Electric field of an electric dipole

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Emanuel84
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Homework Statement


Since the electrostatic field is conservative, show that it is irrotational for an electric dipole, whose dipole momentum is [tex]p[/tex].

Homework Equations


[tex]\nabla \times \mathbf{E} = 0[/tex]

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.. :rolleyes:

Thank you. :smile:
 
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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.
 
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I finally realized Mathematica didn't do all the simplifications! :smile:

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

Thank you, anyway!
 
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