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Vector Equation for Force Between Electric Dipoles

  1. Oct 26, 2012 #1
    (All the variables are vectors, I just didn't feel like fumbling around with the LateX code to make them vectors. Its late and I'm tired a lazy!)
    1. The problem statement, all variables and given/known data
    (This is paraphrased)
    There are two dipoles with arbitrary direction to each other. You know the energy between the dipoles is..

    [tex]W_D=-p \cdot E[/tex]

    What is the force between them? [itex]F_{1,2}[/itex]

    2. Relevant equations
    Force:
    [tex]F=-\nabla W_D[/tex]

    Electric Field of a Dipole:
    [tex]E=\frac{1}{4\pi \epsilon_0 r^3}[3p \cdot a_r-p][/tex]

    a_r is the unit vector from p1 to p2.


    3. The attempt at a solution

    Using the following vector identity..
    [tex]\nabla (a \cdot b)=(a \cdot \nabla)b+(b \cdot \nabla)a+a×(\nabla × b)+b×(\nabla × a)[/tex]

    [tex]-\nabla (-p \cdot E)=\nabla (p \cdot E)=(p \cdot \nabla)E+(E \cdot \nabla)p+p×(\nabla × E)+E×(\nabla ×p)[/tex]

    So I know [itex] \nabla × E =0 [/itex] and [itex]E×(\nabla ×p)=(E \cdot p)\nabla-(E \cdot \nabla)p[/itex]

    So plugging these in, I get.

    [tex]F=(p \cdot \nabla)E+(E \cdot p)\nabla[/tex]

    The prof gives the answer as, [tex]F=(p \cdot \nabla)E[/tex]

    So that means [itex](E \cdot p)\nabla=0[/itex] But I cannot figure out why.
     
  2. jcsd
  3. Oct 26, 2012 #2

    gabbagabbahey

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    Homework Helper
    Gold Member

    Sounds French.:tongue2:

    A vector differential operator does not a vector make. You cannot use the familiar BAC-CAB identity when one of the operands is not really a vector. Instead, what would you expect the curl of a point dipole to be?
     
  4. Oct 26, 2012 #3
    Hmm... I think I would expect it to be zero. But I am not sure. :\
     
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