How to Calculate the Energy Between Two Dipoles?

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SUMMARY

The discussion focuses on calculating the energy between two dipoles, P1 and P2, separated by a vector displacement r. The key method involves using the integral formula 1/(8π) ∫ E² d³x to determine the total electric field, E, of the system while avoiding self-energy terms to prevent infinite results. Participants clarify that the interaction energy can be derived by substituting the electric field of one dipole into the energy equation of the other, confirming that the energy is symmetric between the two dipoles.

PREREQUISITES
  • Understanding of dipole moments and their electric fields
  • Familiarity with integral calculus, particularly in three dimensions
  • Knowledge of superposition principle in electromagnetism
  • Experience with energy calculations in electric fields
NEXT STEPS
  • Study the derivation of the energy of a dipole in an electric field
  • Learn about the superposition principle in electromagnetic fields
  • Explore the concept of self-energy in dipole systems
  • Investigate charge distributions that yield perfect dipole fields
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Physicists, electrical engineers, and students studying electromagnetism who are interested in dipole interactions and energy calculations.

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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? :confused:
 

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    dipoleenergy.PNG
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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
<br /> \frac{1}{8 \pi} \int E^2 \, d^3 x,<br />
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?
 
Physics Monkey said:
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
<br /> \frac{1}{8 \pi} \int E^2 \, d^3 x,<br />
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?
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?
 
Last edited by a moderator:
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?
 
Physics Monkey said:
All you have to do is plug in the particular field of another dipole, your second equation, to find the interaction energy.
Boy do I feel stupid now :eek: . 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:-p
 

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