Coulomb's law remains tricky to test at home

Coulomb's law remains tricky to test at home(chillphysicsenjoyer.substack.com)

14 points by surprisetalk 2 days ago | 11 comments

sobellian 8 minutes ago |

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.

nh23423fefe 1 hour ago |

> 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.

jeremysalwen 7 minutes ago | |

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...

andrewla 1 hour ago | |

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.

NooneAtAll3 1 hour ago | |

you have free constant k in front of the equation

any dimensional analysis gets consumed by its unknown dimensionality

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> 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

Lvl999Noob 1 hour ago |

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.

smallmancontrov 2 hours ago |

Cat + Packing Peanuts

https://commons.wikimedia.org/wiki/File:Cat_demonstrating_st...

tantalor 1 hour ago | |

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

MengerSponge 1 hour ago |

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 1 hour ago |

> 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

nkrisc 45 minutes ago | |

How do you make 3 equal charges by splitting a charge in such a manner?