[Physics] Why does Coulomb’s constant have units

Question

I think of Coulomb's constant as a conversion factor (not sure if this is correct).
Kind of like how you would do calculations in kg and then times it by the conversion constant to convert your answer to pounds.
The conversion factor would be $2.2\: \mathrm{lbs/kg}$.

Since the units for Coulomb's constant is $\mathrm{N \cdot m^2/C^2}$, would it make sense to define the Newton as

$$1\:\text{Newton} = \frac{1}{1/1\: \mathrm{meter^2} \cdot 1\: \mathrm{Coulomb^2}} \, .$$

Would the above definition be valid?

EDIT: So if $k$ is not a conversion factor since the above definition for a Newton is invalid and $k$ is not just a scaling factor, since it has units, then what is it? If its just a proportionality constant to adjust the magnitude then why does it have units? Shouldn't it be a unit less constant?

EDIT: So $k$ is not just a scaling factor (since it has units) and its not a conversion factor since a Newton can't be expressed as the other units. So if its unit just exists so that things cancel out "nicely" doesn't this make dimensional analysis useless since you can add in random constants and units to cancel out whatever you want?

My question is not about the meaning of $k$. Its about its units.

Answer 1

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.