Property says:- [V stands for potential, O for origin]
The value of V at a point $\vec{r}$ is the average value of V over a spherical surface of radius R centered at $\vec{r}$.
Mathematically,
$$V(\vec{r}) = \frac{1}{4\pi R^{2}} \oint_{sphere} Vda\ \ \ \ \ \ ... eq.1$$
Suppose there is a charge q on Z-axis at distance $z_{0}$ from O. I want to calculate the potential at O using above property. If I choose a spherical surface, not including the charge, that is, surface with radius R < $z_{0}$, the answer is
$$\frac{1}{4\pi \epsilon_{0}} \frac{q}{z_{0}}\ \ \ \ \ \ ... eq.2$$
But if a surface is chosen such that it includes the point charge, that is, a surface with R > $z_{0}$, the answer is
$$\frac{1}{4\pi \epsilon_{0}} \frac{q}{R}\ \ \ \ \ \ ... eq.3$$
But isn't this wrong, after all, the potential at O due to a point charge located at (0,0,$z_{0}$) is simply the expression in $eq.2$.
What misconception have I developed?
References : Introduction to Electrodynamics by Griffiths ($4^{th}$ edition) > Chapter 3 (Potentials) > Section 1.4 (Laplace's Equation in 3-Dimensions) and Problem 1
