I am just curious that if constants in cgs units change the answer of an equation. For example, Coulomb's constant, in SI units it equals to $8.98…\times 10^9 \,\mathrm{N\,m^2\,C^{-2}}$. However in cgs units it equals to 1. I think the difference between an answer calculated with Coulomb's constant in SI units and Coulomb's constant in cgs units would be a lot. I don't know if my logic is right.
[Physics] How accurate are constants in cgs units
physical constantsunit conversionunits
Related Solutions
When the electrostatic force was originally being studied, force, mass, distance and time were all fairly well understood, but the electrostatic force and electric charge were new and exotic. In the cgs system, the charge was defined in relation to the resulting electrostatic force (it's called a Franklin (Fr) an "electrostatic unit" (esu or) sometimes a statCoulomb (statC)).
In that system, we express the force on one charged particle by another as $F_E=\frac{q_1 q_2}{r^2}$ where the unit of charge is the esu, the unit of force is the dyne and the unit of distance is the centimeter. In the MKS system (now called SI), we would write $F_E = k_e\frac{q_1 q_2}{r^2}$ where the unit of charge is the coulomb, the unit of force is the newton, and the unit of distance the meter. It would seem that if things are equivalent, then $k_e$ is indeed just a conversion factor, but things are definitely not equivalent.
A little history is probably useful at this point. In 1873, when the cgs system was first standardized, it finally made a clear distinction between mass and force. Before that, it was common to express both in terms of the same unit, such as the pound. So if you think of it, people still say things like "I weigh 72 kg" rather than "I weigh 705 N here on the surface of Earth" and they also say $1 \mathrm { kg} = 2.2\mathrm{ lb}$ confusing mass and weight (the cgs Imperial unit of mass is actually the slug).
This is important, because there is a direct analogy to the issue of units of charge and to your question about the units of $k_e$. The Franklin is defined as "that charge which exerts on an equal charge at a distance of one centimeter in vacuo a force of one dyne." The value of $k_e$ is assumed to be 1 and is dimensionless in the cgs system.
In cgs, the unit of charge, therefore, already implictly has this value of $k_e$ built in. However in the SI units, they started with Amperes and derived Coulombs from that and time ($C=It$). The resulting units of $k_e$ are a result of that choice.
So although the physical phenomenon is the same, it is the choice of units that either gives $k_e$ dimension or not.
See this paper for perhaps a little more detail on how this works in practice.
Since this has been open for a while now, here's the solution. The thing with unit conversions is to remember to be systematic and to double-check for typos and power-of-ten slipups. Even after fifteen years of unit conversion problems I find stupid little mistakes when I write the things out in full like I have below; if I try to "save time" by not writing the conversions out all the way, I make the mistakes but I don't find them.
The density of water is \begin{align} \rho_\mathrm{H_2O} &= 10^3\, \mathrm{\frac{kg}{m^3}} \mathrm{ \times \frac{10^3\,g}{1\,kg} \times\left(\frac{1\,m}{10^2\,cm}\right)^3 } \\&= 1\,\mathrm{\frac{g}{cm^3}}, \end{align} so the density of mercury is $\rho_\mathrm{Hg} = \gamma_\mathrm{Hg}\rho_\mathrm{H_2O} = 13.6\,\mathrm{g/cm^3}$. (This is one of the plusses of CGS units, that the density of water is unity and specific gravities and densities have the same values.)
The acceleration due to gravity is \begin{align} g &= \mathrm{ 9.8\,\frac {m}{s^2} \times \frac{100\,cm}{1\,m} }\\&= 980\,\mathrm{\frac{cm}{s^2} } \end{align}
So the pressure under a 75 cm column of mercury is \begin{align} P = \rho g h &= \mathrm{ 13.6\,\frac{g}{cm^3} \times 980\,\frac {cm}{s^2} \times 75\,cm }\\ &=\mathrm{ 0.9996\times10^6 \,\frac{dyne}{cm^2} \approx 1\,megabarye } \end{align}
Best Answer
I think there's a genuine and interesting physical point to be made here.
Taking a slightly different example, the gravitational acceleration of a massive body on a test particle is $a = GM/r^2$. If you can measure $a$ and $r$ accurately then you can find $GM$ to equal accuracy. But to find $M$ you also need to know $G$, and $G$ is rather difficult to measure. So it's entirely possible in principle to know $GM$ for an astronomical body with better accuracy than $M$, which would make $GM$ a more useful description of the object's mass than $M$, and might make the mass unit in units with $G=1$ more useful than the SI or cgs mass unit. I don't know whether there was any historical era where this was actually the case for any astronomical body, though.
More generally, the measurability/reproducibility of the base quantities of a unit system affects the maximum accuracy of other quantities stated in those units, so some unit systems are actually better than others.
(Edit: according to Wikipedia, "For several objects in the solar system, the value of $\mu$ [= $GM$] is known to greater accuracy than either $G$ or $M$.")