Expanision or what?

  • Thread starter Phymath
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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)
[tex]
\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
[tex]
\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:
Phymath said:
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)
[tex]
\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
[tex]
\vec{E} = \frac{q}{4 \pi \epsilon_0} [\frac{1}{z^2} - \frac{3R^2}{4z^4} + ...]\vec{z}[/tex]

what expansion is this?

let's look at this part:

[tex]
(R^2 + z^2)^\frac{-1}{2} = \frac{1}{z} (1 + \frac{R^2}{z^2})^\frac{-1}{2}.
[/tex]

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

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