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Evaluating a simple electrostatic field integral

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data
    I can find the e-field at point P.
    point p.JPG

    2. Relevant equations
    I get, easily enough, the correct integral solution (for the y-component, Ey - which is all I need to do):
    p4.JPG
    which I can see, informally, evaluates to:

    p3.JPG
    which is the correct answer.
    3. The attempt at a solution
    My question: is it, technically, sufficient for me to simply explain: "well, when evaluated, the infinity in the numerator cancels with sqrt(infinity^2) in the denominator? Or can I offer a more rigorous explanation?

    I tried L'Hôpital's Rule, but it doesn't seem to lead anywhere as far as showing anything definitive.

    Thank you - any hints are appreciated!
     

    Attached Files:

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  2. jcsd
  3. Feb 20, 2017 #2

    blue_leaf77

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    Try factoring out ##x^2## from the squareroot in the denominator, you should know how the resulting ##R^2/x^2## behaves when ##x \to \infty##.
     
  4. Feb 20, 2017 #3
    Hmmm (thanks!). Yes, R2/x2 disappears as x→∞, but that now leaves an un-cancelled x in the numerator that is going to infinity.
     
  5. Feb 20, 2017 #4

    rude man

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    You're not factoring the denominator correctly.
    (x2 + R2)1/2 = {x2(1 + R2/x2)}1/2
    = x(1 + R2/x2)1/2
    etc.
     
  6. Feb 20, 2017 #5
    Ach, yes. Sorry - and thank you!
     
  7. Feb 21, 2017 #6

    rude man

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    Kein Problem!
     
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