The standard explanation in textbooks goes that in the presence of electric field (e.g. external electric field) the free electrons inside the conductor will keep moving until electrostatic equilibrium is reached and that they will create a field which will cancel out external field (e.g. electrons move to one side leaving positive charge on another side, this creating electric field that cancels out the external field).

But what if the external field is so strong that there is just not enough free electrons to cancel out external field? What if there is nothing left to be moved? E.g. I have electric field lines going from left to right, all free electrons move to the left surface, there is nothing left to move, but it's still not enough to cancel out external field. Will there still remain an electric field inside the conductor then? If not, what will happen?

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    $\begingroup$ Related (to the title at least): Veritaseum (a physicist and educator) made a video about the surface-charge model of electric fields in circuits, when current is flowing, creating electric potentials that match the resistive voltage drops through wires. It's quite interesting: youtube.com/watch?v=oI_X2cMHNe0 - the main point of the video is that it explains how soon a switch closing will be seen by an electrically distant but physically close wire, where the electric field starts current flowing in the other part of the wire before speed-of-light time around the wire $\endgroup$ Sep 5, 2023 at 10:26
  • $\begingroup$ Are we only considering a static (non-changing) electric field, or one that may change over time (possibly very quickly)? $\endgroup$
    – user4574
    Sep 5, 2023 at 21:20

4 Answers 4


The "nothing left to move" scenario is impossible with a metal: the electrostatic potential needed to deplete a metal is far beyond what is sufficient to destroy any experiment.

It is, however, possible in a semiconductor, where the density of charge carriers is low enough. The silicon technology that is everywhere these days is based on manipulating depletion regions in silicon crystals, So the answer to your question "what will happen" is that you make something like a diode or transistor, depending on the geometry and the applied potential. It's not as simple as applying a potential and sweeping the charge out: you need a way, with doping, an insulated gate, or a Schottky barrier, to prevent replenishment of the charge via whatever electrodes are applying the potential.

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    $\begingroup$ Thanks, the first paragraph answers my question. Just curious, would would be approximate magnitude of electric potential to deplete a metal? For the 2nd paragraph, i m still to learn this topic :) $\endgroup$
    – Yevgeniy P
    Sep 5, 2023 at 11:05
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    $\begingroup$ Went back and read your other answers. Fantastic stuff. What a deep knowledge of physics you possess. $\endgroup$
    – duffymo
    Sep 6, 2023 at 12:51
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    $\begingroup$ @duffymo Well, you do this stuff long enough, you learn a few things. It's been a fun half century. $\endgroup$
    – John Doty
    Sep 6, 2023 at 12:56
  • $\begingroup$ It shows. Well done. I'm a mere engineer by education. $\endgroup$
    – duffymo
    Sep 6, 2023 at 13:00
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    $\begingroup$ $V/A$ is approximately the thickness of the material. For a 10cm thick piece of copper @Physicist137's equation works out to $153*10^{18} V/m$, or 15 million trillion volts across the piece. $\endgroup$
    – Austin
    Sep 6, 2023 at 21:08

The macroscopic electric field is zero if and only if the conductivity of the medium is infinite. The microscopic electric field is not assumed to be zero but that has no direct relevance to the macroscopic Maxwell's equations. If the conductivity is not assumed to be infinite than both dc and ac fields, voltages/currents can exist inside, see, "skin depth". For infinite conductivity the field intensity is zero despite the fact that there could be a finite current, this is assumed to be the case between the RLC elements when analyzing a circuit using Kirchhoff's equations.

Of course, this is not really physical, just as having infinite conductivity is not physical at least in classical physics (exception noted: superconductors at dc). Said differently, that the internal field intensity is zero when the medium has infinite conductivity is an idealized boundary condition, just as flat surfaces between homogeneous dielectrics and these are no different in principle from assuming that when lifting a weight the Earth stays at the same place without changing its gravity.


The electric field inside a conductor is $0$ only in case of electrostatic conditions.

Consider an external electric field which is making electrons flow in a loop, they cannot now move to attain the electrostatic equilibrium, so no cancellation of electric field takes place. This is what happens in an electric circuit.

If the charges are moving charges, then EF is not $0$.


I was writing a more complete comment, but it became big. So I am going to write an answer instead.

It is not a difficult to calculate what's the field you would need to deplete a metal. If the carrier density is $n$, and the bulk volume of $V$, then, the total charge of the carriers inside is $Q = \rho V = neV$, where $e$ is the charge of the electron (the carrier inside). If you apply an electric field, some of it will go to the surface, which has an area $A$ and an induced density $\sigma$, thus, $\sigma A = Q$. So, the depletion field is:


It was used that $V/A\approx d$, is approximately the thickness of the plate (thanks @Austin).

Take copper which has one valence electron, $n\approx 8.47\cdot 10^{28} m^{-3}$. Take $d = 10cm$, so, a 10 centimeter thick plate of copper, then: $E\approx 1.53\cdot 10^{20} V/m$. Notice this is bigger than the Schwinger limit, thus classical electrodynamics no longer work (maxwell's equations are invalid). So, instead of using Maxwell's Electrodynamics (which is just an approximation for low fields), you would need a more accurate non-linear form of electrodynamics, in this case, Euler-Heisenberg Electrodynamics to calculate what would happen with fields that high.

And all of this doesn't take into account what it would do to the metal itself.. or cold electron emission effects (assuming vacuum), or dielectric breakdown (assuming air), etc.

Of course, the ratio is proportional to the thickness, thus, if you increase $A$ and decrease $V$, say, a really thin sheet of metal, you get a smaller depletion field, but, the fields are still high.

  • $\begingroup$ One question though, as far as i understand n is number of electrons per m^3. But sigma uses Coulombs as far as i know. Can we equate those? Sorry, i m new at this. $\endgroup$
    – Yevgeniy P
    Sep 10, 2023 at 11:25
  • $\begingroup$ @YevgeniyP You're right. There was a mistake. I fixed my answer. Thanks! $\endgroup$ Sep 12, 2023 at 15:31

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