# Why is the electric field across a perfect wire zero? [duplicate]

I’ve looked at the answers given to the previous times this question has been asked, but I still don’t seem to understand how this holds in the case of a closed circuit. Here’s an explanation given before:

“Think of the wire as a horizontal cylinder. If you apply an electric field pointing to the left, the electrons in the wire will move to the right, so that eventually they collect on the right side, and there is a deficit of electrons on the left. This distribution of charge (positive on the left, negative on the right) produces a field of its own, pointing to the right, which works against your applied field. This process will continue, until there is no net field left inside the conductor; the equilibrium is reached once there is no more field and thus the electrons experience no net force.”

This makes enough sense to me if we’re talking about a cylinder, but not a closed loop. Isn’t the whole point of an emf source in a circuit to prevent this sort of cancellation of fields? Instead of allowing electrons to clump up at the positive terminal of a battery, the battery “forces” the charges to the negative terminal to repeat the another cycle through the circuit, so how is it that the electric field in the conducting material of the wire has to necessarily be equal to zero?

I think this also may comes back to a misunderstanding I have about resistance. I’ve always thought of it as this sort of hand wavy property of a material that predicts the ratio of the potential difference through it to the current that runs through it. What part of this property actually allows an electric field to exist to establish a potential difference in a material?

Maybe there is confusion between the static case and the non-static case? The first statement applies to a static situation for any conductor of finite conductivity. For example, if a closed copper circuit is placed between the plates of a capacitor, the charges will move and finally the electric field within the copper will be zero. In the presence of a generator, it is no longer static but possibly in steady state. In this case, the field is nonzero unless the conductivity is infinite.

You can think of voltage as the force the battery is applying to force charges through other things. Ideal wires don't resist the movement of charge at all, so none of the voltage is wasted in the wire: all of it will be applied to whatever other elements are in the circuit, which actually require something to push charge through them.

If you have just a loop of ideal wire, a current can be maintained in that loop without the need of a battery. A battery will cause the current to be infinity (if we're ignoring the inductance of the wire) or to increase over time without bound (if we are not).

• Why is it that none of the voltage is wasted though? Regardless of the presence of resistance in a wire, shouldn’t the work done on an electron by the electric field be Force*distance, implying a voltage drop? Nov 23, 2017 at 19:02
• @Elmer Only if you have an electric field in the wire. In the steady state, all the current moves at a constant velocity through the wire with no need for an electric field. So no electric field means no force means no work.
– Chris
Nov 23, 2017 at 22:07
• How is it possible to create large currents in superconductors then, if no electric field can exist in them? What makes that scenario different from that of a simple circuit with a single resistor and battery? Nov 25, 2017 at 3:12
• @Elmer To be clear, no electric field can exist in an ideal wire, if by ideal we mean "has no impedance of any kind." A superconductor has no resistance, but it still has inductance, and so electric fields can exist in a superconductor. But that's beside the point. The point is that ideal wires can have large currents without electric fields. The electric field is just unneeded. The main difference with superconductors if that you need a battery to start a current, but once you have a current, you can turn the battery off and the current doesn't go away.
– Chris
Nov 25, 2017 at 5:03
• How can larger current be created? Wouldn’t there have to be a field at some point in a superconductor to “accelerate” the current to some desired high number? And how is it that current would start at all for that matter when an ideal wire is connected to a battery? What causes the charges in the material to flow from there electrostatic equilibrium? Nov 25, 2017 at 7:24

If you are imagining a circuit as an ideal voltage source and ideal wires only, then you're correct there's a problem. One component is guaranteeing a voltage difference and the other component is guaranteeing zero voltage difference.

In practice, this isn't a concern. We don't construct circuits consisting only of voltage sources and low-resistance wires. If you did, either the wire or the voltage source would fail in some way.

• But what difference would it make if you added a few resistors to the simple circuit? What physically happens that confines the voltage drops in the circuit to those resistors? In my mind, all that’s necessary for a drop in potential is for the electrons to move any distance along the circuit with an electric force acting on them. Nov 23, 2017 at 21:49
• @Elmer In the case of zero resistance no force is needed to move the electrons so no work is done and there is no potential difference. Jan 6, 2019 at 11:49

My understanding is that the electric field inside a perfect conductor is always zero, even if there is current flowing inside the conductor. Let me try explain.

A loose definition of a perfect conductor could be a medium where:

• Electrons can freely flow without loss of kinetic energy (due to friction/collision effects)
• There are an infinite number of electrons of infinitely small size/mass with infinite space to move.

This, in turn, means that all electrons can be accelerated infinitely fast and to an infinite velocity.

As described in your open-loop example, the electrons will rearrange themselves as to "counteract" the applied electric field. In an ideal conductor, this happens infinitely quickly as the electrons are accelerated infinitely fast due to their negligible mass. This, in turn, means that the resultant electric field is zero, as the new arrangement of the electrons results in an equal but opposite electric field.

Now, let's look at your closed-loop example but with a non-perfect conductor with finite resistance. When the loop is closed, the electric field will propagate through the wire. Electrons will accelerate accordingly, thereby creating an opposing electric field due to the shift in the charge configuration. This opposes impressed electric field and decreases the net field in the conductor.

If the resistance is high, it is "difficult" to accelerate electrons due to collisions with other electrons making them lose their energy quicker. The resultant E-field will therefore be relatively high (and there will be a large voltage drop).

When working with a perfect conductor, the infinite number of electrons will immediately rearrange themselves in attempt to counteract the applied electric field. This arrangement happens instantaneously. The system now enters some sort of equilibrium, where the electrons are in continuous motion and the infinite supply of new electrons that replace the "old" ones leaving the conductor are also accelerated, and the process repeats.

This is just my understanding, and I believe it is sometimes hard to quantify these things when working with "perfect", idealized components. Nonetheless I hope my explanation was helpful in some way!