Factors of $c$ in the Hamiltonian for a charged particle in electromagnetic field I've been looking for the Hamiltonian of a charged particle in an electromagnetic field, and I've found two slightly different expressions, which are as follows:
$$H=\frac{1}{2m}(\vec{p}-q \vec{A})^2 + q\phi $$
and also
$$H=\frac{1}{2m}(\vec{p}-\frac{q}{c}\vec{A})^2 + q\phi $$
with $\vec{p}$ the momentum, $q$ the electric charge, $\vec{A}$ the vector potential, $\phi$ the scalar potential and $c$ the speed of light. 
So basically the difference is in the term $1/c$ multiplying $\vec{A}$, present in the second form (which I use in my lectures) but not in the first one (used by Griffiths in Introduction to quantum dynamics to treat the Aharonov-Bohm effect). Why does this difference exist and what does it mean? And how does the term $1/c$ affect the dimensional analysis (the units) of the problem?
 A: The missing $1/c$ in your first expression is simply a consequence of the units used. The second expression is in Gaussian units while the first one is in either SI units or in natural units. In the latter system of units (natural units) certain constants like $\hbar$ and $c$ have a numerical value of 1, so they can be left out of the equations.$^1$ This is common practice in physics and it doesn't change anything about the dimensional analysis of the problem, as long as you keep in mind that you're working with those natural units.
The same goes for any other system as well. Every system of units $A$ is consistent with any other system of units $B$ as long as you yourself are consistent in their usage and correctly transform everything between $A$ and $B$ when desired.
So there is no fundamental difference between a dimensional analysis in SI, Gaussian or natural units, as long as you keep in mind what units you're working with. The units themselves will (obviously) vary between systems, but dimensional analysis in one system will be entirely consistent with dimensional analysis in another.$^2$

$^1$ Note that this is not the case for SI units. As is rather well-known, the numerical value of $c$ in SI units is about $3\times10^8$ $(\mathrm{m/s})$. The reason for the absence of $1/c$ in SI units is a conventional difference. Wikipedia has a comparison between Gaussian and SI units explaining the major differences here.
$^2$ Perhaps one important note concerning Gaussian and SI units here is that due to the different conventions, it can be more difficult to transform between them. E.g. making an equation dimensionless in SI units, might yield a non-dimensionless equation when transformed into Gaussian units.
One example is when we consider Gauss's law in Gaussian units divided by the free charge density: $(1/\rho)\vec{\nabla}\cdot\vec{E} = 4\pi$. The quantity on the left-hand side is dimensionless in Gaussian units, but not in SI units, where it is $(1/\rho)\vec{\nabla}\cdot\vec{E} = 1/\epsilon_0$. So you have to watch out for that when transforming your equations. Dimensional analysis may therefore also yield seemingly different results in SI or Gaussian units, but there is no problem if you remember the conventional differences and, again, stay consistent.
A: As in any problem in physics you can use whatever units you like the most. In electromagnetism factor like $\frac{1}{4\pi \epsilon_{0}}$ or $\frac{1}{c}$ appear a lot. So you can define a new system of units in which $c=1$ such as the Natural units or the Planck units. See http://en.wikipedia.org/wiki/Natural_units
A: The $\frac{1}{c}$ makes the electric field and magnetic field have the same units. Along with the other answers provided, this makes second equation very useful. 
