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

[Phymath]

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)
$$\vec{E} = \frac{q}{2 \pi \epsilon_0 R^2}(1- \frac{z}{\sqrt{r^2+z^2}})\vec{z}$$

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

what expansion is this?

 

[Brad Barker]

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)
$$\vec{E} = \frac{q}{2 \pi \epsilon_0 R^2}(1- \frac{z}{\sqrt{r^2+z^2}})\vec{z}$$

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

what expansion is this?

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.