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 $\vec{r}$.


$$V(\vec{r}) = \frac{1}{4\pi R^{2}} \oint_{sphere} Vda\ \ \ \ \ \ ... eq.1$$

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


1 Answer 1


You didn't do the surface integral correctly for either case. In each case V varies over the surface, leading to a complicated integral. .

  • $\begingroup$ No! the integrals are quite clumsy but the answers come out to be neat. By the way, the answers are for sure correct. $\endgroup$ Feb 4, 2019 at 16:47
  • $\begingroup$ "The answers are for sure correct." "But isn't this wrong?" Are they right or wrong? $\endgroup$
    – Clem
    Feb 6, 2019 at 12:04
  • $\begingroup$ They are right as per the solution manual $\endgroup$ Feb 10, 2019 at 10:54

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