Short circuit analysis (theoretical)

Question

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.

Answer 1

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.

Answer 2

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.

Credit

and

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

Credit

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).

Answer 3

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.