charges inside a vacuum

BillO

Joined Nov 24, 2008
999
Hey Steve,

With all due respect, I do know Gauss' law. And what it tells us is that the flux through a surface is proportional to the charge contained within. There are some consequences to this, but that all charges within a spherical surface having to be in a in a single, one electron thick, layer is not one of them.

We can have equilibrium in a multilayer case when the force required to place an electron in the outermost layer exceeds the force driving it there. As you pointed out, once we have a sufficient charge density in the outer layer, there is no field within that layer to drive a single electron anywhere. Adding a second electron in that cavity will then provide a mutual repulsion that will force the electrons to the outer shell. Once they get there and get arbitrarily close to the charges there, there will be a potential barrier for them to overcome in order to actually be a part of that shell. That potential is related to driving the existing charges closer together. As the charge density in that outer layer increases so does the potential required to insert another electron. At some point it will be high enough that a small number of electrons will not be able to mutually generate the force required to enter it. As a result, as more electrons are added, not only does the second layer get more populated, but so does the outer one.

If this continues, you will eventually distribute electrons throughout the interior of the sphere.


BTW, I only brought QM into my post to mention it was not significant
 

Thread Starter

wes

Joined Aug 24, 2007
242
The link did help in understanding a little more. I guess for a normal vacuumed hollow sphere, the charge would stay toward the surface of the inner sphere since it would be glass or some other high Dielectric material. Obviously before that the charges would be all over the place but once it became static or as static as can be, they would be situated around the surface. I am still pretty sure they would fill up the vacuum though.


As far as calculating the voltage, I think I might know how too now. A Quote from wiki "
A farad is the charge in coulombs which a capacitor will accept for the potential across it to change 1 volt. A coulomb is 1 ampere second. Example: A capacitor with capacitance of 47 nF will increase by 1 volt per second with a 47 nA input current".

Since a sphere with a radius of 6 inches (.1524 m) has a capacitance value of 16.95pf, then this means 16.95 pA is needed to change it by one volt. Also the Glass dielectric strength is 13.8 MV/m, however the glass will be 2 inches thick let's say. This means the Dielectric strength is only 701 KV for 2 inches of glass.

Now we can calculate the how much charge the sphere can handle.
16.95 pA increases voltage by 1 volt, so 16.95pA x 701KV (glass Dielectric) = 11.88195 uA = 11.88 uC of charge I think (not totally sure but sounds right)

Energy stored in joules is the Voltage x Ampere since 1 watt/second equals 1 joule

701KV x 11.88 uA of charge = 8.32 joules of energy or 8.32 watt/seconds


That is not a lot charge, I think it could defiantly kill you though, lol

So does this sound correct for the total energy storage?
 
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BillO

Joined Nov 24, 2008
999
How do you define when the surface is full? Think about how much charge needs to be there before the electrons are even getting close to each other. This could never happen in a practical case, but hypothetically, at that point I would expect quantum interactions to come in to play.
I define the surface as full when the potential required to add another electron exceeds what is available. Of course, and I eluded to this, the definition of 'full' must change as the number of electrons increases.

I think this would happen much sooner than you are suggesting (again, not going to attempt the math...see below). The electric force is quite strong, and the potential will go up quicker when the electrons are restricted to a surface than they will in a volume simply because of the reduced dimensionality the hence more rapid density increase each time an electron is added.

Why can't the math be done? Thanks to Gauss' law, the calculation is trivial. Think about it. How can electrons on the outer layer repel electrons in the interior?
Gauss' law tells us nothing of what we are discussing here, which is the charge distribution. Every introductory text on the subject warns that Gauss' law holds regardless of the charge distribution and that it tells us nothing about said distribution. The math gets ugly because of two things. First, that we are taking about the 3D distribution of charge, and second, becuse you need to solve for the QM case to know if and when it will be a a factor.

The net effect of the charges in the entire outer spherical shells exactly cancel.
Only in the case of a mysterious, ideal, uniform and continuous charge.

In our case, we are supposed to be talking about electrons. So, once you start getting very close to the outer surface, there will be a field. A fairly complex one at that. However, the complexity can be ignored, but the existence of the field cannot.
 

Thread Starter

wes

Joined Aug 24, 2007
242
So then BillO or anyone that knows, how would you calculate it if it really does fill up the entire volume of the sphere instead of just the inner sphere with a layer of electrons? or is just way to complex for someone like me to even attempt, lol.
 

BillO

Joined Nov 24, 2008
999
So then BillO or anyone that knows, how would you calculate it if it really does fill up the entire volume of the sphere instead of just the inner sphere with a layer of electrons? or is just way to complex for someone like me to even attempt, lol.
Well, its too complex for me to attempt, but I suspect that there would be some 'filling'. But like I said initially, the actual density would depend on the potential you are able to generate.

Of course, if the infinite is at your disposal, all sorts of neat non-linear things are likely, or not, to happen.
 

Thread Starter

wes

Joined Aug 24, 2007
242
well no, not the infinite, more like what we are capable of realistically generating. although it is fun to think about what you might be able to get in there
 
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steveb

Joined Jul 3, 2008
2,436
The idea that there will be an charge distribution extending into the volume (i.e. not confined to the surface) leads to some very absurd results, in the static case. Also, the calculations about this are quite easy, under the usual assumptions we make in classical electromagnetism, and lead to the direct result that this does not happen.

It's strange that a claim is made that there must be an extended charge distribution into the volume, but the form of the distribution isn't provided because the calculation is deemed too difficult. Is it reasonable that such a simple geometry and such a simple situation would be beyond the capability of electromagnetic theory? On the contrary, it is quite trivial, and is dealt with in just about all books on electromagnetism. The application of Gauss' Law, combined with the fact that the charges are "free charges", combined with Newton's laws of mechanics are easily applied to this case and lead to an easy calculation of the charge distribution, which is an impulse function versus radius, located at the surface. One can claim that the outer electrons generate a force to push the inner electrons, but this of course contradicts the Gauss law calculation of the electric field which says electrons will have force pushing outward.

In fact, this situation is so simple, that no calculation is needed, but just simple logic (via reductio ad absurdum) can be applied. Consider that any claim that there is an extended charge distribution directly leads to the result (via Gauss' Law) that there is a nonzero electric field in the interior. Free charges in the presence of an electric field MUST move, and hence this condition is not an equilibrium condition. The only place where there is a force to counteract any electric field force is at the surface of the container. In other words, the only place that the electrons are not fully "free" is at the surface. The bottom line is that the interior electric field is zero once static conditions are established.

I expect this simple logic will be dismissed, but again I must point out that it is the logic to be found in all electromagnetic text books. I'm happy to provide references and even scans of pages from books, if it is thought to be helpful. I could also provide a formal calculation, but I'm leaning toward not doing that because this requires considerable time to put in a form suitable for presentation in a forum of skeptics. There is no offense intended by that statement, and in fact being skeptical is generally a good thing, if not taken too far.
 
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BillO

Joined Nov 24, 2008
999
I expect this simple logic will be dismissed....
Not at all Steve.

I have no time to argue this situation or do the due diligence (the math) so I will just say that you're right and I'm wrong.

In fact, I will no longer participate in any discussion here or elsewhere if there is a question of interpretation. There is just no way to do this sort of thing effectively over the Internet.

No hard feelings.
 

steveb

Joined Jul 3, 2008
2,436
BillO said:
so I will just say that you're right and I'm wrong.
Unfortunately, it doesn't work that way. Usually these types of disagreements are the result of each side having a different mental picture with different assumptions in mind. We may very well both be right from our respective viewpoints. But, by terminating the discussion, the sticking point will not be uncovered.

We are both participating in the thread to help the OP, and the last relevant point the OP made, with regards to the point I was debating, was "I am still pretty sure they (the electrons) would fill up the vacuum though."

So, it appears that we have to terminate the debate with the conclusion that you are right and I am wrong. What the OP walks away with is paramount.

BillO said:
No hard feelings.
Most definitely. But, I do apologize if my method of coming across to make my points was improper.
 

BillO

Joined Nov 24, 2008
999
Unfortunately, it doesn't work that way. Usually these types of disagreements are the result of each side having a different mental picture with different assumptions in mind. We may very well both be right from our respective viewpoints. But, by terminating the discussion, the sticking point will not be uncovered.
I am sorry, but honestly I do not have the time. The next few months are going to be nuts. I understand your sentiment, but I've just run out of time. My free time ends tonight for an indeterminate time-span.

We are both participating in the thread to help the OP, and the last relevant point the OP made, with regards to the point I was debating, was "I am still pretty sure they (the electrons) would fill up the vacuum though."
Well, I can't make a convincing argument without considering the QM case. This might take months or years to work out. So I made some hand-waving statements based on my understanding/knowledge that supported his opinion with caveats. I can't get into more.

So, it appears that we have to terminate the debate with the conclusion that you are right and I am wrong. What the OP walks away with is paramount.
As I said, there is interpretation at play here. It will take more than a brief discussion to work this out. We'd need to discuss initial conditions, definitions of electron expanse, rates of change, classical/QM thresholds, etc... to do any justice to this. Given the depth this subject plumbs, your interpretation is as good as mine at this level. studiot introduced some valid considerations too. It's complex, and I can't do the math to support my assertions given the time I have and the nature of the forum.



Most definitely. But, I do apologize if my method of coming across to make my points was improper.
Not at all. The breakdown here is mine to own and I should have know better knowing the tasks I have ahead of me, beginning this morning.
 

Thread Starter

wes

Joined Aug 24, 2007
242
Well thanks to everyone anyways, I at least now have a somewhat better understanding of how would work.

As far as I can tell though is that there is a reason we don't have vacuum capacitors where we store charge inside a vacuum. It's that normal capacitors are just so much better at storing charge within them. A normal capacitor has opposite polarity charges (protons of the nucleus) to help reduce the interactions between the electrons and thus allows you to store a greater amount of charge because of that. I imagine at least that is part of the reason.

well anyway, thanks to all you for the help.

later
 

studiot

Joined Nov 9, 2007
4,998
Perhaps I should add some more thoughts.

Steve, whilst it is true that there is one stable equilibrium solution to the issue of charge distribution on a sphere as you describe, there is another equilibrium solution.

This is with a single charge at the exact centre of the sphere and the rest evenly distributed around the surface.
Of course, this is unstable equilibrium.
(Wes are you familiar with the difference between stable and unstable equilibrium?)

As to space charge

Well this does occur and appears to be a physical manifestation contradicting the foregoing.

However Steve's classical analysis assumes there are no other forces acting on the charges and that individual charges do not act on themselves. Thus the result (solution) is a balance of all the interactions of the individual charges on each other, with no self interaction.

In the case of space charge there is another (thermodynamic) force acting. There may also be impressed electric fields if the space charge is in an old fashioned vacuum tube or valve.

go well
 

steveb

Joined Jul 3, 2008
2,436
Steve, whilst it is true that there is one stable equilibrium solution to the issue of charge distribution on a sphere as you describe, there is another equilibrium solution.

This is with a single charge at the exact centre of the sphere and the rest evenly distributed around the surface.
Of course, this is unstable equilibrium.
Once again, you are correct sir! I hadn't thought of that, but that is a valid equilibrium solution under the classical assumptions. We could find both classical and quantum reasons why this condition is problematic, but it is interesting and important to identify it.
 

BillO

Joined Nov 24, 2008
999
I occurred to me during the turmoil of my day that we may be talking about a conducting sphere. Is this the case?
 

steveb

Joined Jul 3, 2008
2,436
I occurred to me during the turmoil of my day that we may be talking about a conducting sphere. Is this the case?
That is basically the model I have in mind. Or, maybe it is better to say that people who model the conductors in a simple way have the model of an "electron gas" in mind.

This is a bit of a confusing point, to me, because I like to think of the vacuum as a dielectric, not a conductor. However, a conductor is just a region with free charge carriers in it. So, I think that a solid metal sphere with extra electrons in it (or even the reverse with a deficiency of electrons) has a lot in common with the situation we are talking about.

A solid metal conductor is quite a bit like a bottle that can hold an electron gas because there is a work function (i.e. relating to thermionic or photoelectric emission) to describe the energy needed to remove an electron from the metal surface. The work function acts like the walls of a container, provided that there is not too much heat or light to complicate the situation.
 
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BillO

Joined Nov 24, 2008
999
Well, then I am wrong. There would be no fringe/edge effects with a conducting sphere and the field would indeed be zero everywhere within the sphere no matter how many electrons there were in it, depreciating special cases.

I was under the opposite assumption. That is was a non-conducting sphere.

So Wes, what I was saying does not apply. I guess I should have read the thread more carefully.
 

Thread Starter

wes

Joined Aug 24, 2007
242
Umm, what exactly do you mean by Conducting sphere

I was just talking about a Vacuum with electrons in it, so I guess an electron gas. The sphere material itself was just a dielectric like glass or something high.

Do you mean conducting sphere because of the fact the electrons are inside the sphere and like you said steveb "However, a conductor is just a region with free charge carriers in it" it basicaly would be like a solid metal sphere of metal with free electrons, again like you said, lol. I am pretty sure this what you mean but not 100% sure.

Just a question though, what would a non conducting sphere be?

Well which ever way it works, I definitely don't have the knowledge to argue against it, lol.
 

Thread Starter

wes

Joined Aug 24, 2007
242
Also, to answer your question studiot about the difference between stable and unstable equilibrium. Not really.

The best I can think of is that a stable equilibrium is a state within a system what ever it may be where change has ceased to exist. So take an excess of charge for example and that would be a Unstable Equilibrium because they all want to move away from each other. Once they settle down though to a neutral position with no excess charge, they become more stable in that they don't change or want to change as much. If you were to set a charge next to one that was in a stable equilibrium then it would obviously become more unstable the closer it was but it would I believe inherently want to become stable again by getting as far away as possible again.

All this is just based off my interpretation of the word equilibrium and stable and unstable, lol.
 

steveb

Joined Jul 3, 2008
2,436
Wes,

Here are some quick and crude answers to some of your questions above.

On the issue of stable versus unstable equilibrium (note that unstable equilibrium really should be called marginally stable or marginally unstable), a simple example of a stable equilibrium is a ball resting in a valley. If you wait, the ball is stopped at the bottom of the valley. There is no net force and no movement, so you have equilibrium. It is considered stable because if you push it a little, it comes back to the equilibrium point all by itself. The unstable equilibrium is the case of a ball balanced on top of a hill. If you carefully place the ball at the peak of a hill, you may be able to get it to balance. However, this equlibrium is only marginally stable (or marginally unstable) because the slightest push causes the ball to roll down.

In your example, the charges going to the surface represent equilibrium because all forces are balanced and there is no movement. Studiot pointed out that you could place just one electron at the center and still have an equilibrium. In this case, the charge at the center can't unbalance the spherical symmetry on the surface, and hence the center charge is in a zone with no electric field. The reason that this is unstable is because the slightest movement of the charge away from the center disturbs the symmetry, hence distorting the electron distribution on the surface, and the electron then feels a force that will drive it to the surface. This lone electron is basically like a ball sitting on top of a hill, and it prefers to roll down, which it will inevitably do given enough time.

On the issue of the conducting sphere, the main point is that simple models of charged conductors treat the extra electrons (or even an electron deficiency) like they are an electron gas (or anti-electron gas). So, like you said, the vacuum with free charges can be considered somewhat like a conductor. It's not the same thing as a metal sphere, just as an electron beam is not the same thing as a current flowing in a wire, but there are some very striking and useful similarities, particularly if one is willing to restrict the discussion to very special cases, and to use simplifying assumptions.

As far as the question about non-conducting spheres, I guess a good dielectric material qualifies. I have a vision of a solid styrofoam Christmas ornament, or a crystal ball. If you somehow can get charge on a dielectric (such as by rubbing two surfaces together), those charges are somewhat fixed in place and are not free to move around. This makes it very difficult to predict any particular charge distibution since you can arrange the charges in many stable ways. As an analogy, imagine going to the beach with a bag full of marbles. You can bury those marbles into the sand in any arrangement you like. The sand keeps them where you place them. But, if you go to the ocean and drop them in the water, they sink to the bottom. The water is like a conductor in that the mables are free to move under the influence of gravity, and the sand is like a nonconductor in that the marbles are not free to move under the influence of gravity.
 
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