Electric field of an electric dipole

AI Thread Summary
The discussion revolves around demonstrating that the electric field of an electric dipole is irrotational by showing that the curl of the electric field is zero. The user initially calculated the electric field components in both spherical and Cartesian coordinates, finding a discrepancy in the curl results. Upon further investigation using Mathematica, it was revealed that the software had not performed all necessary simplifications, leading to an incorrect conclusion. After applying the Simplify command, the user confirmed that the curl of the electric field is indeed zero in both coordinate systems. This confirms that the electrostatic field of an electric dipole is conservative and irrotational.
Emanuel84
Messages
13
Reaction score
0

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.. :rolleyes:

Thank you. :smile:
 
Last edited:
Physics news on Phys.org
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.
 

Attachments

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!
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top