Gauss's Law and Electric Dipole

[zorro]

Can we use Gauss's Law to calculate the field distribution around an electric dipole?

 

[Born2bwire]

Not that I am aware of. We can do it for a monopole but that requires us to make assumptions about the nature of the field via symmetry. It allows us to deduce that the field and the Guassian surface are always normal to each other. This makes the flux directly proportional to the field strength.

 

[zorro]

Do you mean considering each monopole of the dipole separately we can find out the electric field distribution?

 

[Born2bwire]

Do you mean considering each monopole of the dipole separately we can find out the electric field distribution?

Well... we could, but only in the sense that it is the same way of finding the field distribution of a charge using Gauss' Law and then using superposition to add the fields from two separated opposite charges to get the dipole field. But I do not think that is in the spirit of what you mean.

 

[kloptok]

This is basically the same thing as Born2bwire said, but in a different formulation:

The problem lies in finding the equipotential surfaces of the dipole field. For a point charge this is easy, it's just a sphere, but what is it for a dipole?

 

[planethunter]

This is basically the same thing as Born2bwire said, but in a different formulation:

The problem lies in finding the equipotential surfaces of the dipole field. For a point charge this is easy, it's just a sphere, but what is it for a dipole?

In theory, just thinking about the most symmetrical "sphere-like" surface you can think of for using Gauss's law on a dipole, couldn't we use a 3-d ellipse surrounding the dipole?

 

[Born2bwire]

No, we can't use an ellipse. Let's say the dipole's axis along the z axis and is centered about the origin. If we look at the fields along the z-axis only, then the fields will point in the +/- z direction. If we chose an ellipse, we can set the major axis to be the z axis and so the tips of the ellipse would be perpendicular to the fields on the z axis.

So that's well and good, but what about on the minor axis? The minor axis would be the x-y plane but the fields on the x-y plane are still parallel to the z axis but the ellipsoid's surface is also parallel to the z axis. So we can quickly see that an ellipsoid is not going to be an equipotential surface.

Take a look at the dipole field from Wikipedia:

You can see from the field lines and arrows that, yes, and ellipsoid would appear to be a good fit until you get to the plane that is normal to the dipole axis and situated in between the two charges. So, even if we could envelope the dipole in a closed equipotential surface, I do not see how we could determine this surface a priori like we can with the single point charge.