capacitor questions

The electrons come from the positive plate which develops an equal and opposite electron deficit. There is an upper limit to the number of electrons that could be shifted - every atom has a finite number of electrons, equal to it's atomic number - hydrogen has only got one, and uranium has 92. Only the outermost electron is available for this sort of exercise, but Avogaro's constant is large - 6.022045x10^23 - and you don't get anywhere near that in real life.

The energy is stored mainly in the electron differential between the two plates. The mechanical compression of the dielectric does store some energy, but several orders of magnitude less than the energy stored in the electron differential - the mechanical distortion shows up in such odd effects as charge soak and forms a part (usually a very small part) of the equivalent series resistance of the capacitor.

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Bill Sloman, Nijmegen

Charge would be pumped until you reached the open-circuit voltage of the solar cell. Why would you think it would go beyond that? (Unless of course the capacitor or something else breaks down before that point.)

If on the other hand you were to shoot electrons at one plate and positive ions at the other, you'd eventually get to the point where the electrons were repelling each other with a force great enough that they would fly off (toward the positive plate), and/or the ions would do similarly. With an air dielectric, you reach a point where the field strength is great enough that the stray electron (there are pretty much always some around) gets accelerated enough by the field that when it hits an air molecule, it knocks off one or more electrons, and pretty soon you have an arc. The lower the air pressure, the longer the mean free path before the electron strikes something, so the lower the electric field required to start an arc, at least up to the point where the mean free path approaches the plate separation.

Capacitors with a vacuum as dielectric don't store energy in polarization of the dielectric, I suppose. The energy is stored very similarly to the energy stored when you lift a weight and place it on a shelf. There is a potential for doing work on the stored charge; the potential is the voltage. If you wish, you can say that the energy is stored in an electric field; but an equally valid way to think of it is that the energy is stored in the fact that two charges (positive and negative) are attracting each other, with a force that can be allowed to act over a distance, just as the gravitational force can be allowed to act on the weight over a distance.

Imagine a capacitor formed of two large plates with some separation which is small compared with the size of the plates, with a vacuum between the plates. Charge the capacitor with charge Q. The plates will see a force pulling them together. Allow the plates to come together half the initial separation. The charge remains the same, but the voltage drops by half, because you have removed half the available energy. At the same time, the capacitance has doubled. Note that this is all consistent with C=Q*V and stored energy = (C*V^2)/2. Also note that if you pull the plates apart, you put energy into the system, and the voltage goes up. (It's how some high voltage generators work...)

Cheers, Tom ===================================

Hi,

if you hooked up a solar cell in space to a capacitor, and let the sun power the solar cell and charge up the capacitor indefinitely, more and more electrons will be pumped to the negative plate of the capacitor, until what limit? Assuming the dielectric can handle the voltage, will all of the free electrons eventually be pumped to the negative side of the capacitor, or is there no limit to the number of electrons that can be pumped leading to a voltage only limited by the dielectric breakdown voltage? If there is no limit to the number of electrons that can be pumped to the negative plate, where do these electrons come from?

I am a bit confused about where the capacitor stores its energy, is it stored as an electron differential in the two plates or is it stored as a polarization in the dielectric? (similar to an inductors core magnetization)

cheers, Jamie

As soon as the capacitor is charged to a voltage equal the the solar cells' voltage, there will be no more current flow.

Every electron you put into a capacitor increases its voltage: V = Cq where V is voltage C is capacitance q is charge

A solar cell provides current like a battery: there's an ideal voltage source that provides electrons through an effective resistance. As you build up charge in the capacitor, the voltage increases until the solar cell can't produce any more current because the photons can't produce electrons with enough energy (read this as voltage) to charge the cap any further. Capacitance stores energy as accumulated charges between plates. Some energy may be stored as a piezoelectric effect which is why you should always beware of high voltage caps that you thing you discharged but this isn't the storage mechanism, it's a side effect of the materials used.

Ok, please replace it with a "current pump" something that doesn't care about the voltage and just pumps electrons using solar energy.

As Bill said then the limit becomes the number of free electrons in the atoms of the positive electrode I guess.. So at that point the current pump would be sucking a vacuum on the positive electrode?

cheers, Jamie

If you're pulling electrons off one plate and whanging them at the other, then yes, theoretically you'd run out. But the donor plate would explode long before then... electrons are what glue things together! A blob of nuclei won't stick together for long. [1]

At numbers like 1e10 v/m field strengths, whole atoms (actually, ions) will fling themselves off a metallic surface. Tomographic 3D atom probes use this principle to disassemble samples one atom at a time.

Plus, the attractive forces between the plates would get absurd. And the negative plate would be spraying/field emitting/tunneling its extra electrons out into space, and they'd find their way back to the positive plate.

In a vacuum, essentially, yes. It took work to haul electrons uphill against an increasing charge gradient, and you can recover that energy by letting them fall back.

Note that energy stored is proportional to v-squared, because as you move electrons, the plates keep charging up and it becomes increasingly hard to get each successive electron from one plate to the other.

If there's a dielectric that's not vacuum, the dielectric stores the excess energy (over/above what you'd have in a vacuum.) You could argue that the plates still store the energy and that the dielectric just makes it easier on them, if you wanted to be tedious.

John

[1] is there a configuration where gravity could hold together a mass of pure nuclei? A proton star?

Keep in mind that for every electron that enters a capacitor, another electron leaves. There isn't a net gain of electrons in the capacitor, just stored energy.

If "solar wind" blew electrons onto a nonconductive surface, you'd end up with a surface charge much like the front of an old TV set. Probably nothing would explode.

Probably. It's a whole lot easier to assemble an electrically neutral star of that size, though. The protons and electrons of normal matter (plasma in particular) are squeezed together over the Coulomb force seperating the bulky ions and they cancel out, forming neutrons in addition to the neutrons already present. Basically, a neutron star is a neutral ion (no protons or electrons) of atomic mass 10^56 or something. Hey, it's Governmentium ;-)

Tim

--
Deep Fryer: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms

Sorry, in our universe, voltage always matters.

In article , John Larkin wrote: [...]

Remember those cartoon magnets that attract knives and horseshoes from miles away. I think anything like enough protons would do the same sort of thing in its quest for electrons. It is also likely that it would spit out positrons. The electrostatic force is much stronger than gravity.

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kensmith@rahul.net   forging knowledge

What if you used a superconducting negative plate so that the electrons paired up once they got there from the positive plate, and then once the voltage was say 100V you quench the negative plate and you get a big voltage boost! :)

cheers, Jamie

capacitor and voltage measurement - a good solution :

the same spamming answer to every post. One for the killfile.