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.

  • $\begingroup$ Without your calculations, it is hard to say what went wrong where. You are correct that the electric field inside of a ring is non-zero. However, I (strongly) suspect that this is not in violation of Gauss's law. Gauss's law has some subtleties when you don't draw a 3D surface. See this: physics.stackexchange.com/q/44515 $\endgroup$ – user363165 Sep 3 '16 at 3:45
  • $\begingroup$ Gauss' law = Coulomb's law; check this answer to know more. $\endgroup$ – Yashas Mar 7 '17 at 14:28

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.

  • $\begingroup$ " 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" My Gauss' Law calculation was simply to enclose the inside of the ring with a sphere of radii very close to the radii of the ring. Thanks for your insight anyways! It's interesting to think about your argument, and it does sort of make sense, although a more robust answer would be better. Any books you'd recommend on the subject? $\endgroup$ – zickens Sep 3 '16 at 18:56
  • $\begingroup$ @zickens I don't quite understand - if the sphere is centered at the center of the ring and has a radius smaller than the ring's radius, then it doesn't enclose any charge, so doesn't Gauss's law just give that the surface integral of the electric field over the sphere is zero? Am I misunderstanding your setup? $\endgroup$ – tparker Sep 3 '16 at 21:51
  • $\begingroup$ That's exactly what puzzles me. Gauss' Law encloses no charge, therefore the electric field results zero. With Coulomb's Law, however, we get that the electric field is non-zero. This difference in result could very well be a mistake in my reasoning, since I've only gone through an introductory course. But still, it puzzles me, and this difference in particular is what motivates my question. $\endgroup$ – zickens Sep 4 '16 at 0:16
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    $\begingroup$ @zickens As, I understand the flaw in your reasoning now. Gauss's law tells you that the surface integral of the electric field over the entire sphere is zero, but that doesn't mean that the electric field is zero everywhere on the sphere. I'm not sure how familiar you are with surface integrals, but the idea is that the electric field makes a positive contribution to the integral when it points outside the sphere and a negative contribution when it points inside the sphere. In your case, the electric field points inward along the sphere's "equator," and outward along its "poles." ... $\endgroup$ – tparker Sep 4 '16 at 0:32
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    $\begingroup$ Gotcha! Read up on the integral version of Gauss's Law and now it makes total sense. E and dS are not parallel for every point on the gaussian surface. Thanks! $\endgroup$ – zickens Sep 5 '16 at 7:16

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.

  • $\begingroup$ "because the opposite bits of the ring hold charge in proportion to distance, NOT the square of the distance" Could you explain this a bit more? It seems to me that the fact that a spherical shell has area has nothing to do with a change in the nature of the electrical field. $\endgroup$ – zickens Sep 3 '16 at 18:52
  • $\begingroup$ There is no workable argument that says a uniformly charged ring has zero field 'inside' the ring. In three dimensions, though, you CAN make an argument for a cylinder. The scaling is part of the argument, but in the ring case it doesn't produce a cancellation against the inverse-square-distance. $\endgroup$ – Whit3rd Sep 4 '16 at 5:20
  • $\begingroup$ In an extreme case, if one chooses a test point very close to the ring, it will have a very similar field to an infinite line of charge (just curved, if you follow it long enough). So, whether inside or outside, next to the ring you should expect an E-field that falls off as 1/r. At sufficiently small distances from the ring, the distant parts are too far away to matter. $\endgroup$ – Whit3rd Sep 5 '16 at 7:13

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.


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