What is $c$ in the Lorentz force expression?

The usual Lorentz force expression I am familiar with is this:

$$\vec F=q(\vec E+\vec v \times \vec B)$$

I have seen some other versions lately that include an extra factor $1/c$:

$$\vec F=q\left(\vec E+ \frac{1}{c} \vec v \times \vec B\right)$$

What is this $c$ and how is it included? I guess other parameters in the expression are also different from the top expression for this to fit?

Example of a text snippet where I have run across this extra parameter: • Where have you seen this? – ACuriousMind May 30 '15 at 17:23
• – Qmechanic May 30 '15 at 17:58

In the second formula, $c$ is the speed of light.
Both formulas use different system of units. The first one uses the SI: $q$ in coulombs, $\vec{E}$ in volts per meter and $\vec{B}$ in teslas. The second one uses gaussian units: $q$ in statcoulombs, $\vec{E}$ in statvolts per centimeter and $\vec{B}$ in gauss (being statvolts per centimeter and gauss dimensionally equivalent).
When dealing with electromagnetism, it is common to encounter various systems of units (SI, gaussian, Heaviside...), in wich equations differ in factors of $c$, $4\pi \mu_0$, etc. It's always convenient to make clear what system of units are you using.
• Come on, this is not a competition. We both answered at the same time. And also I think that you haven't answered explicitly to the OP question: What is $c$? – Bosoneando May 30 '15 at 18:04