# Interaction between two dipoles

1. Nov 13, 2005

### mathlete

I've got to find the energy between two dipoles separated by a vector displacement r. I found what the energy/field of one such dipole is (see attachment), but I don't know how to combine the energy of two dipoles (P1 and P2) using those equations.. any help on a starting point?

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• ###### dipoleenergy.PNG
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2. Nov 13, 2005

### Physics Monkey

Sorry I can't see you attachement yet, so let me know if you're using different units or don't have these methods available yet.

You know the field of the two dipoles (via superposition), right? One way to proceed would be to calculate the energy of the system using
$$\frac{1}{8 \pi} \int E^2 \, d^3 x,$$
where $$\vec{E}$$ is the total electric field of the system. Be sure to throw away the self energy terms otherwise you will get a nonsense infinite answer. This integral is not as difficult as it might look (in fact it is completely trivial if you use the properties of the perfect dipoles). Hint: can you find a charge distribution that gives a perfect dipole field?

3. Nov 13, 2005

### mathlete

Sorry about that, you can see the image here:
http://img368.imageshack.us/img368/3945/dipoleenergy2fe.png

See my problem isn't finding the energy - i'm given the expression for that. So I can easily find the energy of a single dipole, but I don't know how to combine those expressions to account for two dipoles - do I add the electric fields and then plug into my energy equation? Or do I add the energies separately? Or do something different?

4. Nov 13, 2005

### Physics Monkey

Hey, no worries, it is the site that prevents the image from being visible until it's approved. The energy of the dipole in any field is given by your first equation. All you have to do is plug in the particular field of another dipole, your second equation, to find the interaction energy. You can easily check that you get the same answer if you consider the reversed situation i.e. the energy is symmetric between the two dipole moments. You certainly can't use the field of both dipoles because then you get nasty infinities, the self energies. Perhaps you're worried that there should be a factor of 2? Can you convince yourself that this isn't necessary? Think of the analogous situation between two point charges, is there a factor of 2 there?

5. Nov 13, 2005

### mathlete

Boy do I feel stupid now . I was making the problem a LOT more difficult than it had to be - I can't believe I didn't think of that immediately, I don't know what I was thinking. Thanks for your help:tongue2: