Dependence of current on eletric field and vice versa in a circuit

Question

I don't understand the notion "Electric field causes current in a conductor (metal wire)". Many textbooks say that free electrons in a metal wire are moved by elctric field (electrostatic?) - thus forming electric current.

My understanding

1. If there is any electrostatic field outside a conductor, within that conductor that field is zero (= absent).

2. Current in a wire appears only when the circuit is "closed", but electrostatic field exists even if we have only one (say, only positive) charge (positive plug of a power battery), we do not need two charges-sources-of-field (positive and negative).

3. Positive plug of a power battery is a source of electrostatic field (it is charged positively - it is like a "point charge" creating electrostatic field). If we connect a metal rod to it (circuit is not closed) there would be no current in that rod, though we have electrostatic field (positive battery plug) and free electrons within the rod.

4. This "mysterious" electrostatic field that moves electrons in a closed-circuit wire is a composition (superposition) of two separate fields created by two "point-charges" - positive and negative plugs of a car battery. I have problems trying to visualize such a resultant field...

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Experiment

I put my metal rod exactly between positive and negative plugs of a car battery, but with a millimeter gap (rod not touching both plugs). There is almost same electrostatic field - same configuration=places in space of field-source-charges (positive and negative plugs), but there is no current. If the rod touches terminals however then there is current.

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Questions

1. How was there eletric field in my rod without current?

2. How are there current in superconductors without eletric field?

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Preliminary research

This discussion adds to my confusion - in superconductor (ideal conductor) current exists without any electric (electrostatic?) field. And electric field causes current only in non-ideal conductors (not superconductors).

Interesting insight is given here, it attributes electrostatic field inside the conductor to surface charge gradient all along the metal wire.

Answer 1

Most of what I’m going to say can be found on page 297 in Introduction to Electrodynamics by griffiths.

1. The condition that there is no electric field in a conductor/metal is only true in case of electrostatics, i.e., no movement of charges. As you stated, the basic idea is that due to the presence of free electrons, the charges within a metal can rearrange themselves in a way that cancels out any electric field that was supposed to be present in the conductor.

2. When you bring a metal close to the positive end of a battery, it may be true that the positive end is at a higher potential than the metal and so you expect an electric field near that end. The effect this has on the metal is again just to rearrange the electrons in it so that the net electric field in the conductor cancels out. Note that no charges can leave the metal to cancel out the excess positive charge on the positive terminal of the battery as overall the metal must remain neutral.

3. The thing that changes when you also connect the negative end to the other side of the metal is that now you can have a flow of electrons into and out of the metal. The electrons cannot remain in their static rearranged form to cancel out the the external electric field; the positive end of the battery attracts electron out of the metal and the negative end pushes electrons into the metal to compensate for the loss of electrons so that overall the metal is neutral at any given time. The speed at which this flow of electron through the metal happens is determined by its conductivity. The current density $J$ is proportional to the force per unit charge $f$ by $$J=\sigma f$$ where $\sigma$ is a material constant called conductivity. In case of the force being provided by the battery, this $f=E$ for the electric field generated by the battery across the metal wire so we get $$J=\sigma E$$For a material with finite $\sigma$, you can see that you need a finite $E$ to get a current. However in case of superconductors, $\sigma \to \infty$ and $E=J/\sigma \to 0$ so we can have a finite $J$ even though the electric field is 0. Physically all this means is that for superconductors, there is no resistance to the current flowing through the material. Since there is no resistance to the current, you don’t need an external force (like an electric field) to push your electron through the material.

I hope this answers the part of your question on “what happens when you connect a metal wire across the ends of a battery and how it generates an electric field”