Why is the Electric Field inside a hollow sphere zero but not for a ring? I have this simple yet intriguing question that struck me through an introductory Electromagnetism. Being familiar with Gauss' Law and Coulomb's Law, the "Charged Spherical Shell" excercise is something very common to me. A particular thing for this exercise is that if you use any method (at least as far as I know), be it a simple Gaussian or a die-hard spherical coord. Coulomb Law integration, you get the same result. Zero electrical field at any point inside the shell. But one day I thought: "What about charged rings?". 
To my surprise, and if my calculations aren't mistaken, the electrical field inside the ring was zero only in the center, but non-zero everywhere else (the ring's plane). An even more intriguing fact was that now Gauss' Law and Coulomb's Law disagreed for this particular case. I took this question to a professor and he said he didn't know what to tell me, but he did state that I should trust on the Coulomb's Law result, this being an experimental law.
Why is this so? Couldn't I make a "spherical" arrangement of charged rings to make a charged spherical shell? What am I missing? Please do remember I've only gone through an introductory course.
 A: For any point inside a uniformly charged sphere, the sum over all the
sphere's surface results in a zero electric field.   This is because
one can make a symmetry argument, that each force from one tiny bit of the
charged area of the sphere is balanced by a projection of that area through
the point of interest, onto an area on the opposite side of the sphere.
The areas will be in proportion to the square of their distance from
the point, so the pull and push forces (proportional to charge/R**2)
are in balance.   After integrating over the entire sphere, you must get
zero net field.
For a ring, the similar symmetry argument does not hold, because the 
opposite bits of the ring hold charge in proportion to distance, NOT
the square of the distance.   The most-nearly-similar symmetry 
situation, an infinite cylinder uniformly charged, does also get a zero internal field.
Gauss' law gives the same result as the symmetry argument for sphere
and for cylinder.  I don't know how to apply it to a ring.
A: You are correct that the electric field is only zero at the exact center of the ring.  Gauss's law and Coulomb's law always both give the same results in every situation, so if you're getting that they disagree, then you made a mistake.  You almost certainly messed up in the Gauss's law calculation, because I can't think of any way to use Gauss's law to calculate the electric field in this particular situation.
This isn't anywhere close to a rigorous argument, but here's one way to motivate why it's more natural to expect that the electric field might be zero everywhere inside of a sphere than a ring: a sphere cleanly separates space into two different regions, the "inside" and the "outside," so it seems reasonable that the electric field might be qualitatively different in the two regions.  A ring does not, so it would be kind of weird if the electric field vanished everywhere in the plane of the ring and inside it - if you move a tiny bit out of the plane of the ring, are you still "inside" it?  Where exactly is the boundary of the "inside" of a ring?  You would somehow need the field to be identically zero everywhere in the plane of the ring, but start varying wildly in magnitude and direction as soon as you went a tiny bit out of the plane.  That would be weird.
A: Given a sphere or ring with charges on the surface or perimeter but no interior charges, each interior point is on the surface of a sphere of smaller radius concentric with the original sphere or ring. The usual proof with Gauss that $E=0$ inside the sphere, or equivalently that $=0$ across the surface of the interior sphere, relies on the problem's spherical symmetry.
Since a ring lacks spherical symmetry, we cannot prove a spherically symmetric electric field on the surface of such an interior sphere (which extends beyond the ring's plane, but intersects said plane with a circle enclosed in the ring). Nor does the 2D symmetry of the problem prove a 2D symmetry around the "Equator" of such a sphere that would allow a proof the field vanishes on said Equator.
The in-the-ring calculation is discussed here using Coulomb.
