1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Potential of a Spherical Shell of Charge

  1. Jan 9, 2006 #1
    Ok if you have a spherical shell of radius R with an even distribution of charge then outside the shell at a distance r where r>R I get that the shell can be treated as a point charge and inside the sphere (r<R) the electric potential will be constant.
    All my notes cover when the shell has no thickness and I was thinking what happen if the spherical shell did have thickness (say inner radius a and outer radius b)? When r is greater then b can the shell still be treated as a point charge? How about when a<r<b?
     
  2. jcsd
  3. Jan 9, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    If the charge is uniformly distributed throughout the (nonconducting) shell:

    (r > b) Treat as a point charge

    (r < a) potential will be constant

    (a < r < b) the field (and potential) will vary throughout this range; you'll need to integrate. (The field at a radius r depends only on the charge within that radius; the field is that of a point charge, but only the charge within r contributes to the field.)
     
  4. Jan 11, 2006 #3
    Thanks for just clearing that up for me. I've managed to find a question on this (I'm getting practise in before mid-year exams) however it deals in terms of charge density. Is it just as simple as finding Q in terms of the sphere (volume of shell at such and such density).
     
  5. Jan 11, 2006 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Right. You should be able to work with either total charge Q or with the charge density. As an exercise, you might want to find the field as a function of radius for a uniform ball of charge density [itex]\rho[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?