# Neglecting terms (and binomial expansion)

• Jul 2nd 2010, 06:49 AM
Matnater
Neglecting terms (and binomial expansion)
From David J. Griffiths "Introduction to Electrodynamics 2. ed":
Quote:

We're interested in the regime $r\gg s$ (...), so the third term is necligible, and the binomial expansion yields
$(...)\cong\frac{1}{r}\left(1\mp\frac{s}{r}\cos\the ta\right)^{-1/2}\cong\frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta \right)$
"The third term" is further up the page.

What is really going on here? I've been trying to understand it, but I can't figure it out. How does he get $\frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta\right)$?

Edit/context:
This is a step in deriving the formula
$V(P)\cong\frac{1}{4\pi\epsilon_{0}}\frac{qs\cos\th eta}{r^{2}}$
$V_{dip}(P)=\frac{1}{4\pi\epsilon_{0}}\frac{\mathbf {P}\cdot\mathbf{\hat{r}}}{r^{2}}$
• Jul 2nd 2010, 08:29 AM
tonio
Quote:

Originally Posted by Matnater
From David J. Griffiths "Introduction to Electrodynamics 2. ed":

"The third term" is further up the page.

What is really going on here? I've been trying to understand it, but I can't figure it out. How does he get $\frac{1}{r}\left(1\pm\frac{s}{2r}\cos\theta\right)$?

Edit/context:
This is a step in deriving the formula
$V(P)\cong\frac{1}{4\pi\epsilon_{0}}\frac{qs\cos\th eta}{r^{2}}$
$V_{dip}(P)=\frac{1}{4\pi\epsilon_{0}}\frac{\mathbf {P}\cdot\mathbf{\hat{r}}}{r^{2}}$

You can expand the definition of the binomial theorem to non-integer exponents, defining $\binom {\alpha}{k}=\frac{\alpha(\alpha -1)(\alpha -2)\cdot\ldots\cdot (\alpha -k+1)}{k}\,,\,\,k\in\mathbb{N}$.

Thus, $\left(1\mp \frac{s}{r}\cos\theta\right)^{-1/2}$ $=1\mp \left(-\frac{1}{2}\right)\frac{s}{r}\cos\theta+\left(\fra c{3}{8}\right)\left(\frac{s}{r}\right)^2\cos^2\the ta+\ldots$ $\cong 1\pm\frac{s}{2r}\cos\theta$ , which apaprently is justified by the book since r is much larger than s thus from the third summand and on the coefficients are extremely low.

Tonio
• Jul 2nd 2010, 08:41 AM
Matnater
Thank you! This really solved my problem! :)