Coulomb's law remains tricky to test at home

hckrnws

Coulomb's law remains tricky to test at home

(chillphysicsenjoyer.substack.com)

by surprisetalk

nh23423fefe
131d
> 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. amavect
  41d
  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) = K*R(r)*Q(q0,q1) multiply-charges: F(r,q0,q1) = K*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
    31d
    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
    221h
    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.
2. jeremysalwen
  01d
  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...
3. NooneAtAll3
  11d
  you have free constant k in front of the equation
  any dimensional analysis gets consumed by its unknown dimensionality
  ---
  > q^2n = (-q)^2n which know is ruled out by experiment.
  doesn't mean equation can't be using absolute values ("number of electrons/protons") and just applying needed sign at the end
  1. nh23423fefe
    01d
    K isn't free, its bound obviously
    it can't use absolute values, you can't write that equation down
4. andrewla
  41d
  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
    21d
    The universe cares about dimensional analysis because it is invariant under changes of units-of-measure, which are human constructs.
sobellian
01d
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. But no experiment can disprove a=2 for indivisible charges as we would simply obtain charge through new units. For assemblies of charges the forces must add linearly due to conservation of momentum, so there we know a=1.
smallmancontrov
11d
Cat + Packing Peanuts
https://commons.wikimedia.org/wiki/File:Cat_demonstrating_st...
1. tantalor
  01d
  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
Lvl999Noob
01d
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.
gmkiv
01d
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.
MengerSponge
01d
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: https://www.npl.washington.edu/eotwash/torsion-balances
NooneAtAll3
21d
> 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. ted_dunning
  01d
  Charge splitting being equal depends on capacitance which can be hard to make equal.

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) = K*R(r)*Q(q0,q1) multiply-charges: F(r,q0,q1) = K*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.