0
$\begingroup$

Conservation laws take the most seemingly complex of problems and boils them down to a simple abstraction, but they are huge statements to stand on.

I'm just having a really hard time convincing myself that we are certain that when some charge leaves the negative end, that same exact amount of charge is reunited on the other end. It seems as though there would be some minuscule variance of some kind on a molecular level.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ well sure, there is always shot noise, see en.wikipedia.org/wiki/Shot_noise $\endgroup$
    – hyportnex
    Commented Nov 13, 2018 at 17:22
  • 3
    $\begingroup$ Well, suppose it didn't. Then charge would pile up somewhere in the system, since it was going out and not coming back. Pretty soon there would be a hellacious voltage difference that would make itself known by an arc. But we don't see that, so it must all be coming back. $\endgroup$
    – zeta-band
    Commented Nov 13, 2018 at 17:47
  • $\begingroup$ See electronics.stackexchange.com/questions/354705 $\endgroup$
    – sammy gerbil
    Commented Nov 20, 2018 at 1:21

1 Answer 1

Reset to default
2
$\begingroup$

It is BECAUSE of what's happening at the molecular level that we are so sure of the same charge entering a battery as leaving it. Inside the battery, there are two electrodes. A chemical reaction takes place that places electrons on one electrode, and those electrons MUST travel to the other electrode and become attached to a molecule at that electrode in order to keep the chemical reaction going. For details, see Wikipedia regarding electrochemical cells: https://en.wikipedia.org/wiki/Electrochemical_cell

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ When you say "those electrons MUST travel to the other electrode" are you saying the electron exiting one electrode must travel to the other electrode? I'm thinking a metal conductor before being connected to the battery terminals has free electrons throughout. When the connections are made (or switch closed) to the circuit isn’t momentum transferred between free electrons along the conductor so that a free electron enters the other electrode, kind of like a chain reaction? $\endgroup$
    – Bob D
    Commented Nov 13, 2018 at 18:24
  • 2
    $\begingroup$ @BobD, by analogy, the free electrons in the wire act very much like the water in a liquid filled hose. As one electron enters the wire, it pushes an electron on the other end of the wire out, because electrons repel each other, and the wire will seek to remain electrically neutral when current is flowing through it. And yes, I realize that the drift speed of electrons in a wire is exceedingly slow. I should have been a bit more explicit in my posting, but I was seeking to be relatively concise regarding the exact question that the OP asked. $\endgroup$
    – David White
    Commented Nov 13, 2018 at 19:11
  • 2
    $\begingroup$ David, it wasn't meant as a criticism of the answer (I upvoted it), just wanted to make sure my understanding was correct. We are on the same page. $\endgroup$
    – Bob D
    Commented Nov 13, 2018 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.