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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.

banned
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2 Answers2

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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.

jensen paull
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  • 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. – jensen paull Apr 07 '22 at 13:44
  • +1 This question was driving me nuts. Thanks for clearing misconception and showing full picture. – banned Apr 07 '22 at 14:47
  • After reading about It, I don't think this answer is right. This example: https://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. – Marc Barceló Apr 07 '22 at 18:19
  • 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 – jensen paull Apr 07 '22 at 19:45
  • "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 – jensen paull Apr 07 '22 at 19:45
  • @jensenpaull great answer,however clearing a misconception of mine will help greatly, Suppose I have an ideal wire with just a resistance $r$ and battery voltage $V$,now,all the path before the resistor,the wire is resistance less and hence all points are at same potential. Does that mean along all that path electric field $E$ will be $0$ as there is no force required in that resistance less wire. And only through the resistor,the electric field $E$ will be applied? So,in conclusion,$0$ electric field along whole wire and non zero only at resistance? Kindly address this. – a_i_r Nov 08 '23 at 14:22
  • For a circuit with an ideal battery:V=IR, R=0, V=0. This is the potential difference needed to maintain a constant current across the wire. Is the current constant? An electric field exists due to the battery, accelerating the electrons in the wire, ie a non constant current. There is a PD across the wire. However the potential difference needed to maintain a constant current is zero. This last statement is irrelevant to the scenario. Most textbook say that this "potential needed to maintaintain a constant current" is the potential across the wire because in comparison to the pd across the – jensen paull Nov 08 '23 at 18:29
  • Resistor, the pd is negligible and hence using a lumped element model, an approximate way of thinking is "the pd only drops in the resistor" this is false and naive. One can obviously see that the wire has a PD across it by constructing a ideal wire loop and applying KLR. – jensen paull Nov 08 '23 at 18:31
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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.

Marc Barceló
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  • That answer supposes that there is infinite charge available. – Marc Barceló Apr 07 '22 at 09:26
  • Then you can use Ohm's law and the other answer must respond to your question. – Marc Barceló Apr 07 '22 at 09:30
  • Can you explain what happens to terminal of battery? – banned Apr 07 '22 at 09:32
  • Could you concrete more please? – Marc Barceló Apr 07 '22 at 09:37
  • 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. – Marc Barceló Apr 07 '22 at 09:41
  • I think that the answer of 0 electric potential is met only if the battery has internal resistance that accelerates particles with jnitial electric field inside battery. Out of the battery, inertia keep electrons moving at constant speed. If you don't make this assumption, that answer doesn't make sense. – Marc Barceló Apr 07 '22 at 10:03
  • At 0 resistance, the current is undefined. V=I*0, any value of current satisfies this equation and gives zero potential. This isn't due to "infinite current" blah blah, or "infinite charge", ohms law is a STEADY state equation, under the condition that there is zero acceleration. When resistance is zero , zero potential is required to maintain a constant current( 0 acceleration). This definitely doesn't mean the potenti is zero. Nor does it mean the electric field is zero. – jensen paull Apr 07 '22 at 13:21
  • If the potential difference is zero along the wire then the Kirchoff law is not followed (unless there is some resistance inside the battery). As we don't have infinite charges ohm's law can not be used as the circuit must be steady with the arranged charges and as all the charge is sent from one terminal as current is infinity you can not get asteady-statee. – Marc Barceló Apr 07 '22 at 13:43
  • Current isn't infinity applying ohms law. Kirchoffs law will also not be followed even with some resistance in the battery ( good point which I didn't think of actually) .We can have a non constant current and still maintain surface charges/charges, the battery recovers charges from one end of the wire and puts them back. The issue is that people are misunderstanding what "V" means in ohms law, and is only valid with a constant current. – jensen paull Apr 07 '22 at 13:50
  • Kirchoff law is always followed as far as I am concerned. With an internal resistance, all the potential difference is given by that resistance and along the wire, the voltage is equal to the other terminal so it can be followed. Could you bring some sources to show that the Kirchoff law can be broken? – Marc Barceló Apr 07 '22 at 13:58
  • Your misunderstanding what I'm saying. You're correct that Kirchoff law is always followed ( in the absence of changing magnetic fields) , my point was that if the electric field inside the wire was zero, kirchoffs law would break, as there would be 0 PD across the wire, + field inside the battery, which is non zero. Which violates Kirchoff law. Which is proof that there must be an electric field inside the wire – jensen paull Apr 07 '22 at 15:00
  • I agree with you. But, do you think that my answer is incorrect? I don't see the disagreement then. – Marc Barceló Apr 07 '22 at 15:17