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Math Help - Neglecting terms (and binomial expansion)

  1. #1
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    Neglecting terms (and binomial expansion)

    From David J. Griffiths "Introduction to Electrodynamics 2. ed":
    We're interested in the regime r\gg s (...), so the third term is necligible, and the binomial expansion yields
    (...)\cong\frac{1}{r}\left(1\mp\frac{s}{r}\cos\the  ta\right)^{-1/2}\cong\frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta  \right)
    "The third term" is further up the page.

    What is really going on here? I've been trying to understand it, but I can't figure it out. How does he get \frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta\right)?

    Edit/context:
    This is a step in deriving the formula
    V(P)\cong\frac{1}{4\pi\epsilon_{0}}\frac{qs\cos\th  eta}{r^{2}}
    which in turn leads to
    V_{dip}(P)=\frac{1}{4\pi\epsilon_{0}}\frac{\mathbf  {P}\cdot\mathbf{\hat{r}}}{r^{2}}
    Last edited by Matnater; July 2nd 2010 at 07:04 AM. Reason: Adding some context to the problem
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  2. #2
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    Quote Originally Posted by Matnater View Post
    From David J. Griffiths "Introduction to Electrodynamics 2. ed":


    "The third term" is further up the page.

    What is really going on here? I've been trying to understand it, but I can't figure it out. How does he get \frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta\right)?

    Edit/context:
    This is a step in deriving the formula
    V(P)\cong\frac{1}{4\pi\epsilon_{0}}\frac{qs\cos\th  eta}{r^{2}}
    which in turn leads to
    V_{dip}(P)=\frac{1}{4\pi\epsilon_{0}}\frac{\mathbf  {P}\cdot\mathbf{\hat{r}}}{r^{2}}

    You can expand the definition of the binomial theorem to non-integer exponents, defining \binom {\alpha}{k}=\frac{\alpha(\alpha -1)(\alpha -2)\cdot\ldots\cdot (\alpha -k+1)}{k}\,,\,\,k\in\mathbb{N}.

    Thus, \left(1\mp \frac{s}{r}\cos\theta\right)^{-1/2} =1\mp \left(-\frac{1}{2}\right)\frac{s}{r}\cos\theta+\left(\fra  c{3}{8}\right)\left(\frac{s}{r}\right)^2\cos^2\the  ta+\ldots \cong 1\pm\frac{s}{2r}\cos\theta , which apaprently is justified by the book since r is much larger than s thus from the third summand and on the coefficients are extremely low.

    Tonio
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  3. #3
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    Thank you! This really solved my problem!
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