Final answer:
The charges on a grounded spherical conductor induced by a nearby positive point charge distribute uniformly on its surface due to electrostatic equilibrium. The electric potential and field outside the conductor mimic that of a point charge at its center. It is correct to compare the electric potential of this system to that of a single point charge Q, where Q is the accumulated surface charge.
Step-by-step explanation:
When a positive point charge is placed near a grounded spherical conductor, it induces a charge distribution on the sphere. The charges on the conductor will redistribute themselves to maintain electrostatic equilibrium, which means the conductor's surface becomes an equipotential surface, having the same electric potential at every point on it. Due to the mutual repulsion of like charges and the spherical symmetry of the problem, the charges distribute uniformly on the surface of the sphere.
The electric potential inside the grounded spherical conductor is zero, due to it being connected to the Earth, which acts as an infinite reservoir of charge. The outer electric field of the spherical conductor, due to the uniformly distributed induced charge, acts as though it originated from a point charge located at its center. Thus, outside of the sphere, the resulting electric potential and field are identical to that of a point charge situated at the sphere's center with a charge equal to the net charge on the sphere's surface.
The assumption that the accumulated charge on the surface of the sphere is uniformly distributed is correct in the context of a spherical conductor in electrostatic equilibrium. It is incorrect to assume that this distribution is non-uniform because the spherical symmetry and the conductor's properties will naturally lead to a uniform charge distribution on the surface. Hence, the electric potential due to the sphere with the accumulated charge will be the same as that of a point charge at the center of the sphere for all points outside of the conductor.