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A spherical conductor, carrying a total charge $Q$, spins uniformly and very rapidly about an axis coinciding with one of its diameters. In the diagrams given below, the equilibrium charge density on its surface is represented by the thickness of the shaded region. Which of these diagrams is correct? enter image description here

I am not sure if quasistatic approximation of magnetic field applies here since it is mentioned that the sphere is spinning very rapidly. Conductor is mentioned i suppose to say in other words that the charges will reside on surface no matter what.

And i know that magnetic field inside a spinning spherical shell is uniform and points in the direction of rotation vector ${\omega}$ and outside, it is akin to that of a dipole situated at the origin.

The electrostatics part is unclear to me since the charges in one hemisphere cannot interact through electric field with the charges in the other hemisphere since electric field has to pass through the conductor but inside, the conductor it has to be zero.

I am not sure how exactly to use all these facts or just simply say that the centrifugal force is huge due to the high rotation speed and hence charges are just pushed away from the axis in which case the answer without any effort is option b??


EDIT

I may have made a mistake in assuming electric field inside the conductor to be zero since that is only true for a conductor in electrostatic equilibrium.

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Well its certainly not (a) or (c), because there's no reason for electrons to bunch up like that when they still repel each other (assuming no quantum effects, here). So that leaves (d) or (b). When the sphere first spins up, it takes less energy to assume configuration (d) than (b)--so I think there's a brief time it takes on configuration (d). However, after its been spinning at the nominal rate for a while, the electrons are free to absorb more energy and do other things, if they so please. Note that a free electron orbiting in a magnetic field does not change its orbital parameters (except to radiate energy--but that doesn't apply to a classical analysis with uniform charge). So the only things that guide them in the steady state are their repulsion from each other and the centripetal 'force' from their mass. So they will slightly move toward configuration (b).

EDIT for more clarity:

(d) is initially preferred because it takes energy to accelerate an electron, even if it has no mass, because you are creating a magnetic field in the process. There certainly is a magnetic field because of Mach's principle; only if the universe were spinning along with the sphere would there not be a magnetic field. We assume that in the steady state there is no current, because this isn't a superconductor.

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  • $\begingroup$ FYI: sorry, I swapped configuration letters in the first post, but corrected it within a few minutes of posting. $\endgroup$ – Digiproc Oct 23 '16 at 8:50
  • $\begingroup$ I didnt exactly understand how you say that it takes less energy to assume (d) as compared to (b). And what is the energy that you say is being absorbed by the charges (Q) to 'do other things'; please clarify that statement. And what is the magnetic field in which the charges are rotating? Isnt it a little complicated (because of the high speed) so that we cannot rule out the magnetic effects? Also isnt it necessary to account for the word 'conductor'? $\endgroup$ – Prasad Mani Oct 23 '16 at 8:57
  • $\begingroup$ (d) is initially preferred because it takes energy to accelerate an electron, even if it has no mass, because you are creating a magnetic field in the process. There certainly is a magnetic field because of Mach's principle; only if the universe were spinning along with the sphere would there not be a magnetic field. We assume that in the steady state there is no current, because this isn't a superconductor. $\endgroup$ – Digiproc Oct 23 '16 at 9:03
  • $\begingroup$ I've added the above comment as an edit to the answer $\endgroup$ – Digiproc Oct 23 '16 at 9:06
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    $\begingroup$ In (d) the electrons are moving more slowly, therefore they have less energy. I should point out that we haven't calculated how much the effect of migrating toward (d) is countered by the effect to migrate toward (b), I'm just saying the effect is there in the spin up time (but admittedly maybe its overshadowed by the effect to migrate toward (b)). And yes, you are right that the moving electrons are a 'current' in a stationary reference frame, I was speaking of a current in the reference frame of the spinning sphere. $\endgroup$ – Digiproc Oct 23 '16 at 9:21

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