Approximating the force on a dipole Taylor series

[Zack K]

Homework Statement

Show that the magnitude of the net force exerted on one dipole by the other dipole is given approximately by:$$F_{net}≈\frac {6q^2s^2k} {r^4}$$
for $r\gg s$, where r is the distance from one dipole to the other dipole, s is the distance across one dipole. (Both dipoles are of equal length and both have charges of magnitude q).

Homework Equations

$F=\frac {kq_1q_2} {r^2}$
$f(x)=\sum_{n=0}^\infty \frac {f^{(n)}(0)} {n!} x^n$

The Attempt at a Solution

I worked out the net force that a dipole would be acting on another as: $$F_{net}=kq^2(\frac {1} {(r-s)^2}+\frac {1} {(r+s)^2}-\frac {2} {r^2})$$ This equation is right because I plugged in values for r and s given r is much greater than s, and got the same value for if I used the approximated equation at the top.

I just lack the knowledge of using a taylor series to approximate my equation into the desired one.

 

[haruspex]

I just lack the knowledge of using a taylor series to approximate my equation into the desired one.

First step is to get everything inside parentheses into the form 1±α, by suitable division, where α<<1.

 

[Zack K]

First step is to get everything inside parentheses into the form 1±α, by suitable division, where α<<1.

Ok, so then I get $$F_{net}=kq^2(\frac {1} {(1-α)^2}+\frac {1} {(1+α)^2}-\frac {2} {r^2})$$ Do I then expand both the parentheses into polynomials?

 

[haruspex]

Ok, so then I get $$F_{net}=kq^2(\frac {1} {(1-α)^2}+\frac {1} {(1+α)^2}-\frac {2} {r^2})$$ Do I then expand both the parentheses into polynomials?

I wrote "get it into the form, by suitable division", not "arbitrarily change it". It still has to follow from the equation you wrote in post #1.
(Or did you make a mistake in typing out the post?)