# Absence of "charge" as a fundamental unit in natural system

In natural system of units, using speed of light, Planck's constant, and eV (for energy), we express "charge" dimension as:

$[charge]=[Force]^{1/2}[length]$ (from Coloumb's law equation)

There is an implicit assumption that the absolute permitivity of space is dimensionless. How can we ignore a dimensional constant? In CGS system, of course, $k=\frac{1}{4 \pi \epsilon_0}=1$. But there is a dimension in addition to the value.

So, to reconcile with this problem, I guess we must either consider charge or absolute permitivity of space as the 4th fundamental unit.

• The permittivity is not fundamental, it simply defines the unit of electric field. This is analogous to $E \sim k_B T$ in thermodynamics. Commented Aug 2, 2016 at 18:25
• What I mean to say, is how can "charge" dimensionally be equated to powers of Force and length only. Its ridiculous to omit the very important dimensional constant driving Maxwell's equations i.e. $\epsilon_0$ .
– user122637
Commented Aug 2, 2016 at 18:26
• So, is it fine to have the dimensional equivalence as given above?
– user122637
Commented Aug 2, 2016 at 18:28
• Yes, you can set k=1. But that doesn't amount to putting off its dimensionality.
– user122637
Commented Aug 2, 2016 at 18:31
• A "dimensions" is merely the reference to a physical normal. When you say that you are measuring 1m, then you are referring to a metal Stick in paris or a number of wavelengths of a certain optical line or the distance light passes in a certain amount of time. There is absolutely no physics in these random definitions of normals. So what is $\epsilon_0$? It is the constant that tells you the energy density of an electrostatic field. Chose units for energy, length and electric field wisely and it becomes trivial. Commented Aug 2, 2016 at 19:25