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Potential due to a line charge

  1. Sep 14, 2009 #1
    Suppose you have a line charge Q along the x axis from +a to -a. Simple integration should give the potential at any point x as:

    phi(x) = 1/(4pi epsilon) Q/L loge((x+a)/(x-a))

    = 1/(4pi epsilon) Q/L loge((x+xdx)/(x-xdx))

    = 1/(4pi epsilon) Q/L loge((1+dx)/(1-dx))

    So there's no longer any x dependance??????

    I now doubt that the expression is right, even though it seems to be a standard result. Perhaps I've missed out the integration constant
     
  2. jcsd
  3. Sep 15, 2009 #2

    Born2bwire

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    I don't understand what you are doing here, what happened to a and why do you have dx?
     
  4. Sep 15, 2009 #3
    a = x dx. I could rewrite it as a = x b if it makes things clearer. After thinking about it further, the problem is that the charge density is now rho = Q/2a = Q/xdx and so I can no longer take it outside of the integral as a constant.
     
  5. Sep 15, 2009 #4
    If rho is Q/2a then this means the charge density is constant. You need to integrate dq/r where r is the distance from x to the infinitesimal charge dq. dq = (Q/2a)dx' and r = x-x' so your variable of integration is x' not x.
     
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