expanision or what?

**Phymath** · Oct4-05, 06:56 PM

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

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

what expansion is this?

**Brad Barker** · Oct4-05, 07:02 PM

Originally Posted by 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)
$LaTeX Code: \\vec{E} = \\frac{q}{2 \\pi \\epsilon_0 R^2}(1- \\frac{z}{\\sqrt{r^2+z^2}})\\vec{z}$

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

what expansion is this?

let's look at this part:

$LaTeX Code: (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.