# Equlibrium charge density of a charged spinning sphere

I am not sure how to approach this problem. A similar question has gone unanswered before.

If the conductor was at rest, then the charge would have distributed evenly. Now all the charges will feel pushed out due to the rotation of the sphere. However the amount of "push" is proportional to the distance from the axis (centripetal force is proportional to the radius if angular velocity is constant).

But the electrons would not like to accumulate together, so they must be resisting the urge to group together (near the equator).

Have I missed any other electromagnetic effect (other than the Coulombic repulsion)?

Assuming Q is positive and the vertical line through each sketch represents the axis of rotation with the angular velocity vector pointing up, then the moving charge on each side of the sphere produces a magnetic field pointing up on the opposite side of the sphere. This field then pushes the moving charges away from the axis of rotation. As indicated, sketch,b, represents the resulting charge distribution. (If you assume Q is negative, the field is reversed, but the force is still outward.)

You seem to be assuming that the centrifugal (inertial) force on electrons will be significant in this situation. For electrons, inertial forces are very, very much smaller than electromagnetic forces.

The words "very rapidly" gives a hint that this might be a relativistic effect. When two like charges are moving parallel to each other then viewed in the stationary frame the electrostatic repulsive force between them is reduced. We account for this by saying that there is a magnetic field around each, resulting in a magnetic force of attraction between them. This is a relativistic effect and increases with velocity. When the velocity reaches the speed of light the electrostatic repulsion is balanced by magnetic attraction.

The surface charges furthest from the axis are moving at the highest speeds as viewed in the laboratory frame, so they experience the least amount of repulsion. This again leads to answer (b), but not because of the centrifugal force.