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Whenever I see a diagram of a circuit with perfect conductors (e.g usually the wires), the electric field is perpendicular to the conductor. I have learnt that this is because the charges make their way to the surface, so that the electric field is 0 inside of the conductor.

From my understanding, normally, a typical conductive material such as copper, gold, aliminium has resistance which means that means practically this is not possible, and therefore there remains a field within the conductor.

However, in a superconductor there is no resistance, so do the charges make their way to the surface like in the perfect conductor, or does something else happen on a more deeper level. I am assuming the super conductor is in an electro-static state. So for example a piece of wiring put in extremely cold temperatures, since this would mean there is no current flow. Furthermore, if I was drawing one of these conductors within a circuit, do I point the electric field perpendicular to the wire?

EDIT: I am still in search of any texts / diagrams to do with this specific question. I can see there is no resistance, but I cannot find any articles / papers about what happens to the free moving electrons? Do they move to the surface in a super conductor, like they do in an ideal conductor?

FURTHER QUESTION: I think what I am trying to ask, is what would happen if I got a material that could be used as a super-conductor and froze it at extreme temperatures, so it becomes a super-conductor. Would the charges make their way to the surface. In a perfect conductor for example, they'd balance out to make their way to the surface.

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  • $\begingroup$ ‘ copper, gold, aliminium has resistance … therefore there remains a field within the conductor.’ Only if there is a current. $\endgroup$
    – my2cts
    Commented Jul 30 at 22:22
  • $\begingroup$ @my2cts I thought that it is the resistance in the static copper wire which causes internal fields, for example, that stop all of the charges making their way to the surface like in a 'Perfect conductor' $\endgroup$
    – InexperiencedCoder
    Commented Jul 31 at 15:40
  • $\begingroup$ In a conductor ${\vec j} = \sigma {\vec E}$. So ${\vec E}=0$ implies ${\vec j} =0$. $\endgroup$
    – my2cts
    Commented Jul 31 at 15:54
  • $\begingroup$ @my2cts That's internally though? There's still charges on the surface, which have electric fields with a vector perpendicular to the wire. - I may be getting myself confused now with ignorable details. $\endgroup$
    – InexperiencedCoder
    Commented Jul 31 at 17:29
  • $\begingroup$ @InexperiencedCoder On the surface, the Ohm law $\vec{j} = \sigma \vec{E}$ does not hold. So only inside the conductor, zero current implies zero electric field and zero charge density. On the surface, we can have zero current and non-zero electric field and non-zero charge density. $\endgroup$
    – Ján Lalinský
    Commented Jul 31 at 18:04

1 Answer 1

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Yes, regarding electric field and charge density, superconductor in DC context behaves as a perfect conductor. Charge density inside vanishes, and can be non-zero only on surfaces.

This follows from the basic model of superconductor - the Londons' equations. The first Londons' equation is

$$ \frac{\partial \mathbf j_c}{\partial t} = k\mathbf E, $$ where $\mathbf j_c$ is conduction current and $k$ is a (medium-dependent, possibly temperature-dependent) constant. In a situation where the superconductor has constant DC current, $\mathbf j_c$ is constant in time, and so according to this equation, electric field inside vanishes. Using the Gauss law, it follows that net charge density inside vanishes.

Superconductors do not behave as a perfect conductor when it comes to magnetic field and current. Superconductors of type I (the first discovered type, requiring very low temperatures), e.g. tin or lead, manifest the Meissner-Ochsenfeld phenomenon. This means the superconductor rejects magnetic field completely from its insides, so magnetic field inside vanishes. Magnetic field can be non-zero only in a thin layer on the superconductor surface. A theoretical perfect conductor has no such property, it can have any magnetic field inside.

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  • $\begingroup$ Is there any sources for this / diagrams you recommend? $\endgroup$
    – InexperiencedCoder
    Commented Jul 30 at 21:12
  • $\begingroup$ @InexperiencedCoder I've added some explanation. $\endgroup$
    – Ján Lalinský
    Commented Jul 30 at 23:00

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