# Capacitor in a superconducting circuit

Assume we have started a current in a closed circuit which only includes a capacitor. No inductor. For example by having a battery initially, disconnecting and and reclosing the circuit.

For a dissipative conductor, the current would decrease with the timescale of the inverse scattering rate.

What if the circuit were superconducting? Would there be an oscillation of the current with charge sloshing back and forth each plate of the capacitor? At which frequency?

• While you can specify that there is no inductor, you can't actually specify that there is no inductance. Since the circuit encloses a non-zero area, there is necessarily self-inductance even if you assume an ideal capacitor (physical capacitors also have non-zero inductance, i.e., they are self-resonant). Also, you have to look at the radiation resistance which is also non-zero. It could be that the radiation resistance dominates or it could be that the self-inductance dominates. See this related problem. – Alfred Centauri Nov 28 '17 at 22:47
• I know it's not realistic, but what if there were no inductance? So just considering electrostatic repulsion, inertia of the charges and charge accumulation on the plate of the capacitor? – SuperCiocia Nov 28 '17 at 23:07
• "what if there were no inductance?" - what if your capacitor has no capacitance? Honestly, I don't get this line of thought. You're free to stipulate that the self-inductance is insignificant but then you have to check whatever solution you come up with against that stipulation. – Alfred Centauri Nov 29 '17 at 0:29
• OK man thanks. I could say what if the the loop is infinitely big so for the magnetic flux to be ~0, of what if the we twist the wires connecting each plate of the capacitor into a twisted pair so as to have no cross sectional area and hence minimal self-inductance. That's what I mean with no inductance. I mean with the capacitance being the dominant effect. I was actually looking for something along the lines @John Rennie below & his plasma frequency argument. – SuperCiocia Nov 29 '17 at 12:35

I have to disagree with Chris and say there is no reason to require a non-zero inductance to be present (though of course in practice there will always be a non-zero self-inductance). Electrical signals do not propagate instantaneously, so it's perfectly possible to get a stable oscillation even with zero inductance. Your system will oscillate at the plasma frequency of the conductor.

However the capacitor is not really playing any special part in this. Any chunk of metal can sustain plasma oscillations as they are a fundamental property of any electron gas.

• True- it all depends on what non-ideal properties of the superconductor you are assuming dominate. Realistically I suppose you'd likely exceed the critical current density of a real superconductor. And naively I'd assume at the plasma frequency that radiation losses would be large- what is the ballpark plasma frequency for a superconductor? – Chris Nov 29 '17 at 11:13
• @Chris that's a good question and I don't know the answer. For non-superconducting metals the plasma frequency is of the order of optical frequencies - just slightly lower. – John Rennie Nov 29 '17 at 11:31
• Yes this is excatly what I wanted. So there will be an oscillation anyway. Which will die because of radiation resistance I guess? – SuperCiocia Nov 29 '17 at 12:38
• Hello? How does radiation loss come into play? – SuperCiocia Jan 18 '18 at 0:10
• @SuperCiocia to be honest I don't know how to calculate the radiation losses. I guess the system is in effect an aerial so aerial theory would apply. – John Rennie Jan 18 '18 at 5:14

To take your question at face value: what if we somehow connected the two ends of a charged capacitor with an ideal wire, that somehow avoids having self-inductance or radiation losses or anything?

Well, we can write a differential equation using Kirchoff's voltage law: $$\frac{Q}{C}=0$$

Whoops! That's nonsense- our initial conditions are not $C=\infty$ or $Q=0$, so we have a contradiction. Which means our assumptions are bad. Basically the problem we run into is that you can ignore the inductance of a wire when the circuit is dominated by other effects, but the more ideal you make your circuit components, the more whatever inevitable non-ideal properties the other components have stand out.

In particular, any loop of wire necessarily has a self-inductance. If the self-inductance is $L$, ignoring radiation losses, then our equation becomes the much more sensible $$L\ddot{Q}+\frac{Q}{C}=0$$

which gives the solution $Q(t)=Q_0\cos(\omega t)$, where $\omega=\frac{1}{\sqrt{LC}}$.

For small radiation losses, this formula is still approximately correct if $Q_0$ is replaced with a slowly varying function of time.

So, yes, in the limit that the radiation losses are small, the charge does slosh back and forth, with a frequency determined by the size of the inductance.

As John mentions in his answer, this is by no means the only possibility. While you can never decrease the inductance to zero, you can make it small enough that some other effect takes over.