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Approximation of electric field of uniform charged disk

  1. May 30, 2008 #1
    Hi,

    1. The problem statement, all variables and given/known data
    The electric field of a uniform charged disk at a point on its axis at a distance x from the disk is given by
    [tex]E = 2k_e\pi\sigma(1-\frac{x}{\sqrt{x^2+R^2}})[/tex]
    where R the radius of the disk and [tex]\sigma[/tex] the surface charge density.
    In my notes it says that when [tex]x\gg R[/tex], that is when the distance x to the disk is much bigger than the radius of the disk, then
    [tex]E\approx k_e\frac{Q}{x^2}[/tex]
    with the Q the total charge on the disk. How do they come to that result?

    2. Relevant equations

    3. The attempt at a solution
    When [tex]x\gg R[/tex], then [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}=x[/tex]. But then I get that [tex]E=0[/tex], which is not obviously not correct.
    I guess the approximation that [tex]\sqrt{x^2+R^2}\approx\sqrt{x^2}[/tex] is wrong. But how is the approximation then?

    Thank you.
     
  2. jcsd
  3. May 30, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi yoran! :smile:

    Always divide so that you get a "1" at the beginning:

    √(x²/(x² + R²)) = 1/√(1 + R²/x²) ≈ 1 - R²/2x². :smile:
     
  4. May 30, 2008 #3
    Also Yoran your mistake was that basically you did a 0th order expansion!

    If you are not comfy with binomial expansions, you can always MacLaurin expand the function in R/x. You are still using the same thing though-- expanding in R/x instead of x because you want a variable that is small i.e. much less than one (so you can truncate the expansion and still have it legitimately approximate the original function), which R/x satisfies, while just straight up x is very large.
     
  5. May 31, 2008 #4
    Allright then thanks a lot!
     
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