What is the voltage at every pair of points along a ideal wire that is connecting the two terminals of a battery?

Question

If you stick a wire across a battery, obviously it is going to discharge the battery. This will be due to a voltage difference between the two terminals. However, the equation $V = IR$ tells you how to calculate the voltage change due to circuit elements, plug in $R=0$ for a perfect conductor and it would seem that the voltage should stop immediately after entering the wire (take a infinitesimal chunk of wire as your element and apply the equation). Of course real wires have some negligible resistance, but that doesn't seems like it should compensate enough, even if it compensated in the correct directions, but if you think about it resistance would cause impediment so shouldn't it keep the electrons -yet again- right at the terminal?

Another effect of electrodynamics to consider would be the electron repulsion, imagine all the electrons moving out of the positive terminal and into the wire in the first instant of time. In the next instant yet another group of electrons comes out to occupy the same space. To prevent that the electrons from before scuffle over to the next chunk of wire?

I'm probably just misunderstanding some equation, maybe the $V=IR$. I'm remembering now it's the voltage drop, so the voltage drop is zero in that first infinitesimal, so the voltage between the negative and each point is the same? This seems contrary to the loop laws and the idea of electron crowding. The voltage being due to electrons crowding would cause a continuously varying voltage because the closer you get to the source the more crowded it would get. Kirchoff's voltage laws state that, for a closed loop, the sum of the voltage differences is zero. $V=IR$ tells you that the voltage drop everywhere is zero. The only thing left to analyse is the battery itself, how can anything cancel out the battery off the whole wire was a flat zero? Perhaps this is one of the cases where that law breaks. Therefore it is expedient to revert back to first principles.

The battery works by creating a charge imbalance, chemical reactions ferry electrons through a soup via ions. In the end your wire is fed electrons on one end and electron holes on the other. The resulting emf causes current to flow so that the charge imbalance can correct itself via Coulomb's law force is proportional to the charges over the square of radius. Since radius comes into play it's clear that the length of the wire must be considered. So why isn't length a part of any of the voltage laws? Ok I give up on that approach, I can't see any relation between that and voltage.

Edit: Ok so here's another way to look at it, the terminals are just studs of metal. If you think of two terminals as being really long so that the chunk of wire under consideration is right between them the voltage across it is just the voltage of the battery. But if you do the same thought experiment with all the other chunks of wire then they all have a voltage drop equal to that of the battery which is nonsense. Basically how can I define voltage as a property of small chunks of the circuit?

Answer 1

For a truly ideal conductor, the voltage is identical at all points on the conductor in a steady state. So if you attached this to a real battery, the battery would shove charge along "trying" to maintain the voltage between the terminals. It wouldn't be able to do this. A very powerful battery might heat up trying to do this so much it damages itself or catches fire. A weak battery (like a coin cell) will simply be drained and heat up a bit.

Note I said steady-state (where $V=IR$ is true). Even in a perfect conductor, there will bet some inductance in the circuit. The inductance at points will resist current changes. So while the current is changing, there may be voltage differences between different parts of the ideal conductor.

Answer 2

With a perfect conductor, there is no resistance in the wire. However, that does not mean there is no resistance in the circuit, as the battery itself has some internal resistance. The current through the wire will then be limited by that resistance, i.e. $$I=E/R_i$$

The internal resistance is quite low. It has to be as otherwise its voltage would drop too much under normal loads. However, under heavy loads you do need to take it into account, as it means the effective battery voltage is lower at heavy load than at light load.

What will happen to a shorted battery depends on the battery type, whether it has built-in protection circuitry, etc. Without protection circuits the battery will definitely overheat and might even catch fire.

Answer 3

You are right for an ideal voltage source and an ideal wire Ohm's law doesn't work anymore. You can easily see that mathematically for fixed $V$ and $R=0$ the equation $V=IR$ has no solution for $I$.

So Ohm's law doesn't work here, but since there is no ideal voltage source in the real world why bother looking for some generalized law that includes this case. For example if you replace your ideal voltage source with a combination of an ideal voltage source with internal resistance (which models a real battery pretty accurately) Ohm's law works again.

Your problem is not that Ohm's law gives you garbage but that Ohm's law gives you no answer at all for this scenario, it's simply not applicable. If you want your simulation to be able to simulate this case you have to use some other law instead of Ohm's law but this law can't be derived from nature since this scenario doesn't exist in nature.

And as others have pointed out $V=IR$ has other limitations it is certainly not a general theory for electrodynamics. If you use a physics theory/equation you should always be aware of its assumptions and limits.