# Calculating potential due to point charge using properties of Harmonic functions (Solutions of Laplace's Equation )

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