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Dipole Moment interaction

by raintrek
Tags: dipole, interaction, moment
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May24-08, 12:29 PM
P: 76
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:



2. Relevant equations


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!
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May24-08, 02:59 PM
P: 76
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]


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:


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!
May24-08, 04:26 PM
P: 76
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...

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