The solution before revision #18524 of , by Luisito. This is not the current version.

Statement

6.3.7. [Insert the problem statement]

Solution

As the statement of the problem doesn't clarify if the sphere is conducting or non-conductive, let's consider both cases.
Case 1) Nin-Conductive sphere:
Let's assume that charge is uniformly distributed in the volume. Let electrical potential be:
(1)
for the center, let's consider , or .
Applying Gauss Law:
For ,

but the enclosed charge into a sphere of radius r is related to the charge distribution per unit of volume,


so,

and for , the enclosed charge is Q, then,
(2)
This mean that function has two behaviors, depending on values of . According (1) and assuming ,
Extra \left or missing \rightV(r) = -\left(\int_{\infty}^{R} \vec{E} \cdot d\vec{r} + \int_{R}^{r} \vec{E} \cdot d\vec{r} (I)
developing,

as tends to zero,
(3)
Case 2) Conducting sphere
In this case, charge stays on sphere surface, so the electric field inside of it is null (), so potential is constant inside the sphere and coincides with the value of it on the surface. According to (1) and (2),

(4)
Finally, if sphere is non-conductive depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it \textbf{doesn't depend} on it. If the charge is non-uniformly distributed, the potential on the surface \textbf{does change} with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge, Q, as if it were a point charge at the center.

Answer

[Insert a concise answer or boxed result]