> But it's not at all obvious to me why the exponent a is 1 in nature
i think we can rule out any exponent just by dimensional analysis if you allow powers of q, then K has ambiguous units. Same reason you cant exponentiate unitful quantities
More specifically I think we can rule out even exponents by anti-symmetry of charge. That is q^2n = (-q)^2n which know is ruled out by experiment.
Why would the universe care about dimensional analysis? Besides, the outside constant would do the unit conversion from whatever the right-hand side produces to units of force.
This is making assumptions about the nature of charge and its relation to force that are what is under discussion! We get no guarantee from nature that the force is linear in the amount of charge; if it turned out to be sqrt(charge) then we'd still have to deal with it. We could redefine charge to the sqrt of the previous quantity, but then other aspects of linearity (like conservation of charge) would have to take a different form.
We're lucky (and maybe only in the linear domain, which is more accurately the case) that these ended up being the same.
But if it were sqrt(charge) then we'd define Culoumbs constant to have commensurate units to translate sqrt(charge) * sqrt(charge) / distance^2 to force units.
Partitioning a charge can't change the physics. If i have a bar of charge Q generating a force F on q which is proportional to sqrt Q then by partition i have n bars of charge Q_i = Q/n producing forces F_i
sqrt Q ~ F <> sum F_i = sum sqrt(Q_i) = sum sqrt(Q/n) = n * sqrt (Q/n) = sqrt (Qn)
Partitioning charge would lead to infinite forces.
nh23423fefe · · focus · HN ↗
i think we can rule out any exponent just by dimensional analysis if you allow powers of q, then K has ambiguous units. Same reason you cant exponentiate unitful quantities
More specifically I think we can rule out even exponents by anti-symmetry of charge. That is q^2n = (-q)^2n which know is ruled out by experiment.
andrewla · · focus · HN ↗
mitthrowaway2 · · focus · HN ↗
andrewla · · focus · HN ↗
We're lucky (and maybe only in the linear domain, which is more accurately the case) that these ended up being the same.
But if it were sqrt(charge) then we'd define Culoumbs constant to have commensurate units to translate sqrt(charge) * sqrt(charge) / distance^2 to force units.
nh23423fefe · · focus · HN ↗
Partitioning a charge can't change the physics. If i have a bar of charge Q generating a force F on q which is proportional to sqrt Q then by partition i have n bars of charge Q_i = Q/n producing forces F_i
sqrt Q ~ F <> sum F_i = sum sqrt(Q_i) = sum sqrt(Q/n) = n * sqrt (Q/n) = sqrt (Qn)
Partitioning charge would lead to infinite forces.
mhh__ · · focus · HN ↗