A question about electric potential

Question

I want to ask about the interpretation of electric potential as it relates to this simple problem : " "A $0.800$ mm diameter ball bearing has $1.50 \times 10^9$ excess electrons. What is the ball bearing's potential?"

The electric potential is the amount of work needed to move a unit of charge from one point to another within an electric field, without producing an acceleration. What does it then mean to ask, 'what is the potential of an object'? In this case, the potential is easily calculated by $U=Fr$, where $r$ is the radius of the ball. Is this the amount of work needed to move a charge from the center of the sphere to the edge, from $x=0$ to $x=r$? Nothing is said regarding charge distribution in the ball, so I don't think we can say much about the electric field either.

My second interpretation, was that this $U$ is the amount of work needed to bring a total charge $Q$ from infinity into this ball of charge. But since we are not given any information regarding the charge distribution within the ball, I don't know how to develop this further.

Answer 1

To answer your main question, it makes sense to ask about the potential because the potential inside a pecfect conductor is constant and you can calculate it as the potential at the surface (radius $R$).

For a bounded charge distribution like the one you say, the potential at $\vec{r}$ is the work that you have to do against the electric field to bring a positive unit test charge from infinity to that location. The force that you would have to do is $-\vec{E}$, so the potential of the ball bearing is $$V_0=\int_{+\infty}^R(-\vec{E})\cdot d\vec{r}$$ where $\vec{E}$ is the electric field created by the ball bearing outside of it (it creates electric field because it is charged). The excess of electrons create a net charge $Q=-Ne$, ($N=1.50\times10^9$, $e$ the elementary charge). If you assume this charge to be evenly distributed within the ball bearing then the electric field produced outside is the same as if all the charge were concentrated at the center, so it would go as $$\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}$$

Answer 2

Well, everywhere on the ball bearing, the potential is same. So potential of the bearing is a justified statement, referring to anywhere on the sphere. Though we cannot say potential. atba point, and only potential differences make sense, we usually take potential at $\infty$ to be zero and potential on any point on the bearing is just $\frac{kQ}{r}$ $ where$ $k=9×10^9$

Note : only for a constant Electric field, potential $U = \vec E.\vec r$. Its because the Integral of $\frac{1}{r^2} is $\frac{-1}{r}$ that it seems so.