We have an uniform electric field E in the vertical direction andwe have a rod of superconducting material in the field. Since there is an electric field there will be a current in the material till the field inside the material becomes 0, at which point it stops.

Since the material is superconducting it should not produce any heat. However if I take the same scenario for a conducting slab (by slab I mean that the seperation between the vertical ends is much smaller than the horizontal expanse), we can do an approximate calculation of the heat loss due to the current as follows:

Since the separation between the vertical end points $l$ is small, we can approximate the ends as a near infinite plate with area A. The electric field between the ends induced by charge accumulation at the ends is $\sigma/\epsilon$ where $\sigma$ is the surface charge density. The current will keep flowing till this $\sigma/\epsilon$ field cancel the external field E. Assuming resistance $R$, the current is $(E - \sigma/\epsilon) l/R$. This current builds up the charge density $\sigma$ at the end terminals giving $$ \frac{(E - \sigma/\epsilon)l}{R} = I = A \dfrac{d\sigma}{dt}$$

Similarly the heat produced is $$ \dfrac{dQ}{dt} = I^2R = \frac{(E - \sigma/\epsilon)^2l^2}{R} = (E - \sigma/\epsilon)lA \frac{d\sigma}{dt} $$

From this we see that the heat loss is independent of R, which is surprising as that means even with a superconductor (R=0) there would be some heat generated.

This seems to be in opposition to the idea that current in superconductor wont cause heat loss. So I want to make sense of this result. Will heat be generated or not? If no, what is wrong with the calculation? If yes, how given R = 0? Does it happen via some other mechanism?

  • $\begingroup$ Nice, but can you add an actual question? Like "am I correct?" or "is this good approach?" or "what is the deeper explanation of this?" Otherwise people don't know what do you want them to tell you. $\endgroup$
    – user46147
    Apr 26, 2023 at 17:41
  • $\begingroup$ thanks!! i just want to make sense of the result. edited the question $\endgroup$
    – ssj009
    Apr 26, 2023 at 17:44

2 Answers 2


I think the short answer is that you also need to think about the magnetic field, as Penguino alluded to, and other loss mechanisms like radiation.

Suppose that the time variation is slow enough that we can make the quasi-static approximation when dealing with magnetic fields, i.e. changing E-field do not considerably affect B-field. Writing Faraday's law along the blue loop and using Ohm's law $E_\text{inside}=\frac{RI}{\ell}$, $$RI+(V_+-V_-)=-\frac{d\Phi}{dt}=-L\frac{dI}{dt}$$ where $\Phi$ is the magnetic flux through the loop and $L$ is the self-inductance of the slab, defined by $\Phi = LI$. With $V_+ - V_-=(E-\sigma/\epsilon)\ell=E\ell-\frac{\ell}{\epsilon A}q$, $$RI-\frac{\ell}{\epsilon A}q+L\frac{dI}{dt}=-\ell E.$$ Note that we recover your equation for $L=0$ and $q=\sigma A$.

At this point I'd like to propose an analogous problem that is easier for me to reason through: a series RLC circuit.

The loop equation is $$RI-\frac{Q}{C}+L\frac{dI}{dt}=0 $$ which is the same as the above equation if we let $C=\epsilon A/\ell$ and $Q=q$, apart from a static "background solution" of $q = A\epsilon E$ in the original problem. If $L=0$, the same apparent paradox arises as $R\rightarrow 0$.

When the non-zero inductance is considered and $R=0$, instead we have an LC circuit that never reaches steady-state where the current oscillates forever with frequency $1/2\pi\sqrt{LC}$. However, even in a circuit with no ohmic resistance, there would at least be radiation losses that can be modeled by a radiation resistance: this would eventually cause the current to decay to zero. In an actual superconductor, there might be other loss mechanisms in the presence of transient fields that would result in heat generation. Perhaps someone more familiar with superconductivity can chime in on that.

  • $\begingroup$ Yes - there are other loss mechanisms in actual superconductors. Not only is there a limit to the current they can carry, but time-varying currents do typically result in heat being deposited into the material. In addition, the original post doesn't consider the time over which the electric field rises to its steady-state value, or the manner in which the slab is inserted into it - which would have a bearing on what happened. $\endgroup$
    – Penguino
    Apr 27, 2023 at 1:57
  • $\begingroup$ So I guess the answer to heat generation is yes; and even with superconductors there will be some loss not associated with the operating device. I guess all resistance is usually combined into a single representative R, and ohmic resistance is the big component; but with superconductors we do need to take other things into account. But interestingly, the heat is independent of R $\endgroup$
    – ssj009
    Apr 27, 2023 at 15:21
  • $\begingroup$ @ssj009 Yes, the energy loss is just the capacitor energy: it is independent of $R$. However, if the radiation resistance is not negligible, not all the energy is dissipated as heat, it is partially radiated with the slab acting like an antenna. I don't know enough about semiconductors to say how the ohmic resistance would compare to the radiation resistance. $\endgroup$
    – Puk
    Apr 27, 2023 at 15:56

It isn't a valid conclusion to make for superconductors. At the R=0 limit, you would predict that charge transfer results from a nominally infinite current acting for za nominally zero length of time. That is clearly non-physical for a number of reasons, for example:

  • (Special relativity) This doesn't allow for an instant displacement of charge
  • (Maxwells equations) The current would generate an infinite magnetic field
  • (Quantum physics) Action completed in less than the Plank time
  • (Mathematics) Despite your cancelling of R, it looks suspiciously like a case of dividing by zero.

In practice, you would have to look at a number of things - I would start with the self-inductance of the plate resulting from any flowing currents.


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