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Gauss's law for a charged ring

  1. Oct 5, 2015 #1
    Hello physics world,

    I am having a hard time understanding a rather simple thing. Let's consider the electric field produced by a uniformly charged ring of radius R, at a position $z$ along the ring's axis. From Coulomb's law in every textbook, we know that E_z∝Qz/(R^2+z^2). That is, there is a net field produced by the ring along its axis.

    Now, if we consider the same problem with Gauss's law, I run into a conceptual problem. Let's say we take a Gaussian sphere with radius r, smaller than R, whose center is placed at the center of the ring. In that case, there is no charge enclosed within the Gaussian surface. So Gauss's law should say that the electric field must be zero. But we know that the electric field is finite along $z$ and does not cancel out from $z$ and $-z$. Instead it seems to me that the flux would be 2E_z.

    There must be something wrong with my understanding of Gauss's Law, because as I understand it, if there is not charge inside the Gaussian sphere, the electric field flux through that sphere is 0. Any help clarifying this would be sincerely appreciated.

    (I know that using Gauss's law for this problem is not a good idea because the field will not be constant in the surface integral. My question is thus rather conceptual.)
     
  2. jcsd
  3. Oct 5, 2015 #2

    Meir Achuz

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    Gauss's law applies to the surface integral of E, not E at every point.
    It can only give E at each point if there is enough symmetry to say that E is constant on the surface.
    Although E is not zero within your sphere, its integral over the surface of the sphere is zero.
     
  4. Oct 5, 2015 #3

    jtbell

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    Or more visually, if you can imagine what the field lines look like for this situation: over some parts of your Gaussian sphere, the field lines enter the sphere, but over other parts, they leave the sphere. Field lines cannot simply "disappear into thin air", and there is no charge inside the sphere, so any field line that enters the sphere must leave the sphere somewhere else.
     
  5. Oct 5, 2015 #4
    Excellent responses!! Thank you very much for clarifying the situation.
     
  6. Oct 12, 2015 #5
    Conceptually or otherwise, the point is that: Gauss'slaw says that if the net charge is zero inside a Gaussian surface, the total flux over the surface is zero. It does not say that the electric field is zero. Generally, Gauss's law is always true, but as a calculational tool, it has limited usefulness. You must match the symmetry of the field, with the symmetry of the Gaussian surface, if you need to calculate the field. In your case, the field of the ring has a very different symmetry compared to the symmetry of the spherical Gaussian surface you chose.
     
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