Correct answer: (B) uniformly over the outer surface
Explanation
At electrostatic equilibrium the field inside a conductor is zero. Any excess free charge moves to its outer surface rather than remaining in the conducting bulk.
The sphere is geometrically symmetric and no external asymmetric field is specified, so the surface charge density is uniform. The added electrons therefore spread **uniformly over the outer surface, B**. The amount of charge changes the potential, not this symmetry argument.
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