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Thus one Planck charge may be equal to any real number.

I worked out the Planck charge and found that any real value of the Planck constant exponent is consistent. Thus one Planck charge may be equal to any real number. I set:ε^{a}_{0}ħ^{b}c^{c}G^{d} = 1 Coulombto get this result. Leaving out "G" leads to an inconsistency.Therefore something other than the particle should determine its charge.

No, the Planck Charge is a constant

It looks like you don't understand dimensional analysis.

You get a more "natural" number if you multiply by Ap.

It looks like you don't understand dimensional analysis.Still.

Quote from: Bored chemist on 23/12/2021 13:23:23It looks like you don't understand dimensional analysis.Still.I know how to calculate it: set:ε_{0}^{a}ħ^{b}c^{d}G^{e} = x Coulomb.

Set the constants as shown, then calculate a, b, d, e by equating the exponents of the dimensions in the constant raised to the power a, b, d, e to the corresponding exponent of the dimension of the left side. Then you get 4 equations in 4 variables. a = 1/2 you get from ε_{0} which must have the exponent of it's C dimension = 2a = 1. Further you get:(2) d+2b-3a+2e = 0(3) -d-b+2a-e = 0(4) b-a+e = 0you get d = 1/2 andb+e = 1/2. Just set b+e = 1/2 in (2) and (3). You get (2), (3) and (4) from a knowledge of what units the constants are made up of.We have that b and e is left as b+e =1/2, so b and e can be chosen as any real numbers so long as their sum is 1/2.

What do you set kgm^{2}s^{-2} equal to?

b and e can be chosen as any real numbers so long as their sum is 1/2