A spinning dielectric sphere

1. Dec 1, 2014

Nikitin

1. The problem statement, all variables and given/known data
http://web.phys.ntnu.no/~ingves/Teaching/TFY4240/Exam/Exam_tfy4240_Dec_2013.pdf
http://web.phys.ntnu.no/~ingves/Teaching/TFY4240/Exam/Solution_tfy4240_Dec_2013.pdf

problem 2g
2. Relevant equations

3. The attempt at a solution

Hi, this is taken from problem 2g in the problem set above.

Assume you have an electric field $\vec{E}$, and in there you have a polarized dielectric sphere. Then the sphere will have a polarization density that varies continuously in the direction of $\vec{E}$.

Now, if you start spinning this sphere about the direction of $\vec{E}$, will a magnetic field be induced? I.e. will the spinning surface charges induced on the sphere constitute a current that will by maxwell's laws produce a magnetic field?

It's kinda logical it should since you have moving charges, but according to my professor it won't. Reason being since the polarized surface charge distribution exhibits axial symmetry around $\vec{E}$, apparently. But what does that have to do with anything? You still have moving charges!

Maybe he means that since these are POLARIZED charges, the netto charge is approximately zero everywhere and so the "currents" kill each other?

2. Dec 1, 2014

Staff: Mentor

I agree with you. The sphere will have different charges on the two hemispheres, so you have a current there. It won't lead to a magnetic field everywhere, but at some (most) places you get a magnetic field in the same way a spinning charged ring gives a magnetic field.

3. Dec 2, 2014

Nikitin

Hmm I've been thinking. While you do have a surface charge density, its from polarization charge and not free charge. Then wouldn't it be like moving lots of ideal dipoles around? Those dipoles provide no net current i think.

4. Dec 2, 2014

Staff: Mentor

It is polarization charge, but it appears like a net charge distribution as the polarization is not uniform (trivial, it is limited to the sphere).
You can take the limit of a "perfect" dielectric material, then you get an actual charge.

5. Dec 2, 2014

Nikitin

What do you mean "it appears like a net charge distribution"?

I disagree with you:

In the limit the polarization at the surface being made up of ideal dipoles, then each movement of each dipole charge will be accompanied by its dipole partner. Now, since these two partners are of opposite but equal charge, then it must follow that the net current is zero everywhere.

6. Dec 2, 2014

Staff: Mentor

Exactly that. If you replace it with a conductor, you even get a "real" charge distribution.
The dipole partner is at a different position, that's the point of a dipole.

7. Dec 3, 2014

Nikitin

Hmm? In the limit of an ideal dipole they are in the same point.

8. Dec 3, 2014

Staff: Mentor

If they would be at the same point, they would need "infinite" charges. The product of charge and distance is the same on both cases.
And your dipoles have the size of molecules.

9. Dec 3, 2014

TSny

This is an interesting question. For an ideal conducting sphere, would the induced charge on the sphere rotate with the sphere or remain stationary?

10. Dec 4, 2014

Nikitin

from a macro POV a molecule can be considered a point in space, though. no?

11. Dec 4, 2014

Staff: Mentor

You don't have isolated molecules, the whole space is filled with molecules.
Imagine a chain OO OO OO OO where the left side gets some positive charge and the right side gets some negative charge: +- +- +- +- +-. This is equivalent to a macroscopic charge at one end and an opposite charge at the other end.

@TSny: It is a relevant question, but I don't see a reason why they should not. Not rotating would give a current in the material, and I think we can exclude that for the (pseudo)static case.

12. Dec 4, 2014

TSny

I actually misread the question. I was picturing the sphere as rotating about an axis perpendicular to the E field and so the induced charge would be sort of like the tidal bulge of the earth (which, of course, does not rotate with the earth). I agree with your assessment for the actual question. I need to be more careful.