Electric field of an electric dipole

[Emanuel84]

Homework Statement

Since the electrostatic field is conservative, show that it is irrotational for an electric dipole, whose dipole momentum is $$p$$.

Homework Equations

$$\nabla \times \mathbf{E} = 0$$

The Attempt at a Solution

I know that the components of the electric field in spherical coordinates are:

$$E_r = \frac{2 p \cos \theta}{4 \pi \epsilon_0 r^3}$$

$$E_\theta = \frac{p \sin \theta}{4 \pi \epsilon_0 r^3}$$

$$E_\phi = 0$$

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

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:

$$E_x = \frac{p}{4 \pi \epsilon_0} \frac{3 x z}{r^5}$$

$$E_y = \frac{p}{4 \pi \epsilon_0} \frac{3 y z}{r^5}$$

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

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.

 

[Emanuel84]

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.

 

[Emanuel84]

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!