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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.

Please share your views on this problem.

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    $\begingroup$ The permittivity is not fundamental, it simply defines the unit of electric field. This is analogous to $E \sim k_B T$ in thermodynamics. $\endgroup$
    – knzhou
    Commented Aug 2, 2016 at 18:25
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    $\begingroup$ 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. $\epsilon_0$ . $\endgroup$
    – user122637
    Commented Aug 2, 2016 at 18:26
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    $\begingroup$ So, is it fine to have the dimensional equivalence as given above? $\endgroup$
    – user122637
    Commented Aug 2, 2016 at 18:28
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    $\begingroup$ Yes, you can set k=1. But that doesn't amount to putting off its dimensionality. $\endgroup$
    – user122637
    Commented Aug 2, 2016 at 18:31
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    $\begingroup$ 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. $\endgroup$
    – CuriousOne
    Commented Aug 2, 2016 at 19:25

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Charge does not need to be a fundamental unit because it's really just a raw number. 1 Coulomb just stands for a large number of electrons (minus some number of protons(quarks)), much like Avogadro's number.

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  • $\begingroup$ 1C should equal large number of protons (minus some number of electrons) since 1C is positive. $\endgroup$
    – Anubhav Goel
    Commented Aug 3, 2016 at 6:59
  • $\begingroup$ @jwimberley, I do understand your point, however it seems to be not entirely convincing. By your argument, we could replace units of physical quantities with "dimensionless" numbers by comparing with the "amount of that quantity" contained in a fundamental particle(or some other fundamental thing), which is exactly what units are meant to do(namely, to hide the absolute information within itself and allow to work with such quantities using the "ratios" or numerical values only). $\endgroup$
    – user122637
    Commented Aug 3, 2016 at 10:58
  • $\begingroup$ @jwimberley, so by your logic there still exists a unit, and that is :Charge= 10 n(say), where n is the charge of the fundamental particle,that is not questioned). My point it even when we do ordinary counting, we would have to resort to units if we want to talk about counting different kinds of things. $\endgroup$
    – user122637
    Commented Aug 3, 2016 at 11:01
  • $\begingroup$ @Subhobrata Chatterjee I agree with your first comment. However, I challenge you to find another unit for which there is a fundamental particle representing an indivisible amount of that unit in terms of which all measurable quantities are integer multiples. The only other such units are the charges of the other forces. $\endgroup$
    – jwimberley
    Commented Aug 8, 2016 at 21:31

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