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

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? – Elmer Nov 23 '17 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 '17 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? – Elmer Nov 25 '17 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 '17 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? – Elmer Nov 25 '17 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. – Elmer Nov 23 '17 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. – my2cts Jan 6 at 11:49