is the interior part of the shell theorem.

  • $\begingroup$ I understand your point. But by using spherical symmetry in this case, we can get the electric field with r < R and this electric field does not depend on the charge outside. How can this be? $\endgroup$
    – Yiyang Zhi
    Aug 6, 2019 at 1:13
  • $\begingroup$ I’ve amended my answer, since I was confused about what you were asking. $\endgroup$
    – G. Smith
    Aug 6, 2019 at 1:53

Your calculation assumes spherical symmetry at the surface. That happened when you assumed that the electric field was perpendicular and had the same value at each point.

If that’s not true because of some external charges not spherically arranges, then your calculation doesn’t give the correct field: the real field might not be symmetrical.

If the external charges are arranged in a spherically symmetric way, your calculation is still right, though, and the field is still symmetric.


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