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Why is electric field zero in a wire with 0 resistance given nonzero voltage and infinite charge inside a battery?

It is true that for a wire with $0$ resistance there will uniform voltage across the wire. But comparing with an electron in an empty space it seems to be different. Imagine in a space I reference $0$ potential energy very far away from electron($e_o$). Now if we insert an electron in the system and place it near the electron let's call it $e_i$ then shouldn't $e_i$ lose potential as it is freed? It is counterintuitive that if there is an ideal wire like this then there is no change in potential energy thus no potential difference. And for ohm's law $V/R=I$ for resistance equal $0$ current is undefined so what is happening in conductor? Is electron even moving? I think there is also another way to reason which is equipotential surface but it is not clear to me how with $0$ resistance the wire just become equipotential.

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    $\begingroup$ Does this answer your question? Why is the voltage drop across an ideal wire zero? $\endgroup$
    – joseph h
    Commented Apr 7, 2022 at 7:13
  • $\begingroup$ Your answer is wrong. If a potential difference is applied. There is an electric field. The fact that E is zero in a perfect conductor relies on the surface being in an equipotential. This isn't the case. $\endgroup$ Commented Apr 7, 2022 at 13:24
  • $\begingroup$ This can also be seen by the fact that is the absence of forces, if E is zero, how can there be an INITIAL current. The E=0 in a conductor is due to equipotential surface, this definitely isn't the case if we apply a potential to the wire $\endgroup$ Commented Apr 7, 2022 at 13:27

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It's not zero.

This is a missaplication of ohms law.

Ohms law states the potential needed to maintain a constant current under a resistive force.

It is a steady state solution of a differential equation, when the applied electric force equals the resistive force

Ie, the condition that $\vec{a} = 0$

V=IR

When R=0, V=0

Why does this equation give zero? Because in the absence of resistive forces, what is the potential needed to maintain a CONSTANT current?

Clearly zero potential is needed to maintain a constant current as in the absence of resistive forces, the current will continue to move at a constant rate. Ie, zero potential is needed to maintain it.

This is all ohms law is saying, it is a steady state solution under the assumption there is no acceleration.

This is also why using ohms law at 0 resistance, we can say "I" Is anything, as when potential is zero, all currents satisfy the condition that the current is constant.

If a potential is applied to a superconducting wire, 0 potential is needed to maintain a constant current.

Does this mean the potential is zero? Obviously not. The potential will be whatever the applied potential is, and thus there obviously IS an electric field.

I apply a potential under zero resistance, the current is changing, and thus using ohms law in this way to say the potential is zero is false. This is only the case when I is constant, which would be in the absence of an applied potential difference, which only occurs, when I don't apply a potential difference.

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  • $\begingroup$ Don't confuse this with the electroSTATIC condition that E=0 in conductors, this only works if the surface is an equipotential, which isn't the case here. $\endgroup$ Commented Apr 7, 2022 at 13:44
  • $\begingroup$ +1 This question was driving me nuts. Thanks for clearing misconception and showing full picture. $\endgroup$
    – banned
    Commented Apr 7, 2022 at 14:47
  • $\begingroup$ After reading about It, I don't think this answer is right. This example: physics.stackexchange.com/a/179386/307674 shows us that there is inevitably a self distance that acts as a resistance to avoiding infinite current. I think that in your case the electrons would arrange themselves so the electric field is constant as in a usual circuit. $\endgroup$ Commented Apr 7, 2022 at 18:19
  • $\begingroup$ This is inductance, the limit of current is a current that would cause the velocity of the charge to be the speed of light. Inductance is another factor In slowing down the currents increase. Inductance on its own cannot cause a constant current. Ofcourse electrons will configure itself to make the electric field constant, I never said that they wouldn't. My point was that there exists an electric field in the first place. The only thing that I now withdraw from my answer is that when I said $\endgroup$ Commented Apr 7, 2022 at 19:45
  • $\begingroup$ "This is also why using ohms law at 0 resistance, we can say "I" Is anything, as when potential is zero, all currents satisfy the condition that the current is constant.". I should have been more careful when discussing relativistic velocities of currents. My main argument still stands, and the concept of inductance doesn't has no bearing on the missapplication of ohms law $\endgroup$ Commented Apr 7, 2022 at 19:45
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If resistance is zero, the current becomes infinity as current density does. This is derived from Ohm's law $V=IR$. Analogously, current density is proportional to electric field as $\mathbf{J}=\sigma \mathbf{E}$. Then you could say that the electric field becomes infinity as both $I$ and $\mathbf{J}$ are proportional. But does having an infinite current make sense? What would it mean?

Indeed it would mean an infinite amount of charge going through the wire per unit of area. But, in reality, you don't have an endless number of charges, so all the charges available will instantaneously be sent through the wire as soon as you close the circuit and you would run out of charge.

Nevertheless, the two variants of Ohm's law: $V=IR$ and $\mathbf{J}=\sigma \mathbf{E}$, can only be used when the current is constant and as long as you don't run out of charge. That is because the electric field along the wire is not only created by the voltage difference from the battery, but also from some steady charges placed at the wire that create extra electric fields. These steady charges are set if and only if the current is constant and you don't run out of it. So, you can not use Ohm's law to find the electric field from the current density ($\mathbf{J}=\sigma \mathbf{E}$) because you run out of charges instantaneously and these charges are not capable of placing themselves.

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  • $\begingroup$ That answer supposes that there is infinite charge available. $\endgroup$ Commented Apr 7, 2022 at 9:26
  • $\begingroup$ Then you can use Ohm's law and the other answer must respond to your question. $\endgroup$ Commented Apr 7, 2022 at 9:30
  • $\begingroup$ Can you explain what happens to terminal of battery? $\endgroup$
    – banned
    Commented Apr 7, 2022 at 9:32
  • $\begingroup$ Could you concrete more please? $\endgroup$ Commented Apr 7, 2022 at 9:37
  • $\begingroup$ In addition, the electric field in the other answer you linked comes from the assumption that there is a constant current. But with V=IR you have that I is infinity. $\endgroup$ Commented Apr 7, 2022 at 9:41

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