I believe this question has been asked $>666$ times already, however I am trying to analyse it from a different perspective.

Consider the completely theoretical case of an ideal voltage source ($0 \,\Omega$ internal resistance) having its terminals connected to one another through an ideal (perfectly conducting, zero resistance) wire. For simplicity, lets assume that positive charge carriers constitute the current (conventional current).

Now at $t=0$, when the circuit is just completed, the (positive) charge carriers experience a force due to the electric field of the source and they accelerate for some time. After a few milliseconds or so, the net field inside the wires becomes zero. (There exists no field in the material of a perfect conductor). After this point, the charge carriers are not subjected to any force: Their speed becomes constant.Let this speed be $u$.

Assuming uniform cross section area, of the wire ($A$), and assuming $n$ charge carriers per unit volume, each having a charge $e$. We will then have, in small time $dt$, an amount $dq= nedV=neAdx$ of charge flowing. We will then have $I=\frac{dq}{dt}=neAu$, which seems to be a finite (constant)quantity. So do we have a steady value of current flowing even with zero resistance?

Now I believe we cant use Ohm's law for a case where $R=0$. We can clearly see that the model that derives $V=IR$ takes into account collisions (and therefore collision time $\tau$) which aren't happening at all in this case. However, We can "extrapolate" for a fixed $V$. As $R$ decreases, $I$ increases. So surely, you can't have a finite, steady value of current for $R=0$.

Which leads me to the question: Which term in the expression I derived: $I=nqAu$ causes problems? Or does something happen before the field inside wires becomes zero?

Alfred centauri points out that the charges would redistribute instantaneously. If this is the case, Then the current should be exactly zero, and again,This doesnt comply with the extrapolation.

I believe theres some confusion regarding the "electric field becoming 0". I must confess, This was based on the following answer: Why is the voltage drop across an ideal wire zero?. And I believe it is likely to be correct, As it is the only explanation for the linked question.

  • $\begingroup$ I'm afraid I cannot make much sense out of the second paragraph. Are you assuming that the protons inside the battery start moving in the loop. Also, why would the field become zero if the circuit is complete? $\endgroup$
    – Elendil
    May 25, 2020 at 8:30
  • $\begingroup$ @Krishna Edited, Stupid(and incorrect) phrasing. However, The net field inside the material of a conductor is zero at steady state. This is a property of conductors, And the circuit doesnt need to be "open" for this to happen. It happens due to rearrangement of charges inside a conductor. $\endgroup$
    – satan 29
    May 25, 2020 at 8:49
  • $\begingroup$ Question: in this theoretical setup, are you thinking about it in the limit of infinitesimal size (an assumption that belongs to ideal circuit theory)? If not, then keep in mind that you've formed a loop that bounds a surface with non-zero area which means that the self-inductance of this loop dominates (since the resistance is zero), i.e., the (ideal) battery will see an inductance rather than a zero ohm wire. $\endgroup$ May 25, 2020 at 13:48
  • 1
    $\begingroup$ @AlfredCentauri I am obviously happy with them. $\endgroup$
    – satan 29
    May 25, 2020 at 17:32
  • 1
    $\begingroup$ satan 29, I've updated my answer to address both your comments to my answer and your recent edit. $\endgroup$ May 25, 2020 at 19:34

3 Answers 3


After a few milliseconds or so, the net field inside the wires becomes zero. (There exists no field in the material of a perfect conductor).

There exists no field in the material of a perfect conductor at steady state. In your scenario, there is no steady state to reach. A field will remain in the conductor indefinitely. The charges continue to accelerate and current grows without bound.

  • $\begingroup$ How exactly can we conclude that the net field in this case never becomes zero? $\endgroup$
    – satan 29
    May 25, 2020 at 8:59
  • 1
    $\begingroup$ Because the wire is connected to an ideal voltage source. There must be a field between the voltage source or there would be no voltage. In a normal circuit, the current would increase so that 100% of the voltage drop occurs outside the zero-resistance wire. In your scenario, that is impossible, so the field must remain within. $\endgroup$
    – BowlOfRed
    May 25, 2020 at 9:00
  • $\begingroup$ Um, What i meant was "net field". The field that the voltage source provides gets nullified by the rearrangement of the charges In a perfect conductor, at steady state. $\endgroup$
    – satan 29
    May 25, 2020 at 9:03
  • $\begingroup$ And you are claiming that this doesnt happen here. $\endgroup$
    – satan 29
    May 25, 2020 at 9:06
  • $\begingroup$ I have edited the question to explain what i mean by the "net field becoming zero" $\endgroup$
    – satan 29
    May 25, 2020 at 9:07

If you're imagining that this hypothetical system is extended in space, i.e., that the wire connected to the battery forms a loop bounding a surface with non-zero area, then your transient analysis must include the self-inductance (since you've stipulated that the resistances are zero, you cannot ignore the self-inductance even if it is very small).

The full battery voltage will exist across the ends of wire but the electric field inside the conductor will be zero. Why?

If there is a changing current through the wire, there is a changing magnetic flux threading the loop formed by the wire and battery. This means that there is an induced electric field (associated with the changing magnetic flux) within the wire.

Because the perfect conductor does not allow a net electric field within, it must be that the electric field due to the charge separated onto the terminals of the battery is precisely cancelled (within the wire) by the induced electric field associated with the changing magnetic flux.

Thus, the current must change at precisely the rate consistent with the condition that the electric field within the conductor is zero.

Since there is no resistance in your hypothetical system, the current must start increasing from zero at $t=0$ and continue increasing linearly with time (until something breaks).

Now, you might wonder: how can the current (charge flow) increase when there is no electric field inside the wire? Recall that you stipulated that the wire is perfectly conducting and this implies that the charge carriers can instantaneously redistribute within the conductor such that the field inside is zero. Sure, it's not physical but it's your hypothetical.

Update to address comments from OP:

First, OP stipulates that the conductor is perfectly conducting, zero resistance. This may seem redundant but, in fact, there's subtle difference.

A perfect conductor has, by definition, zero electric field inside always. Here are two descriptions I've found that support this:

Let us consider for a moment what time dependent EM fields look like at the surface of a ``perfect'' conductor. A perfect conductor can move as much charge instantly as is required to cancel all fields inside.



The charges inside a perfect conductor are assumed to be so mobile that they move instantly in response to fields, no matter how rapid, and always produce the correct surface charge density in order to produce zero electric field inside the perfect conductor


OP is understandably perplexed at how there can by any electric current at all through a perfect conductor if there is always zero electric field inside.

Indeed, it seems contradictory to even state that charge would instantly redistribute to keep the electric field zero because there seems to be a chicken and egg situation - what, if not a non-zero electric field, is the cause for the redistribution of charge (which is in itself a current - a 'screening' current)?

This is a good question and the best way I can think of (at this time) to ask one to think about it at this introductory level is to use a limit process. Start by thinking about this with some characteristic time constant that takes into account the inertia of the charge carriers such that the redistribution of charge is not instantaneous.

See that, even with zero resistance (no voltage required to sustain a current through), such a conductor is not a perfect conductor. Finally, take the limit as this time constant goes to zero and see that this does not imply that the current must be zero.

Alright. So , in this situation, The charges redistribute instantaneosly, nullify the battery field, and so the current should be zero! No current so there's no magnetic field , and thus no flux, and thus no induced electric field.

I wanted to address this here rather than in the comments. The above electrostatic situation is not a possible solution. Consider:

(1) There is a non-zero voltage across the source terminals

(2) The wire is connected across the source terminals

(3) There is zero electric field through the wire

Now, if I integrate the electric field within the wire from one terminal to the other, the result is zero by (3) above. Thus, there is zero potential difference between the ends of the wires.

But this contradicts (1) combined with (2), i.e., there is a potential difference between the ends of the wires.

So this isn't a possible solution (and there is no electrostatic solution).

  • $\begingroup$ Wait: I am now confused after reading your last paragraph. For a Theoretically perfect conductor, If the charges redistribute instantaneously, then....We have a net field =zero inside the wire,at t=0. So why is there any current at all? The last para raises this question but doesnt answer it... $\endgroup$
    – satan 29
    May 25, 2020 at 17:29
  • $\begingroup$ And , by the way, Are "zero-resistance" and "perfectly conducting" synonymous? $\endgroup$
    – satan 29
    May 25, 2020 at 17:32
  • $\begingroup$ @satan29, almost. The point I made above about the charge carriers instantaneously redistributing is the key, i.e., the charge carries have no inertia. Think about that a bit before replying. $\endgroup$ May 25, 2020 at 17:43
  • $\begingroup$ Wait: charge carriers redistributing instantaneously would imply that they are massless?? $\endgroup$
    – satan 29
    May 25, 2020 at 17:46
  • $\begingroup$ That surely shouldn't be implied If we assume a perfect conductor. $\endgroup$
    – satan 29
    May 25, 2020 at 17:49

If you have no resistance in the conductor, all electrical charge carriers from battery are pushed away. As the battery is ideal, we can assume there is an infinite number of charge carriers. So $n \rightarrow \infty$, hence $dq \rightarrow \infty$ and you will get infinity current as follow from Ohm's law.

  • $\begingroup$ Martin, I've downvoted your answer because it doesn't seem (to me) to address the essential question posed by OP, and because your conclusion that the current is infinite isn't justified $\endgroup$ May 25, 2020 at 15:10

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