How Do You Calculate the Potential Energy Between Two Dipoles?

[raintrek]

Homework Statement

Consider two dipoles with moments u1 and u2 arranged as in the following diagram. Each dipole is depicted as two charges of equal magnitude separated by a distance d. The centre-to-centre separation of the two dipoles is the distance r. The line joining the two dipole centres makes an angle theta with the lower dipole (ie. q1 and -q1). Derive an expression in terms of u1, u2, theta and r which describes the potential energy of interaction of these two dipoles which is valid when d<<r. In the spirit of the hint below, your answer should not consider any (d/r)^n terms where n is greater than 2:

http://ds9.trekcore.com/dipole.JPG

Hint:

\frac{1}{\sqrt{1-ax}}\approx1+\frac{1}{2}ax+\frac{3}{8}a^{2}x^{2}

Homework Equations

U(r)=\frac{kQQ}{r}

The Attempt at a Solution

I've been trying to solve this for the past hour without any luck. It centers around getting an expression for the separation between q1 and -q2, and -q1 and q2. I'm fairly certain the expression should be from Pythagoras given the hint (ie, I need to take a square root of r at some point), but I can't find one which involves d/r as also specified in the hint. If anyone could offer any pointers, I'd be most appreciative. Thanks!

 

[raintrek]

OK, I've got a little bit further (this seems brutal!)

I've been able to see that the separations of the charges mentioned above are:

q1 & -q2: \sqrt{(d+rcos\theta)^{2}+(rsin\theta)^{2}}
-q1 & q2: \sqrt{(-d+rcos\theta)^{2}+(rsin\theta)^{2}}

Simplifying:

q1 & -q2: \sqrt{d^{2}+r^{2}+2rdcos\theta}
-q1 & q2: \sqrt{d^{2}+r^{2}-2rdcos\theta}

However this quickly makes the dipole-dipole interaction energy horrible:

U(r)=\frac{q_{1}q_{2}}{4\pi\epsilon_{0}}\left(\frac{2}{r}-\frac{1}{\sqrt{d^{2}+r^{2}+2rdcos\theta}}-\frac{1}{\sqrt{d^{2}+r^{2}-2rdcos\theta}}\right)

From that point I see no way to simplify the last two terms to get to a point where I can apply the Taylor expansion in the hint. I really am pulling my hair out over this now, if anyone can suggest anything I'd be ever grateful!

 

[raintrek]

LOL, ok probably talking to myself here. Still playing around with this, taken it further, although I'm pretty sure my final answer here is wrong...

http://voy.trekcore.com/working.jpg