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Mutual energy of two dipoles.

  1. May 26, 2007 #1
    i need to show that the mutual energy between two dioples p1 and p2 (not necessarily parallel to eachother) is [tex] U=-\frac{p_1\cdot p_2}{|r|^3}-3\frac{(p_1\cdot r)(p_2\cdot r)}{|r|^5}[/tex]
    where r is the vector from p_1 to p_2. (the p's are moments of diople).

    i tried using this equation: [tex]U=\int dV \rho_2 * \phi_1 [/tex]
    and also this :[tex]\phi=\frac{p\cdot r}{|r|^3}[/tex]
    (phi is the potential and rho is the density).
    [tex]\rho_2=-1/4\pi\nabla^2\phi_2=-1/4\pi[\frac{1}{|r|^2}@/@r(r^2@\phi/@r)+\frac{1}{|r|^2*sin(\theta)}@/@\theta(sin(\theta)@\phi/@\theta)][/tex]
    where @ stands for peratial derivative, but i didnt get to the desired answer.
    any pointers?
     
  2. jcsd
  3. May 26, 2007 #2

    Pythagorean

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    What about

    [tex]U=-{m\cdot B}[/tex]

    where m is the vector for one of the dipoles in field B from the other dipole.
     
  4. May 26, 2007 #3

    Pythagorean

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    you'll have to dig the equation for the field (B) of dipole, too, I reckon.
     
  5. May 26, 2007 #4
    by B, you mean the magnetic field, well i havent learned it yet (i mean we havent touched it in class as of yet, i myself read it form purcell), i pretty much sure i don't need here to use B, perhaps something else?
     
  6. May 26, 2007 #5

    Pythagorean

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    So these are electric dipoles then? I guess p usually denotes electric dipole and m is for magnetic dipole.
     
  7. May 26, 2007 #6

    Pythagorean

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    anyway, same equation, just:

    U = -p.E

    instead of -m.B

    and dig up the E field for an electric dipole
     
  8. May 26, 2007 #7
    well for E i found already, shouldn't i prove that it equals p.E? or in other words how to derive it?
     
  9. May 26, 2007 #8

    Pythagorean

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    If you want to find a derivation for U = -p.E, you might also remember that potential Energy is

    the integral of F.dl

    and the Force from a dipole is

    F = (p.grad)E
     
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