poobah said:
It is only "during" that energy can be lost..so, that has to be it..the alternative, is that energy is lost arbitrarily during the transfer via series resistance and the flash energy. Arbitrary loss will not produce the half voltage result obtained which is consistent with charge conservation.
poobah said:
I had considered it, but there are some details that need to be worked out.
1. The flat plate capacitors have to be superconducting. They have to be of sufficient current capacity that a quench does not occur during the transfer.
2. An inductor must be used to limit the possibility of quenches. The inductor must be toroidal, as external fields will couple to the environment.
3. A superconducting shield around the system is needed to prevent external transmission. This raises an additional problem, which is persistent currents. This will require raising the temperature slowly after the experiment, to measure the helium heat rise when the persistent currents quench out. Note that this problem will also affect the toroid.
4. With an ideal inductor and two ideal capacitors, the oscillation will proceed unhindered forever. The only entity which will slow it down would be eddy currents in the surrounding volume.
5. The capacitor will have to be rather large to provide any reasonable capacity..Liquid helium dielectric constant is not very high, and other materials are not well characterized at that temperature.
6. Current temperature sensors are only capable of a millikelvin or so of resolution, so the heat capacity of the helium may render temperature rise measurements of persistent current energy below the system resolution.
7. Silicon switches are very odd in liquid helium...Regular PN junctions have forward voltage drops of between 8 and 30 volts, you have to warm them to 20 kelvin or so before they begin to conduct again.
Cheers, John
Sheesh, a lot of writing involved in this chop-busting post..😉
You ain't bustin my chops. You just threw in the prerequisite L (or R) that I have been maintaining is the sucker punch in all of this.
😀
😀
LOL, i got a pretty strong hunch it was a sucker-punch-brain-teaser when i found the exact experiment in the Dutch 🙄 wikipedia.
It was under the category "anomalies in physics" 😀
Gotta run guys. Got a worp-core breach on my hands 😀
Klaas
It was under the category "anomalies in physics" 😀
Gotta run guys. Got a worp-core breach on my hands 😀
Klaas
Hi Klaas,
I cannot access wikipedia from here. ...would be glad if you could send me that wikipedia essay by mail.... Oehem... only dutch? urghs..!
I cannot access wikipedia from here. ...would be glad if you could send me that wikipedia essay by mail.... Oehem... only dutch? urghs..!
long thread, i take it everything has come up roses?
another related, an more beleiveable form of this problem (that doens't violate the charge sharing stuff or involve infinate current) is:
charge 2 caps in parallel to 100V each.
remove both.
attache these in series.
another related, an more beleiveable form of this problem (that doens't violate the charge sharing stuff or involve infinate current) is:
charge 2 caps in parallel to 100V each.
remove both.
attache these in series.
theChris said:
It is not exactly the same.
Two caps in series, 1 farad each, 10 volts.
E = 1/2 C V2
in parallel,
C = 2, V = 10, E = 1/2 *2 *100, or E = 100 joules
then, put them in series
C = 1/2, V = 20, E = 1/2 * 1/2 * 400 or E = 100 joules
No loss.
(I was thinking of this scenario also...great minds think alike, right?😉
Cheers, John
But John, you know that on inspection because the charges can't move.
If you want to have fun, you can make the energy "disappear" by connecting them back-to-back.
If you want to have fun, you can make the energy "disappear" by connecting them back-to-back.
SY said:
yup, I simply pointed out the conservation of energy.
It was stated that ""another related, an more beleiveable form of this problem "", I pointed out the difference.
SY said:
Apparently we have different ideas of what constitutes "fun"😕 😕 😀
Cheers, John
I've been called worse by better people..and I've been thrown outta better places than this..😀
Cheers, John
Cheers, John
has this been answered yet?
i'm thinking that energy is lost to the circuit when the charge is redistributed on the device.
eg, it takes energy to change the charge density from Q/A to Q/2A -- intuitively makes sense -- the charges on the edges are _forced_ to move to the other end of the capacitor, which is some distance away.
i'm thinking that energy is lost to the circuit when the charge is redistributed on the device.
eg, it takes energy to change the charge density from Q/A to Q/2A -- intuitively makes sense -- the charges on the edges are _forced_ to move to the other end of the capacitor, which is some distance away.
Yes... the energy is lost to the resistance or L that must be in a realizable ciruit... sucker punch... not an anomaly.
😉
😉
no, not what i meant.
i meant even without a series resistance, the energy is lost because thats what happens to move the charge.
from the definiton, i guess that would be a "resistive" effect, but I find it important to clarify that it has nothing to do with "practical" circuits and limitiations thereof.
inductance has no place at all, and really, neither does resistance as the concept of resistance would be very poorly applied to this. I'm talking about basic electrostatics.
edit -- the concept of resistance is poor because time isn't relevent. it is assumed that after some time the charge redistributes, but the actual amount of time isn't important. the energy lost to a "mathimatically equivilent" resistance would have a time dependence.
basically i'm saying the conditions of the problem introduce a condition where energy is transfered, but not obviously. basically you say "now charge on each cap decreases by a factor of 2". well, to do this requires energy, but the statement doesn't seem to explicitly state it!
i meant even without a series resistance, the energy is lost because thats what happens to move the charge.
from the definiton, i guess that would be a "resistive" effect, but I find it important to clarify that it has nothing to do with "practical" circuits and limitiations thereof.
inductance has no place at all, and really, neither does resistance as the concept of resistance would be very poorly applied to this. I'm talking about basic electrostatics.
edit -- the concept of resistance is poor because time isn't relevent. it is assumed that after some time the charge redistributes, but the actual amount of time isn't important. the energy lost to a "mathimatically equivilent" resistance would have a time dependence.
basically i'm saying the conditions of the problem introduce a condition where energy is transfered, but not obviously. basically you say "now charge on each cap decreases by a factor of 2". well, to do this requires energy, but the statement doesn't seem to explicitly state it!
theChris said:
To transfer charge from one cap to another requires current flow.
If you assume for a minute, zero resistance, there will always be inductance present. With no resistance, the system will oscillate with no bounds....cept one...
The definition of "inductance" is the relationship between the current in a system, and the energy stored in the magnetic field of the system.
E = 1/2 L I2
Any change in current, which the oscillating system will indeed be doing, will cause oscillation of the magnetic field of the system. That oscillation will broadcast energy out of the system.
This is the energy bleedoff that is being used to explain the energy loss.
Cheers, John
Yep,
It is simple enough, and perfectly valid, to explain the "lost" energy via charge-through-a-field math... the "before and after" approach.
What is lost here is the understanding that charge through a field says resistor, inductor, or a flock of electrons smashing into a conductor (plate in a tube). It lies at the very definition of those things.
Consider the zero resistance thing: these capacitors have plates, plates require area, area requires distance for charge to flow. What distance can be non-inductive? Don't get tricky and try to show the net current is zero within a volume of both plates that has no thickness... the distance between the plates cannot be zero.
If the "must have R or L theory" does not set well, then flip the problem. Show where the energy goes and by what means... it must be stored or radiated (or create mass...hopefully oil). Yes, the apple fell from the tree... but something happened to the apple, Newton's head (yeah, yeah), and the air it passed through.
🙂
It is simple enough, and perfectly valid, to explain the "lost" energy via charge-through-a-field math... the "before and after" approach.
What is lost here is the understanding that charge through a field says resistor, inductor, or a flock of electrons smashing into a conductor (plate in a tube). It lies at the very definition of those things.
Consider the zero resistance thing: these capacitors have plates, plates require area, area requires distance for charge to flow. What distance can be non-inductive? Don't get tricky and try to show the net current is zero within a volume of both plates that has no thickness... the distance between the plates cannot be zero.
If the "must have R or L theory" does not set well, then flip the problem. Show where the energy goes and by what means... it must be stored or radiated (or create mass...hopefully oil). Yes, the apple fell from the tree... but something happened to the apple, Newton's head (yeah, yeah), and the air it passed through.
🙂
I still think that my idea is good and i have for you.
I also talked to some friends of mine, one of them is studying physics, and they all have the same oppinion about this problem as I do.
Another thought:
The energy wasted on a resistor between the two caps is E=I^2*R*t.
If you half the R, you'll also half t and you'll double I, so the energy remains constant.
You continue to half R until you get to a very small value which is near 0.
Well..this is an oversimplified point of view...but at least you can try it 😀
I also talked to some friends of mine, one of them is studying physics, and they all have the same oppinion about this problem as I do.
Another thought:
The energy wasted on a resistor between the two caps is E=I^2*R*t.
If you half the R, you'll also half t and you'll double I, so the energy remains constant.
You continue to half R until you get to a very small value which is near 0.
Well..this is an oversimplified point of view...but at least you can try it 😀
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