1. The problem statement, all variables and given/known data 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: Hint: [tex]\frac{1}{\sqrt{1-ax}}\approx1+\frac{1}{2}ax+\frac{3}{8}a^{2}x^{2}[/tex] 2. Relevant equations [tex]U(r)=\frac{kQQ}{r}[/tex] 3. 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!
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: [tex]\sqrt{(d+rcos\theta)^{2}+(rsin\theta)^{2}}[/tex] -q1 & q2: [tex]\sqrt{(-d+rcos\theta)^{2}+(rsin\theta)^{2}}[/tex] Simplifying: q1 & -q2: [tex]\sqrt{d^{2}+r^{2}+2rdcos\theta}[/tex] -q1 & q2: [tex]\sqrt{d^{2}+r^{2}-2rdcos\theta}[/tex] However this quickly makes the dipole-dipole interaction energy horrible: [tex]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)[/tex] 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!
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...