I just began studying electrostatics in university, and I didn't understand completely why the electric potential due to a conducting sphere is
$$
V(\vec{r})=\begin{cases}
\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}, & \text{if $r \le R$}.\\
\\
\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r}, & \text{if $r \gt R$}.
\end{cases}
$$
Where Q is the total charge and R is the radius of the sphere (the sphere is located at the origin).
I only understand the second part of this equation (when $r > R$).
I know Gauss Law. know the charges go to the surface. I know the electric field strictly inside it must be zero. (I also know the electric field is not defined for a point that lies exactly in the surface). And I know $\vec{E} = -\nabla{V}$.
Therefore, I know the electric potiential inside the sphere must be constant. Let $C$ be this constant.
It seems that
$$
C = \lim_{r \to R^+} V(r) = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}
$$
But why is this true? My textbook says: because the electric potential must be a continuous function. But why? I am hoping for a non-experimental reason.
Please be precise when mentioning $r<R$ or $r\le R$. Those are different and I get easily confused when people misuse those.
Best Answer
Imagine you have a point charge inside the conducting sphere. Obviously, since the electric field inside the sphere is zero (as you state), there is no force on the charge, so no work done. Therefore the potential is constant. So far so good.
Now as we approach the boundary, we can imagine moving an infinitesimal amount to go from $r = R - \delta r$ to $r = R + \delta r$. As long as the electric field is at most some finite amount $E_{shell}$, then the work done moving from just inside to just outside is $E_{shell}*2\delta r$; as $\delta r \rightarrow 0$, the work done will also tend to zero. The only way this would not be true is if the electric field at $r=R$ was infinite - which it is not.
This means that the potential is continuous across the shell, and that in turn means that the potential inside must equal the potential at the surface. Whether we mean by "at the surface" as $R$ or $R + \delta r$ doesn't matter since the difference vanishes as $\delta r$ becomes sufficiently small.