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Expanision or what?

  1. Oct 4, 2005 #1
    ok i got the famous electric field of a flat circular disk a distance z above the center of the the disk ok easy enough to find the E-field (btw the disk has uniform charge q and radius R)
    \vec{E} = \frac{q}{2 \pi \epsilon_0 R^2}(1- \frac{z}{\sqrt{r^2+z^2}})\vec{z}[/tex]

    however my question asks to show the E-field can be expressed as
    \vec{E} = \frac{q}{4 \pi \epsilon_0} [\frac{1}{z^2} - \frac{3R^2}{4z^4} + ...]\vec{z}[/tex]

    what expansion is this?
    Last edited: Oct 4, 2005
  2. jcsd
  3. Oct 4, 2005 #2

    let's look at this part:

    (R^2 + z^2)^\frac{-1}{2} = \frac{1}{z} (1 + \frac{R^2}{z^2})^\frac{-1}{2}.

    now do the binomial expansion on the part in the parantheses, and there you go.
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