(Physics) Capacitors and superconducting wires and oh my [finished]

L|ama

Two identical capacitors of arbitrary capacitance C are charged to voltage V, and then connected in parallel with superconducting wire.

a) Calculate the total energy before and after the capacitors are connected

b) Is energy conserved?

c) Account for the energy of the system before and after the capacitors are connected


My thoughts:

a) energy stored beforehand = .5*CV^2 in each capacitor, CV^2 total. After they are connected, the charge on each plate flows through the wire and equalises, since the capacitors are identical they have the same charge on them and it exactly cancels out, so there is no voltage over either capacitor and so the energy is zero in each of them.

b) Yes, it always is

c) This is where I'm stuck. I first thought that the energy would be used up in making the charge flow, but since it's a superconducting wire there is zero resistance. I also thought that it depends on how the capacitors are connected - if the negative plate of one is attached to the negative plate of the other and positive to positive, the charge can't equalise and so it will just spread out uniformly along each half of the circuit, but the question doesn't specify how they're connected.

dispatch.o

L|ama wrote: »

Two identical capacitors of arbitrary capacitance C are charged to voltage V, and then connected in parallel with superconducting wire.

a) Calculate the total energy before and after the capacitors are connected

b) Is energy conserved?

c) Account for the energy of the system before and after the capacitors are connected


My thoughts:

a) energy stored beforehand = .5*CV^2 in each capacitor, CV^2 total. After they are connected, the charge on each plate flows through the wire and equalises, since the capacitors are identical they have the same charge on them and it exactly cancels out, so there is no voltage over either capacitor and so the energy is zero in each of them.

b) Yes, it always is

c) This is where I'm stuck. I first thought that the energy would be used up in making the charge flow, but since it's a superconducting wire there is zero resistance. I also thought that it depends on how the capacitors are connected - if the negative plate of one is attached to the negative plate of the other and positive to positive, the charge can't equalise and so it will just spread out uniformly along each half of the circuit, but the question doesn't specify how they're connected.

This may be a physics question, and I couldn't answer it. It sure doesn't seem practical though. If they were connected negative to positive they would discharge immediately, and if they were positive to positive nothing at all would happen.

L|ama

Well it's not quite instantaneous, but on the order of nanoseconds, and if the wires weren't connected originally the excess charge would spread out along the wire too to become uniformly distributed.

ProPatriaMori

I heard about this in an intro to EE class. Pretty sure the energy leaves as EM. Or you could use up some energy blowing the capacitors.

L|ama

Our lecturer prefaced it with this



and said it was easier than that, so at least it's not a fourier transform.



Also thinking about the +/+ plates being connected or +/-: + connected to + would be parallel and + to - would be series, wouldn't it?



edit: Could it just be that the energy goes into the magnetic field caused by a moving charge? But then where does that energy go once the current hits zero...

I'm currently tossing up between that and huge current explodes the capacitors.

CycloneRanger

Assuming the capacitor structure is also superconducting, you'll get a great big magnetic field, which will soon induce a great big current in the wire, which will recharge the capacitors, which will discharge creating a big magnetic field...

Eventually induced currents in something nearby will damp the system to zero potential. That, or it'll radiate its energy as photons.

That's my guess based on a minute or two of Googling. It's also possible that the current may rise high enough to create a magnetic field powerful enough to destroy the superconductivity of the material.

shadydentist

Although the superconducting wire carries no resistance, any circuit will still have some inductance, which limits the rate at which the current in the wire can change. The powered stored by the magnetic field in a coil is equal to (1/2)LI^2, where L is the inductance and I is the current.

So, thats where the energy goes, at least. As for radiation, I'm reasonably certain that in a closed loop, energy lost to radiation is negligable at any reasonable distance, as it falls off much faster than distance squared. Of course, I'm an engineer, not a physicist, so we're trained to ignore things that probably won't affect the result much.

Letarian

If the two capacitors are joined in parallel, then the two negative plates are joined to eachother, and the two positive plates are joined to each other.

Yes energy is conserved, and the total energy is CV^2^ (i.e. 2 x 0.5CV^2^).

I believe part 3 is refering to the fact that when you connect capacitors in parallel, the total capacitance is:
C(T) = C(1) + C(2) +.... etc
Therefore the total capacitance of your system is C + C = 2C

The thing about the equivalence is that is if you assume that V is constant, then the energy after connecting the capacitors together = 0.5 x (2C) x V^2^ = CV^2^ - which seems to work.

The superconducting wire is a little bit of a red herring here, at least for a lot of people - the intention of the questioner is actually make you dismiss things like inductance and resistance - i.e. no energy will be lost through the process of connection or something like that (basically I think people are overthinking the purpose of the wire). Note that there should be no current in the wire, as both capacitors should have the same number of electrons (and holes), so there is no need for any redistribution of holes and electrons.

It is ten years since I did this stuff, but I am fairly sure this is what is going on (both in my answers, and in what the question is actually asking).

physi_marc

As a physics grad, I'm fairly certain Letarian has it right. The only purpose of the superconducting wire is so you can ignore the resistance and other energy losses that may occur in a regular wire.

nuclearalchemist

I would just be careful about whether or not these are charged beforehand. If they are charged, and then removed from a potential source, then you have to use;

E = 0.5*Q^2 / C

Per capacitor. Then the total energy is:

Etotal = Q^2 / C

If they are joined in parallel, then the two plates of the capacitor add together to form a new capacitor, changing the effective area of each of the plates. What you can do then is that originally, you know the charge on each of the capacitors (or at least one of the plates), and you can show that the energy should remain the same. Essentially, you just have to be careful with factors of 2.

But yeah, I think everybody else here is right.

L|ama

Yeah, after sleeping on it (and thinking about some of the stuff he put in questions during the lecture), I've decided this lecturer just has a thing for red herrings, which is going to make the exam... interesting.


edit: okay I wrote the question down wrong and only one of the capacitors was charged initially, giving 1/2 the energy afterwards and it dissipates from the resonance of the circuit and EM radiation. whoops.