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I What is the electric field just at the shell?

  1. Oct 25, 2015 #1
    Here's the question I am thinking about:
    If there is a spherical shell with uniform charge on it, I know the electric field inside the shell is zero and that outside the shell is calculated just the same with a single charge at the center of the sphere, but how about the field just at the shell (r=R)?
    I try to solve this problem by Gauss' Law. I can draw a gaussian surface just above and below the shell, but when I try to draw just at the shell, how many charges does the surface enclose? The charges are assumed to be infinitely small. When they are just at the gaussian surface, are they enclosed?
     
  2. jcsd
  3. Oct 25, 2015 #2

    jbriggs444

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    Undefined. There is a discontinuity in the field strength right at the shell.

    Physically, this is no problem because there is no such thing as a uniformly charged, infinitely thin spherical shell. Charge is discrete, not uniform. In addition, quantum mechanics imposes limitations on the extent to which charge can be localized to such a sharply defined shape.
     
  4. Oct 25, 2015 #3

    jtbell

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    A zero-thickness shell is an idealization. In practice any shell has a finite thickness, across which the electric field varies continuously from one side to the other. It's a common exercise to calculate the electric field for a charged spherical shell of finite thickness, inside the shell (r < a), within the thickness of the shell (a ≤ r ≤ b), and outside the shell (r > b).

    Even with a conducting shell, for which all the charge resides on the surface in electrostatics, the charge layer has a small but nonzero thickness, although we usually idealize it as zero thickness. In that case the electric field is described by a step function.
     
    Last edited: Oct 25, 2015
  5. Oct 25, 2015 #4
    Thank you for all your replies! I understand now.:smile:
     
  6. Aug 6, 2016 #5
    And can you tell me exactly how to fild field in case of a conducting shell of some thickness. specially the a<r<b case
    thanks
     
  7. Aug 6, 2016 #6

    jtbell

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    Staff: Mentor

    Use the integral form of Gauss's law, similarly to any other spherically symmetric charge distribution. Take care to find the "enclosed charge" appropriately for r in the various regions: r < a, a < r < b, and r > b.

    If you know Gauss's law but have trouble with the details, post your attempt in our Introductory Physics Homework forum and someone will probably be able to help.
     
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