Electrostatic charge distribution questions

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A conducting sphere can have a non-uniform charge distribution inside while maintaining a uniform charge distribution on the outside due to the properties of conductors. The inner surface charge rearranges itself to cancel the electric field from the internal charges, ensuring that the potential remains constant throughout the conductor. This results in the outer surface charge distributing evenly, as charges on the outer surface only interact with each other. Even with a non-uniform surface charge, the conductor remains an equipotential surface because any potential difference would cause charges to move until equilibrium is reached. Thus, the inner and outer charge distributions can differ without affecting the uniformity of the external field.
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If charge distribution inside sphere is non-uniform, how is distribution on outside of sphere uniform? (Think spherical shell with point charge anywhere inside)

And how can a non-uniform surface charge be still equipotential. If there was a strong enough induced charge, wouldn't the behavior become as if it were a dipole?
 
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If the charge distribution inside a sphere is non-uniform, then the distribution outside the sphere would be non-uniform. Unless the sphere is conducting. I don't understand your second question, can you clarify it?
 
kanato said:
If the charge distribution inside a sphere is non-uniform, then the distribution outside the sphere would be non-uniform. Unless the sphere is conducting. I don't understand your second question, can you clarify it?

Yes, the sphere is conducting... sorry... so how can that be?

My second question relates to how a nonuniform surface charge can still be equipotential on the surface. If there were a very strong induced charge, the surface charge would become as if it were a dipole, right? And the potential on the surface of a dipole like that would vary because the electric field is not exactly the same in each direction.
 
In electrostatics, a conductor is always an equipotential surface. If the conductor had different parts at different potentials, then the charges in it would move from areas of high potential to areas of lower potential until the potential is equalized everywhere in the conductor.

If you consider the sphere alone, then yes it would be like a dipole. But remember that the charge distribution in a conductor is a response to the charge distribution elsewhere, so when you look at the whole situation the conductor will be an equipotential surface. This is true even if there is a non-uniform surface charge distribution on the conductor; even though the electric field at the surface will vary the conductor will still be at a constant potential.
 
kanato said:
If the charge distribution inside a sphere is non-uniform, then the distribution outside the sphere would be non-uniform. Unless the sphere is conducting. I don't understand your second question, can you clarify it?

Thank you for answer... but in the first question the sphere is conducting so how can conducting spherical shell have a non-uniform inner charge yet uniform outer charge?
 
The inner charge and the outer charge don't "see" each other. The charge on the inner surface of the shell exactly cancels the electric field from whatever non-uniform charge distribution is inside the shell. The charge on the inner surface arranges itself to equalize the potential on that surface so that there is no electric field inside the conductor. If that does not require equal amounts of positive and negative charge, then the the "balancing" charge that makes the sphere neutral will go to the other surface, and those charges will only see the other charges on the same surface, so they will distribute themselves evenly over the outer surface. Remember, in electrostatics there is no electric field inside a conductor, so there is no way for charges to see each other through a conductor.
 
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