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Coulomb's law remains tricky to test at home

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smallmancontrov · 1790262005 · focus · HN ↗
Cat + Packing Peanuts
<a href="https://commons.wikimedia.org/wiki/File:Cat_demonstrating_static_cling_with_styrofoam_peanuts.jpg" rel="nofollow">https://commons.wikimedia.org/wiki/File:Cat_demonstrating_st...
1. tantalor · 1790262789 · focus · HN ↗
  Good idea, you could do a monte carlo method where you throw many packing peanuts at the cat (or many cats!) and count how many stick, divide by how many you threw, et voila
nh23423fefe · 1790263068 · focus · HN ↗
> 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.
1. andrewla · 1790263791 · focus · HN ↗
  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.
  1. mitthrowaway2 · 1790269700 · focus · HN ↗
    The universe cares about dimensional analysis because it is invariant under changes of units-of-measure, which are human constructs.
    andrewla · 1790279166 · focus · HN ↗
    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.
    nh23423fefe · 1790286942 · focus · HN ↗
    This isn't true.
    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.
  2. mhh__ · 1790271744 · focus · HN ↗
    Because we aren't dealing with the universe but rather ruling out possible forms of models of it
2. NooneAtAll3 · 1790264074 · focus · HN ↗
  you have free constant k in front of the equation
  any dimensional analysis gets consumed by its unknown dimensionality
  1. nh23423fefe · 1790276518 · focus · HN ↗
    K isn't free, its bound obviously
    it can't use absolute values, you can't write that equation down
3. jeremysalwen · 1790269179 · focus · HN ↗
  I thought it was "obvious" based on the principle that two charges at the same location should have the same force as one combined charge at that location. Of course this immediately brings up the question of the self-force of a point charge...
4. amavect · 1790269897 · focus · HN ↗
  Show by experiment that the force of charge0 against charge1+charge2 equals the force of charge0 against charge1 plus charge0 against charge2. Induce an additive-homomorphic property F(r,q0,q1+q2) = F(r,q0,q1)+F(r,q0,q2). Then, exponent 1 follows.
  I figured this out by listing a bunch of mathematical properties. I couldn't see how the author jumps from zero-preserving to multiply-charges, and I still don't know how, but we can call it out of scope lol
  r : distance between p and q q0 : charge 0 q1 : charge 1 F : coulomb force function charge-commutative: F(r,q0,q1) = F(r,q1,q0) zero-preserving: 0 = F(r,q0,0) additive-homomorphic: F(r,q0,q1+q2) = F(r,q0,q1)+F(r,q0,q2) homogenous-degree-1: F(r,q0,n*q1) = n*F(r,q0,q1) multiplicative-separability: F(r,q0,q1) = R(r)*Q(q0,q1) multiply-charges: F(r,q0,q1) = R(r)*(q0*q1)^a Given F(r,q0,q1) = K*R(r)*(q0*q1)^a, charge-commutative, zero-preserving, additive-homomorphic. Induction using additive-homomorphic proves homogenous-degree-1. (For example, F(r,q0,2*q1) = F(r,q0,q1+q1) = 2*F(r,q0,q1)) Equational proof follows from homogenous-degree-1: K*R(r)*(q0*n*q1)^a = n*K*R(r)*(q0*q1)^a (q0*n*q1)^a = n*(q0*q1)^a n^a*(q0*q1)^a = n*(q0*q1)^a n^a = n n = 0 or a = 1 n≠0, therefore a=1.
  1. nh23423fefe · 1790285647 · focus · HN ↗
    Can you use the homomorphism to get that?
    F(r,q0,q1) = F(r,q0,q1+0) = F(r,q0,q1)+F(r,q0,0) = F(r,q0,q1) + 0
    amavect · 1790351458 · focus · HN ↗
    Actually, yes! We can use homogenous-degree-1 (provable from the homomorphism) to prove it.
    F(r,q0,q1) = F(r,q0*1,q1*1) = q0*q1*F(r,1,1)
    Very simple! It remains to show that F(r,1,1) = K/r^2, as intended.
    nh23423fefe · 1790353425 · focus · HN ↗
    I feel like that should be gettable from rotational symmetry and somehow mentioning L2 norm to get the square. Either that or conservation of E-field flux.
    amavect · 1790361872 · focus · HN ↗
    Rotational symmetry or L^2 norm only matter in vector formulations. I assumed a scalar formulation.
    Conservation of E-field flux certainly implies additive-homomorphism (addition of charges equals addition of forces). But that seems a bit ahistorical because Maxwell would develop field theory 70 years after Coloumb. Either way, the axiom you choose requires empirical justification, and I think a home hobbyist could more easily demonstrate by experiment that adding charges will add the forces.
NooneAtAll3 · 1790264019 · focus · HN ↗
> Supposedly, one way to make two equal charges is to charge one thing and then put it in contact with the other thing, so that the charge splits by symmetry. But then how would we check that indeed we have two equally charged things?
same as with making guaranteed flat surface - you make 3 and measure each pair
1. nkrisc · 1790266913 · focus · HN ↗
  How do you make 3 equal charges by splitting a charge in such a manner?
2. ted_dunning · 1790271100 · focus · HN ↗
  Charge splitting being equal depends on capacitance which can be hard to make equal.
MengerSponge · 1790265078 · focus · HN ↗
It's probably easier to verify Gauss's Law, which then gives you Coulomb's law for free.
Direct verification of inverse square laws is hard! For electromagnetic interactions (ie Coulomb's law) you can use scattering. If you want to do a static experiment (Coulomb's Law the hard way or gravity) you probably need a torsion pendulum experiment. AFAIK the best in the world at that are at UW in the Eöt-Wash group: <a href="https://www.npl.washington.edu/eotwash/torsion-balances" rel="nofollow">https://www.npl.washington.edu/eotwash/torsion-balances
Lvl999Noob · 1790265720 · focus · HN ↗
Could you use an electromagnet and a metal ball to create a fixed charge? If you change the current in the electromagnet, that would change the intensity of the magnetic field. With the metal ball fixed in place nearby and a careful curve for the change of current, you could potentially have a constant charge remaining on the ball.
sobellian · 1790269128 · focus · HN ↗
For two different indivisible charges, this is a bit slippery but I think a=1 by definition. How do we measure charge, practically speaking? By how much force is measured between it and a reference charge. So we take a=1 as a convention. For assemblies of charges the forces must add linearly due to conservation of momentum.
gmkiv · 1790271190 · focus · HN ↗
Oh, he's so close! He's already constructed an electroscope. Now he just needs to use that to perform Cavendish's experiment, and he'll have verified Coulomb's law.

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