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Potential of charged disk for x R

  1. Jul 25, 2008 #1
    I understand the answer to this part of the problem intuitively but there's definitely something in the math that I'm missing--

    1. The problem statement, all variables and given/known data
    Along the axis through the center of a charged disk, for a point P at a distance x from the center, the potential will be

    [tex]\frac{Q}{2\pi\epsilon_{0}R^{2}}\left[\sqrt{R^{2}+x^{2}}-x\right][/tex]

    2. Relevant equations
    For x>>R, I'm supposed to understand that this reduces to

    [tex]V\approx\frac{Q}{2\pi\epsilon_{0}R^{2}}\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right]=\frac{Q}{4\pi\epsilon_{0}x}[/tex]

    3. The attempt at a solution
    As I said, I understand that this must work out, but I'm misunderstanding something about the math. It seems as though the term under the radical above ought to become just [tex]x^{2}[/tex] for x>>R, and thus inside the brackets x-x=0 and then V=0, which is untrue. Could someone please explain how [tex]\left[\sqrt{R^{2}+x^{2}}-x\right][/tex] becomes [tex]\left[x\left(1+\frac{1}{2}\frac{R^{2}}{x^2}\right)-x\right][/tex] ?
     
  2. jcsd
  3. Jul 25, 2008 #2
    Bring the x^2 term outside the radical, then apply the general binomial expansion formula for (1+y)^(1/2) = 1 + (1/2)y - ....
     
  4. Jul 25, 2008 #3
    Binomial approximation will get you there.

    [tex]\sqrt{R^2+x^2}=\sqrt{\frac{R^2 x^2}{x^2}+x^2}[/tex]

    Now use the approx
    [tex]x \sqrt{\frac{R^2}{x^2}+1}=x(1+\frac{1}{2} \frac{R^2}{x^2})[/tex]
     
  5. Jul 25, 2008 #4
    *thumps forehead* thanks.
     
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