I learned electrostatics in SI units. In SI, the electrostatic potential due to a point charge $q$ located at $\textbf{r}$ is given by

$\Phi(\textbf{r}) = \frac{q}{4 \pi \epsilon_0 |\textbf{r}|}$.

Now, the Griffiths electrodynamics textbook says, "Converting electrostatic equations from SI to Gaussian units is not difficult: just set $\epsilon_0 \rightarrow \frac{1}{4 \pi}$."

So, in Gaussian/CGS units, apparently

$\Phi(\textbf{r}) = \frac{q}{|\textbf{r}|}$.

However, one textbook (Understanding Molecular Simulation, by Frenkel and Smit) says that the potential due to a point charge is

$\Phi(\textbf{r}) = \frac{q}{4 \pi | \textbf{r} |}$.

Did I make a mistake, or did Frenkel and Smit?

Thank you.

  • $\begingroup$ FYI: In my version of Frenkel and Smit, this equation is (12.1.4), page 295. $\endgroup$ Commented Apr 27, 2012 at 21:07
  • $\begingroup$ Same for me. This equation is (12.1.4), page 295. $\endgroup$
    – Andrew
    Commented Apr 27, 2012 at 21:35
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/1673/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Aug 15, 2013 at 14:01

2 Answers 2


Frenkel and Smit definitely make a mistake. Eq. (12.1.3) page 294 is: $$-\nabla^2 \phi(\mathbf{r}) = 4\pi \rho(\mathbf{r}) $$ then immediately afterwards, Eq. (12.1.4) is "the solution of this equation" "for a single charge $z$ at the origin": $$\phi(\mathbf{r}) = \frac{z}{4\pi |\mathbf{r}|}$$ This is a mistake: Eq. (12.1.4) is definitely not "the solution" to Eq. (12.1.3). In fact, see here $$-\nabla^2 \frac{z}{4\pi|\mathbf{r}|} = z\delta(\mathbf{r}) = \rho(\mathbf{r}) \neq 4\pi \rho(\mathbf{r})$$ Eq. (12.1.4) would be correct with Lorentz-Heaviside units (for example). Eq. (12.1.3) would be correct with Gaussian units.


Gaussian is one of several CGS dimensional systems. It could be the authors are using the Lorentz–Heaviside CGS system, or something else. There is a useful explanation of the taxonomy of CGS subsystems (with a table and everything) located here:



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