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Homework Help: Electric potential problem

  1. Feb 6, 2007 #1
    1. The problem statement, all variables and given/known data
    derive an expression for the electrical potential at a distance x measured along the axis from the centre of a circular ring of radius R on which a charge Q is uniformly distributed.hence derive an expression for the electric field strength at this point


    2. Relevant equations
    to be honest im not too sure where to go after:
    V= -⌠E.dl
    any guidance at all would be much appreciated!!!thanks:biggrin:


    3. The attempt at a solution
     
  2. jcsd
  3. Feb 6, 2007 #2
    It looks like a start, what if the "ring" were simply two charges, each of 1/2Q, and -r,0 and r,0 relative to a test charge at 0,x. What would that look like?
     
  4. Feb 6, 2007 #3
    This simply involves a lil integration.

    Q is distributed uniformly throughout the ring. This linear charge denstity on it would be lambda= Q/2pi*R.

    Let dV due to each infinitesimally small element (dl) on the ring = (1/4pi*epsilon)*(lambda*dl/(x^2+R^2)^1/2.

    Then just integrate from 0 to 2pi*R.
     
  5. Feb 6, 2007 #4
    oh i see, so its ok to treat them as 2 separate point charges and then sum the electric potentials at the end.
    doing this i got an expression:
    V= - Q/4*pi*ε(x^2 + R^2)^1/2

    and E= dV/dx
    E = - [Qx]/[4*pi*ε(x^2 + r^2)^3/2]
     
  6. Feb 6, 2007 #5
    no, what i meant was look at it first as 2 point charges, then 4, then an infinite number spread around the ring, but Mr 4 points the way I was hinting at in post above directly.
     
  7. Feb 6, 2007 #6
    Quite so. I think your logic would be more useful in some questions. Thanks for that denverdoc.
     
  8. Feb 6, 2007 #7
    Still getting the hang of helping without doing the work, ie trying to help posters conceptualize w/o telling them how to pursue directly. Sometimes I think I just add to the confusion:grumpy:
    J
     
  9. Feb 7, 2007 #8
    Nope. I get you loud and clear! Maybe its coz I'm just as confusing!
     
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