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Question on Electric Potential Energy

  1. Feb 9, 2006 #1
    I can't get this question, can someone give me a hand? I will reprint the question below as it appears in the text:


    A Solid sphere of radius R has a uniform charge density p(roh) and a total charge Q. Derive an expression for the total electric potential energy.(Suggestion: imagine that the sphere is constructed by adding successive layers of concentric shells of charge dq=[4(pi)r^2 dr)p and use dU=V dq)


    I tried it a few times but I couldn't come up with a reasonable answer. Any help would be greatly appreciated. Thanks.
     
  2. jcsd
  3. Feb 9, 2006 #2

    Chi Meson

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    Start with an initial point charge. The work required to bring that charge to that region of space is zero. Now bring in a new charged "shell" of thickness dr. How much work is required to put this shell of radius "r" and thickness "dr" outside the charge that is already there?

    try changing the "dq" statment you're given into a "dr" statement: what's the volume of that shell?
     
    Last edited: Feb 9, 2006
  4. Feb 9, 2006 #3
    Ohhhh, that makes a lot of sense and I wasn't looking at it that way. I will go try that and report back with the result. Thank you kindly.
     
  5. Feb 9, 2006 #4
    O.K. I tried that for a white, but I am still getting nowhere . Do I have to integrate dV first then integrate it again to get U?

    For the dr statement you were talking about, I get either dr=dq/((4 PI r^2)p), or I can get dr=(r^2)/3 if I play around with dq=p dV. Can you offer me any more help? Thanks a lot!
     
  6. Feb 9, 2006 #5

    Chi Meson

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    Sorry, I was out.

    Let's go straight to the source for EPE:

    U=(kQq)/r [Q= one charge, q = other charge, k = 1/4pi epsilon]

    You are integrating U from 0 to R.
    For each successive shell of thickness dr, you are bringing in charge q "from infinity" to the previous volume of point charge Q.

    the shell with thickness dr will have charge "dq" =(4 PI r^2)p dr [this is essentiall what you had.

    dU will then be [(kQ)/r]dq
    Q will be the volume of the sphere (that's already there when dq arrives) times charge density. Radius of the sphere and the shell are essentially "r"

    you're going to get an integral that ends in "r^4 dr"
     
  7. Feb 10, 2006 #6
    Aha! Now I see! Thank you very, very much. What a tricky question.

    :)
     
  8. Jul 21, 2010 #7
    lol Could you explain this one more time , sub in the volumex charge density for Q , then sub in the dq definition , which gives me a (4*p^2 *pi/3Eo) int r^4 dr , i know im messing up somewhere , i just cant wrap my head around where.

    thx for any help :)
     
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