Expansion of E-Field of a Uniformly Charged Flat Circular Disk

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SUMMARY

The electric field of a uniformly charged flat circular disk at a distance z above its center is given by the equation \vec{E} = \frac{q}{2 \pi \epsilon_0 R^2}(1- \frac{z}{\sqrt{r^2+z^2}})\vec{z}. To express this in the form \vec{E} = \frac{q}{4 \pi \epsilon_0} [\frac{1}{z^2} - \frac{3R^2}{4z^4} + ...]\vec{z}, a binomial expansion is applied to (R^2 + z^2)^{-\frac{1}{2}}. This involves rewriting the term as \frac{1}{z} (1 + \frac{R^2}{z^2})^{-\frac{1}{2}} and performing the expansion.

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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?
 
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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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