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Electric field of an electric dipole

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Since the electrostatic field is conservative, show that it is irrotational for an electric dipole, whose dipole momentum is [tex] p [/tex].


    2. Relevant equations
    [tex] \nabla \times \mathbf{E} = 0 [/tex]


    3. The attempt at a solution
    I know that the components of the electric field in spherical coordinates are:

    [tex] E_r = \frac{2 p \cos \theta}{4 \pi \epsilon_0 r^3} [/tex]

    [tex]E_\theta = \frac{p \sin \theta}{4 \pi \epsilon_0 r^3} [/tex]

    [tex] E_\phi = 0 [/tex]

    so applying the curl is just a matter of calculus, and it's easy to show that
    [tex] \nabla \times \mathbf{E} = 0 [/tex].

    Otherwise, using cartesian coordinates, if I choose the z-axis oriented as the dipole and set the origin in the dipole's center, the components of the electric field are:

    [tex] E_x = \frac{p}{4 \pi \epsilon_0} \frac{3 x z}{r^5} [/tex]

    [tex] E_y = \frac{p}{4 \pi \epsilon_0} \frac{3 y z}{r^5} [/tex]

    [tex] E_z = \frac{p}{4 \pi \epsilon_0} \left( \frac{3z^2}{r^5} - \frac{1}{r^3} \right) [/tex]

    and the curl is different from 0, as one can easily prove, in contradiction with the previous result!

    So, my question is:

    Did I mistake or miss something? I really can't see what's wrong with this problem, at this time.. :uhh:

    Thank you. :smile:
     
    Last edited: May 6, 2007
  2. jcsd
  3. May 18, 2007 #2
    Here is a quick computation I made with Mathematica regarding this problem.

    As you can clearly see, in one case the curl is 0, in the second one is different from 0.
     

    Attached Files:

  4. May 18, 2007 #3
    I finally realized Mathematica didn't do all the simplifications! :rofl:

    By using Simplify command it comes up that curl(E)=(0,0,0) even in cartesian coordinates, as it should be.

    Thank you, anyway!
     
    Last edited: May 18, 2007
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