electrostatics - Electric potential due to a point charge in Gaussian/CGS units

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.

Answer

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.