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Electric Field over a charged cylinder using only Coulomb's Law

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the electric field at a point on the central axis of a solid, uniformly charged cylinder of radis R and length h?

    2. Relevant equations

    Well, I've set up the triple integral and have gotten to this point:

    [itex]\bar{E}_{z}=\frac{2\pi\rho}{4\pi\epsilon_{0}}\int^{h/2}_{-h/2}[1-\frac{z'-z}{\sqrt{(z'-z)^{2}+R^{2}}}]dz[/itex] , z' is the location of the test point

    When I integrate this I get a constant term and a mess of logarithms. I know that the field should be 0 when z' = 0, but it doesn't check out. What's wrong?
     
  2. jcsd
  3. Oct 20, 2013 #2

    TSny

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    Re-evaluate how you got the 1 inside the brackets in the expression above. I suspect that you got this from assuming that ##\frac{z'-z}{\sqrt{(z'-z)^2}} = 1##. But, recall that ##\sqrt{x^2} = |x|##.
     
  4. Oct 20, 2013 #3
    Alright. Now that I've changed that term, it vanishes when integrated, but the second term in the integrand still turns into

    [itex]\frac{1}{2}ln| (z'-z) + \sqrt{(z'-z)^{2}+R^{2}} |^{h/2}_{-h/2}[/itex]

    Which, as far as I can tell, can't be reduced to 0 when z' = 0
     
  5. Oct 20, 2013 #4
    Nevermind, integrating this absolute value function is a bit more work than I thought it was.
     
  6. Oct 20, 2013 #5

    TSny

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    Also, I don't see how you are getting a logarithm expression in the result of the integration.
     
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