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Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Hi all,
I wonder if someone here ever thought about something like this:
When we calculate capacitance between objects, we always count just their surface and distance. Still there's a finite number of electrons in the system and total amount of charge I can move between spheres is limited!
Movable charge (number of free electrons) would depend directly on mass as I think.
Imagine two conductive spheres at some distance in vacuum, and a current source between them. As I move charge from one sphere to another, won't I come to the point where there's no more electrons left on positive sphere and there's nothing to move anymore?
Can anybody explain what exactly would I observe while approaching and at this state?
It is obvious that it's not really some significant macroscopic effect.
If I had a sphere made of 1mg of aluminium I would only have 6.7*10^19 electrons to move, which is about 10 coulombs.
That is 10 amps for 1 second, doesn't look like a lot, and I really don't see a limit how much can I reduce mass.
Now, I can look at it this way, if I had 1 nF between these spheres, I could charge them to 10 gigavolts before juice is depleted. Ok, maybe I should have reduced mass a bit more :p
Now is there some physical effect preventing this from happening, or it just affects equations as we know them?
If there is no charge left no current can flow anymore - I'm just curious how would this look like.
Registered Member #964
Joined: Wed Aug 22 2007, 12:39AM
Location: Stockton, CA
Posts: 134
This is all ignoring charge leakage and corona effects..
I would imagine that you would need some humongous anchor to stop the two heavily charged spheres from banging together with a force of.. a lot of tons. Got any magical space-fixating anchors hanging around? It would be interesting to see how they deform from the high-velocity impact....
Another thought comes up: If one sphere had TWICE as many electrons as normal, and the other has NO electrons, then the spheres would explode quite phenomenally from electrostatic repulsion... would be something like a low-power nuke I imagine, only no fireball, just a rapidly expanding cloud of aluminum particles sything away from each other at around the speed of sound..
Registered Member #540
Joined: Mon Feb 19 2007, 07:49PM
Location: MIT
Posts: 969
Wouldn't it be much faster than the speed of sound because it's an explosion. There is also a lot of energy behind it. Also if it were in a vacuum then it could go even faster than in air because there is no (I know there is no perfect vaccum) air resistance. I guess
Account deactivated by user request on 6/11/2009. Registered Member #1071
Joined: Fri Oct 19 2007, 02:13AM
Location:
Posts: 44
At some point the electric field on the negatively charged sphere would be sufficiently high to where you would begin to get field emission of electrons. This is where you have electrons leaving the surface of the material because of the modified potential in the presence of the huge electric field. This would happen way before you ever got to 10 GV. It is governed by Fowler-Nordhiem if you are interested in the explanation. This would prevent the Nuke like explosion as mentioned above. There is likely an analogous description for ions in a sufficiently high electric field, such that they are forced out of the matrix, but I dont know off the top of my head.
However, if you are interested in a nuke like explosion you might want to check out Coulomb explosions, this happens when an ultrahigh intensity laser is passed through a thin foil. The ultrahigh electric field is able to remove all the electrons from the atoms in the foil in essentially an infinitely small amount of time. That leaves only positive charge at the density of the metal which subsequently explodes due to coulomb repulsion (they dont have lasers this intense yet BTW, need 10^23 Wcm^-2).
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Yes Nik, I only put valence electrons into upper equation. Would it be possible though, to further ionize the atoms this way?
At some point the electric field on the negatively charged sphere would be sufficiently high to where you would begin to get field emission of electrons. This is where you have electrons leaving the surface of the material because of the modified potential in the presence of the huge electric field. This would happen way before you ever got to 10 GV. It is governed by Fowler-Nordhiem if you are interested in the explanation. This would prevent the Nuke like explosion as mentioned above. There is likely an analogous description for ions in a sufficiently high electric field, such that they are forced out of the matrix, but I dont know off the top of my head.
Yes, but would field emission necessarily occur regardless of anything?
If I made the walls of the spheres atom-thin I could have very little mass and much smaller electric field (since it depends on surface and the charge only?).
More that attraction and explosion I'm interested how would this look from electrical point of view. As I start running out of electrons, would the capacitance between spheres would decrease?
Would this start looking like something similar to magnetic saturation?
From the point of EM theory it would be very strange, it would look like permitivitty of the space between spheres is dropping down. Any thoughts on this?
Registered Member #29
Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500
If you could somehow keep the electrons "apart" from your sphere, you could deplete all valence electrons to the extent that removing another electron (from more tightly bound states) would begin to disrupt the solid structure of your metal.
However, you need not be so radical in removing every electron. In fact, in semiconductors there is a region where all mobile electrons are "depleted"; that is , removed. This is the region close to the junction of p and n type semiconductors. The depleted zone, for all practical purposes is no longer conductive and extremely high static electric fields exist there. Carriers (electrons or holes) must be "injected" (like in the emitter/base junction of a transistor or "spontaneously generated" by application of some external energy (photons as in solar cells and massive particles/photons in radiation detectors or by spontaneous thermal generation).
Removing all mobile electrons means that an electric field can exist inside the "depleted conductor", because there are no mobile charges to redistribute themselves to cancel out the internal field within it (as in usual metals, etc.). If a charge appears inside of this depleted zone, it is immediately swept out of the depleted zone by the field. This is what happens in all junction semiconductors. (MOS devices are somewhat different..)
Attempting to remove "every single electron" in the sense you describe would be like trying to create a perfect vacuum. You would reach a point of severely diminishing returns as the electrons become scarcer and harder to "pump out"... Plus, your "vessel" would leak: emission would certainly spoil any chance of actually doing this on any large scale.
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Hi waverider, thanks :)
WaveRider wrote ...
However, you need not be so radical in removing every electron. In fact, in semiconductors there is a region where all mobile electrons are "depleted"; that is , removed. This is the region close to the junction of p and n type semiconductors. The depleted zone, for all practical purposes is no longer conductive and extremely high static electric fields exist there. Carriers (electrons or holes) must be "injected" (like in the emitter/base junction of a transistor or "spontaneously generated" by application of some external energy (photons as in solar cells and massive particles/photons in radiation detectors or by spontaneous thermal generation).
Yes that is one of things I hope to understand better through this.
But, you guys still haven't told me the first answer - if what I'm saying is right, then how does mass, or, more accurately, total number of valence electrons relate to, for example, capacitance of a plate capacitor?
u0*ur*A/l would not be accurate anymore if mass of the plates matters.
Removing all mobile electrons means that an electric field can exist inside the "depleted conductor", because there are no mobile charges to redistribute themselves to cancel out the internal field within it (as in usual metals, etc.). If a charge appears inside of this depleted zone, it is immediately swept out of the depleted zone by the field. This is what happens in all junction semiconductors. (MOS devices are somewhat different..)
So, capacitance of my sphere would indeed simply decrease as it is getting depleted?
From equations I don't see why. How can capacitance decrease if no dimensions are changed, and charge is uniformly depleted from the surface of the sphere?
Something is from the other side telling me that simply it's ohmic resistance would increase?
Hope you can clear that a bit,
PS.
Plus, your "vessel" would leak: emission would certainly spoil any chance of actually doing this on any large scale.
yes... but maybe I could use some very large 'vessel', like earth, to distribute the electric field more evenly?
Registered Member #29
Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500
Capacitance is a field effect. As you pull electrons out, your "space charge" region gets larger (the depletion zone, remember). This is like pulling apart the plates of your capacitor, hence the decrease in capacitance (Varactors/varicaps use this phenomenon for changing their capacitance).
Mass has little to do with it (other than the mass of electrons transferred)..relativistic effects will generally be negligible.
Registered Member #89
Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145
Mass has little to do with it (other than the mass of electrons transferred)..relativistic effects will generally be negligible.
Sorry, sir, but have you understood what I was trying to say?
Simply, a solid metal ball will have far more free electrons than same diameter ball with wall one atom thick, and I'll have to remove far more charge to deplete it. So how then it isn't a function of mass?
Why is that always ignored? No relativistic effects, just the total number of available free electrons.
Capacitance is a field effect. As you pull electrons out, your "space charge" region gets larger (the depletion zone, remember). This is like pulling apart the plates of your capacitor, hence the decrease in capacitance (Varactors/varicaps use this phenomenon for changing their capacitance).
yes, but spheres do not pull apart... does the one which I'm removing electrons from appear to ''shrink''? How can that be if charge is always distributed on it's surface? That's pretty confusing.
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Capacitance, electrostatics question
