# electricity problem two capacitors

Meneer Dries
Posers and Puzzles 05 Nov '09 15:31
1. 05 Nov '09 15:31
My electricity teacher asked me this question:

Imagine a circuit with two equal capacitors, two switches and a DC source
(The o's don't mean anything, the forum just won't let me put blanks in between the 'wires'ðŸ˜‰
|----| |------|
|ooooooooo|
|ooooooooo /
|----| |------|
|ooooooooo|
|ooooooooo|
|-----DC----|

Now, if the source is turned on the bottom capacitor will charge up to Q=CV
This means an energy of 1/2CV² is stored in that capacitor
The upper capacitor won't do anything.
Now if the DC is source is removed by means of a switch and if then the other switch is closed , the bottom capacitor will partly discharge into the upper capacitor.
|----| |------|
|ooooooooo|
|ooooooooo|
|----| |------|
|ooooooooo|
|ooooooooo /
|-----DC----|
There will now be Q/2 charge on each capacitor and a voltage of V/2 over each one. That means a total energy of 2*1/2*C*(V/2)²= 1/4C*V
The energy has been halved!
Where did the other half go?
2. PBE6
Bananarama
05 Nov '09 16:03
Originally posted by Meneer Dries
My electricity teacher asked me this question:

Imagine a circuit with two equal capacitors, two switches and a DC source
(The o's don't mean anything, the forum just won't let me put blanks in between the 'wires'ðŸ˜‰
|----| |------|
|ooooooooo|
|ooooooooo /
|----| |------|
|ooooooooo|
|ooooooooo|
|-----DC----|

Now, if the sourc ...[text shortened]... nergy of 2*1/2*C*(V/2)²= 1/4C*V
The energy has been halved!
Where did the other half go?
I never took electricity, but this sounds wrong...energy should be conserved in this system as there are no inputs or outputs when the circuit consists of the two ideal, indentical capacitors, so the energy should be divided equally between the two capacitors, not the voltage, no?

The initial energy provided by the voltage source is 1/2V²C, so the energy of each of the two capacitors would be (1/2)*(1/2V²C) = 1/4V²C, not 1/4VC. Is that right?
3. 05 Nov '09 16:321 edit
I made a mistake there: of course it has to be 1/4 CV².
But this is the total energy over the two capacitors, not the energy per capacitor.
If each capacitor has to contain this energy each should have sqrt(2)/2Q charge on it, thus extra charge has to be created from somewhere, which is not possible.
I know energy should be conserved, or clearly wasted somewhere, question is: where is the mistake in the given calculation?
4. 05 Nov '09 17:12
The calculation is fine. The energy is lost as heat due to resistance in the wires.
5. 05 Nov '09 17:14
Originally posted by mtthw
The calculation is fine. The energy is lost as heat due to resistance in the wires.
And before anybody asks - in the theoretical case where there is zero resistance (and so no loss of energy) the system never actually settles down to have half the charge in each capacitor - instead, you'd get a permanent oscillation.
6. PBE6
Bananarama
05 Nov '09 18:51
Originally posted by mtthw
And before anybody asks - in the theoretical case where there is zero resistance (and so no loss of energy) the system never actually settles down to have half the charge in each capacitor - instead, you'd get a permanent oscillation.
Aha! Very interesting. So the assumption that energy is conserved does not apply in this case. I found this on a physics forum, and it seems to corroborate your solution:

7. 06 Nov '09 09:45
Originally posted by PBE6
Aha! Very interesting. So the assumption that energy is conserved does not apply in this case. I found this on a physics forum, and it seems to corroborate your solution:

Here's another source I found - uses an analogy with two water tanks to explain it (lots of electricity problems have got equivalents in hydraulics).

http://www.iop.org/EJ/article/0031-9120/33/5/018/pe8510.pdf
8. 08 Nov '09 16:19
So, its a bit like asking: If I swing a pendulum, its oscillations gradually decrease and when it comes to a stop it has less energy, where has the energy gone?

The capacitors / mechanics anology has: All electrons in one capacitor = pendulum one way, all electrons in other capacitor, pendulum other way, electrons evenly distributed = pendulum in middle. In the capacitor system "momentum" = current.

In the case of the capacitors, the oscillations in current lose energy in the form of radio waves as well as through resistance in the wires. Maybe in the pendulum system, the oscillations in momentum lose energy by heating the string and surrounding air and also in the form of gravity waves?
9. sonhouse
Fast and Curious
11 Nov '09 02:18
Originally posted by mtthw
Here's another source I found - uses an analogy with two water tanks to explain it (lots of electricity problems have got equivalents in hydraulics).

http://www.iop.org/EJ/article/0031-9120/33/5/018/pe8510.pdf
They want 30 bucks to access the article. Are you saying the energy going between the two caps, supposing the entire system is superconducting, capacitor plates and wires, then when the one cap with a charge is hooked up to the other cap with no charge, the charge will more or less completely discharge one and put most of that charge in the other and then the vice versa effect goes on an that newly charged cap finds a place with less electron energy (the other cap) and goes back and recharges that cap, ad infinitum? I don't think the energy loss in say, silver wires and capacitor plates would be so great as to stop that kind of oscillation or at least it should go on for several cycles, enough to be seen with an oscilloscope or some kind of voltmeter.

In regular oscillator tank circuits, you have energy say first in a cap that then discharges into it's opposite, an inductor and energy can bounce back and forth between the two different kind of energy storage media.

Have superconducting experiments actually been done on the two capacitor circuit shown? Do you have links?
10. 11 Nov '09 10:37
Originally posted by sonhouse
They want 30 bucks to access the article. Are you saying the energy going between the two caps, supposing the entire system is superconducting, capacitor plates and wires, then when the one cap with a charge is hooked up to the other cap with no charge, the charge will more or less completely discharge one and put most of that charge in the other and then t ...[text shortened]... nducting experiments actually been done on the two capacitor circuit shown? Do you have links?
Ah, sorry. Where I work must have an institutional subscription.

Yes, that's pretty much what I'm saying. It probably could be detected - I've not seen the experiment, though. The charge passes back and forth until the oscillation is damped down.

The article I linked to used the parallel of two linked storage tanks, one initially full and the other empty. Again, you end up with the water split evenly between the two tanks. But the energy (gravitational potential energy in this case) is less than before. But in this case it would be easy to see the water sloshing backwards and forwards, until the excess energy is lost due to friction.
11. 11 Nov '09 23:03
Originally posted by mtthw
Ah, sorry. Where I work must have an institutional subscription.

Yes, that's pretty much what I'm saying. It probably could be detected - I've not seen the experiment, though. The charge passes back and forth until the oscillation is damped down.

The article I linked to used the parallel of two linked storage tanks, one initially full and the other em ...[text shortened]... e the water sloshing backwards and forwards, until the excess energy is lost due to friction.
Lost to friction + (perhaps) a small amount of gravity waves.
12. 12 Nov '09 09:59
Originally posted by iamatiger
Lost to friction + (perhaps) a small amount of gravity waves.
No, probably not.

Unless you mean this sort of gravity wave: http://en.wikipedia.org/wiki/Gravity_wave, which would be the case in a stratified atmosphere.
13. sonhouse
Fast and Curious
12 Nov '09 17:312 edits
Originally posted by mtthw
No, probably not.

Unless you mean this sort of gravity wave: http://en.wikipedia.org/wiki/Gravity_wave, which would be the case in a stratified atmosphere.
If the material used in the capacitors and wires were both superconducting, it looks to me like there would be no losses so the oscillation would go on forever. How would you calculate the time constant if there were no losses? The regular formula is RC which is Resistance in ohms and C capacitance in farads. If Either R or C is zero then the time constant according to that simple formula would also be zero.

Therefore that formula must be modified to account for superconductivity. My guess is the actual frequency would be the product of the actual size of the system, shorter wires=quicker electron flow and faster response, if electrons flow at C in a superconductor, then it would look like one nanosecond per foot so it would be 1000 Mhz if the wires were half that, 6 inches leading to a path length of one foot, 6 inches coming, 6 inches going. It must be something like that, eh.

One loss venue would be RF radiation. If the path length was 6 inches, and that leads to a frequency of 1000 mhz(1 gigahertz) which is just my finger in the air estimate, then the wires might be close in length to a 1 ghz antenna. A dipole at that frequency at least for normal wires is about 7 cm long, 70 mm, so 6 inches would be about twice the length of an antenna at that frequency. This would be an antenna out of resonance and therefore not a perfect radiator but still would represent some loss in the form of RF radiation from the original energy given to the system, so it would be another path to dampen the oscillation. The exact amount I cannot be sure of but it looks like it would radiate some if steps were not taken to reduce that loss.

There is a almost lossless means of transmission where the two conductors are parallel at a certain distance, which relates to the impedance of the line, the size of the wire V the distance between them. Amateur radio operators use among other things, a transmission line called 'ladder line', two main varieties, 450 ohm impedance, wire separated by about 20 mm, and 600 ohm impedance, wires separated by about 100 mm.

In either case the transmission line radiation is very low, not perfect of course but it would greatly reduce EM radiation from the wires. The plates of the cap would still radiate I would think so there would still be some radiation from the system unless it was in a 1000 mhz resonator designed to suppress radiation from the whole system. It would be an interesting experiment if we could get our hands on real superconductors. I contacted American Superconductor about a project I wished to pursue using superconducting wire to study superconducting antennae but they thought the project too mundane to sell me any.

I thought a superconducting antenna would give a real benefit for radio communications by being itself an extremely narrow bandwidth system. The Q of an antenna is the impedence divided by the wire resistance, Rxl/R so if R approaches zero, the Q of the antenna goes way way up and I wanted to find out just what the Q would be, it looked to me like it would reduce total system noise, allowing communications with much less actual transmitter power. I am waiting till superconductors become widely available to actually perform that experiment.