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Electric field of a uniformly distributed rod

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data
    rod -a-------|---------a on x-axis, for x>-a has -Q charge, for X<a has +Q charge, find the electric field at point (x,0) x is positive


    2. Relevant equations
    lambda = Q/a linear charge density dQ = (lambda)(ds)
    E=KQ/r^2

    3. The attempt at a solution
    i make dE=kdQ/(x-s)^2
    integrate kdQ/(x-s)^2 between -a and a ( i can make it 0 to a and multiply by two, symmetry)
    i get the result 2KQ/a[1/x^2 - 1/(x-a)^2] for E sub-x

    now if i want to find x>>a i need o somehow change [1/x^2 - 1/(x-a)^2] to the 2nd taylor expansion term but how? And is my integration correct?
     
  2. jcsd
  3. Sep 25, 2007 #2

    Astronuc

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    Staff: Mentor

    I suggest checking the assumption about symmetry.
     
  4. Sep 25, 2007 #3

    learningphysics

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    Homework Helper

    I don't understand how you multiplied by 2... you have the integral:

    [tex]\int_{-a}^{0}\frac{k\sigma ds}{(x-s)^2} + \int_{0}^{a}\frac{-k\sigma ds}{(x-s)^2}[/tex]

    where sigma is Q/a
     
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