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Homework Help: Electrical Potential of a Uniformly Charged Ring

  1. Feb 16, 2006 #1
    I am stuck on this problem:

    I started this problem by looking at the electric potential of a ring, which is:

    So then if it varies in the thickness of the ring, would it be reasonable to have it be:

    Or would I need to use the equation for the electric potential of a uniformly charged disk with radius R_out, then subtract the electric potential of the inner disk of radius R_in.

    Thanks for your help!

    P.S. Hope I did the LaTeX right
    Last edited: Feb 16, 2006
  2. jcsd
  3. Feb 16, 2006 #2


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    No, that is not correct. You can't just substitute R_in and R_out.
    One way of doing this, is finding the contribution of a small ring and integrating to get the contribution of the disc.
  4. Feb 17, 2006 #3
    Would it be something more like this?:


    I'm not sure if this is the right integral for the equation?
    Last edited: Feb 17, 2006
  5. Feb 18, 2006 #4


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    No, that's not right.

    First, consider a small ring of thickness [itex]dr[/itex] with a radius of [itex]r[/itex].

    Now, what is the potential due to this ring at point at a height 'z' at the axis? It is,


    where, dQ is the small charge on the ring. How do you find this dQ?
    You can find the uniform surface charge density [itex] \sigma [/itex] from the question right? (Hint:Total charge/Total area). Since the ring has a small thickness [itex]dr[/itex], the small area will be [itex] 2 \pi r dr [/itex]. From this, you can find the charge "dQ" on the small ring.

    How do you find the potential of the disc? Notice that, if you sum the contributions of such small rings, you will get the potential of the disc. So you integrate the contribution the small ring where the radius varies from [itex]r_{in}[/itex] to [itex]r_{out}[/itex]

    Can you put this in mathematical equations and post the answer here?
    Last edited: Feb 18, 2006
  6. Feb 19, 2006 #5
    Alright, I see how it works now, thanks so much for the help. The final solution is:

    Last edited: Feb 19, 2006
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