# Math Help - Partial fractions of an electric field

1. ## Partial fractions of an electric field

Hello. I've posted an imageshack link to this question because I haven't figured out how to enter in math fonts.

http://img193.imageshack.us/img193/6...question20.jpg

In this question q and d are constants representing charge and the distance between charges, but z is a variable. I've been trying to use partial fractions after simplifying the equation. I wound up with a constant multiplied by z/[(4z^2-d^2)^2], and using partial fractions I would up with z = (Az + B)(4z^2 - d^2) + Cz + E. Trying to solve for that has lead me nowhere though. Am I at least going in the right direction, or is there some other way to solve this that I'm entirely missing?

(I posted this in the math forum because, despite this coming from a physics text, it doesn't require any actual physics)

2. Originally Posted by bnay
Hello. I've posted an imageshack link to this question because I haven't figured out how to enter in math fonts.

http://img193.imageshack.us/img193/6...question20.jpg

In this question q and d are constants representing charge and the distance between charges, but z is a variable. I've been trying to use partial fractions after simplifying the equation. I wound up with a constant multiplied by z/[(4z^2-d^2)^2], and using partial fractions I would up with z = (Az + B)(4z^2 - d^2) + Cz + E. Trying to solve for that has lead me nowhere though. Am I at least going in the right direction, or is there some other way to solve this that I'm entirely missing?

(I posted this in the math forum because, despite this coming from a physics text, it doesn't require any actual physics)
You have:

$E=\frac{Qd}{2\pi \varepsilon_0z^3}\left[ \frac{1}{(1+(\frac{d}{2z})^2)^2}\right]$

So you need the first two terms of the binomial expansion of what is in the square brackets:

$\frac{1}{(1+(\frac{d}{2z})^2)^2}=1-2\left( \frac{d}{2z} \right)^2+..$

(this is the form valid for large $z$)

CB