In electrostatics, the surface of a conductor $S$ is always at a constant potential $\phi _{0}$, where the aforementioned potential is a scalar function $\phi (x,y,z)$ defined as :
$$\phi (\textbf{x}) =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert {\textbf{x}-\textbf{x}'\lvert\lvert }} da'$$
where $\sigma(\textbf{x})$ is the surface charge density on $S$. As mentioned earlier, the surface of a conductor is equipotential, so if we choose an $\textbf{x}_{0} \in S$, we will have:
$$\phi (\textbf{x}_{0})=\phi_{0} =\int _{S} \frac{ \sigma(\textbf{x}')}{\vert\lvert {\textbf{x}_{0}-\textbf{x}'\lvert\lvert }} da'=\int _{S}g(\textbf{x}')\sigma (\textbf{x}')da'$$
Is it possible to use the mean value theorem to rewrite the above surface integral as :
$$\phi_{0} = g(\textbf{x}_{1})\int _{S}\sigma(\textbf{x}')da'$$
where $\textbf{x}_{1}$ is a constant?
I assume that $\sigma > 0$ and that the surface $S$ is compact. Then the function $g(y) = \frac 1{\|x_0-y\|}$ is continuous outside of any small neighborhood of $x_0$ and thus has and attains a minimum $c>0$ there. Since $g(y)$ is extremely large in small neighborhoods of $x_0$ we conclude that the image of $g$ on $S\setminus\{x_0\}$ equals $[c,\infty)$. In particular, $g(y)\ge c$ for all $y\in S$. This implies $g(y)\sigma(y)\ge c\sigma(y)$ for all $y\in S$ and thus $\int_S g(y)\sigma(y)\,da'\ge cI$, where $I := \int_S\sigma(y)\,da'$. Thus, $\frac 1I\int_S g(y)\sigma(y)\,da'\ge c$ and $g$ attains this value, i.e., there exists $x_1\in S$ such that $g(x_1) = \frac 1I\int_S g(y)\sigma(y)\,da'$, which was the claim.