Capacitor Charging- Where are all the electrons? 11

[WhMcC]

Greetings,

I am hoping that someone may have some insight here:

When a capacitor is charged, the electrons that are held in the electric field within the device must be physically distributed in some manner- how and where? Do the additional electrons fill normally vacant positions in the valence orbits of the material that comprises the plates of the capacitor? Or, is there some other mechanism by which the electrons are located and held in the field?

Thanks for your thoughts on this one!

 

[IRstuff]

Cbarn,

unfornately, that's incorrect, since capacitors work perfectly fine in a vacuum, which is technically a dielectric with a constant of 1, but is composed of no material, so there is no place to store charges.

The number of electrons, metal or otherwise, is constant only for a neutral material.

The same atomic principles must hold true regardless of the material, so if that was the effect, you would not be able to charge up glass rods and combs, since their electron/hole number should be constant as well.

From personal and shocking experience, it is trivially easy to charge metal objects to very high voltages and have them discharge themselves into your body. You can be quite certain that this concept will work perfectly fine in a vacuum as well.

TTFN

 

[IRstuff]

Jmarko,

See my post above, capacitors work in a vacuum, therefore the charges must reside on the plates. All capacitor dielectrics are chosen for non-conductivity, therefore, they cannot remove the charge from the plates.

The concept you are describing is actually dielectric dipole re-alignment, similar to magnetic dipole re-alignments in magnetic materials. The key here is to think about what it means to have a dielectric constant greater than 1. This essentially means that the effective spacing is smaller than the physical spacing; we'll ignore the reduction in the speed of EM waves definition, since we're talking statics. So, if we imagine a dielectric that looks like two plates close to the external plates with a wire in-between, the effective capacitance is significantly higher, and does not require actual movement of charge from the external plates. Just like magnetic dipoles, the dielectric dipoles have a positively charged and negatively charged ends, so the field generated by the charges on the plates are terminated within the dielectric by the first layer of dipoles and the opposite ends of those dipoles are terminated by another layer of dipoles, etc, until you get to the other plate. Since the dipoles chew up the physical spacing of the plates, but do not participate in the potential drop between the plates, the effective plate spacing is reduced and the capacitance is increased and the dielectric constant of the material, given by the ratio of the actual capacitance to the equivalent vacuum dielectric capacitance can be quite large, although the useful ones are not much more than 15, since larger constants correspond to increased conductivity and a conductive capacitor is prit near useless.

This link has a pretty nice graphic showing dielectric dipole re-alignment:

http://www.mse.vt.edu/faculty/hendricks/mse4206/projects97/group01/solidstate/dielect.htm

As mentioned above, Chapter 26 in Halliday and Resnick discusses capacitors in general, and sections 26-3 through 26-5 discuss the physics of dielectrics. As to the energy storage in the electric field, this is simply the manifestation of the separation of the positive and negative charges by the gap between the plates and the effective energy stored is given by the EE equation 1/2CV^2. This discussion can be found in section 26-6 of Halliday and Resnick.

As to the net charge, you are correct that the capacitor as a whole is neutral, but all physics texts talk about the charge on one plate, to be consistent with the electron current model. Even in Halliday and Resnick, while the discussion talks about charge in terms of negative charges, the figures for capacitors all show postive and negative charges on the plates.

To answer a previous posting, positive charge current consists of holes, not protons. Protons are too large to move within most materials and are usually held in place through atomic bonds. While it seems silly to talk about holes, they are fundamentally different in the effective mass and effective mobility, that's because the electron moves in the conduction bands, while holes move in the valence band, thus experiencing stronger interactions with the electron shells around the atoms.

TTFN

 

[jmarko]

Dear IRstuff,
thank you for yor opinion and I agree with it, it helped me to express better; I think that we are only describing the same picture with different words and that we are coming to final solution of what WhMcC answered: how are electrons distributed in charged capacitor. Also I think that different models tend to simplify things and sometimes distort real picture. (I am not a native english speaker and I also sometimes use confusing words and examples, as in this case). And I find example with water sphere also confusing.

Dear WhMcC,
sorry, I have maybe mislead you; I think this is correct answer:
At vacuum/gas (ε=1) capacitor all charge carriers are on the plates.
And also, with dielectric (with ε >1) capacitor, all charge carriers are on plates, and polarised molecules in dielectric helps to store ε times more electrons (and holes) on plates; almost no free electrons appear in dielectric (only thermal noise induced, and their number is irrelevant at room temperature). Free electrons appear in dielectric only a few moments before breakdown (and after, but then is capacitor usually already destroyed), if voltage is increased more than construction of capacitor allows. Voltage is linearly distributed from one plate(+) thrue (homogenous) dielectric to other plate(-). Molecules in dielectric are evenly distributed, and they contribute in charging with their orientation, not with lack or surplus of electrons or holes. That effect takes place in semiconductors.

Charging procedure of capacitor:
-with discharged capacitor, with no voltage applied, molecules in dielectric are randomly oriented;(or in vacuum capacitor there are no molecules between plates);
-when we apply voltage to plates, current flows thru wire for a moment : electrons (from negative battery pole) travel to one plate and the lack of electrons (holes)(from positive battery pole) travels to other plate;
-molecules in dielectric change orientation, they are forced to makes rows (+ - + -) from one plate to another; with that they allow additonal number (ε times more comparing vacuum) of electrons (and holes) to stack on plates; in vacuum there is no such activity ;
-if we then remove voltage (open circuit), electrons and holes stays separated on plates (and wires: density is evenly); and they hold molecules in dielectric (which tend to take random position again) in rows : capacitor is charged;
- if we then connect short (or via resistor) both plates, electrons and holes tend to make equilibrium in conductor(in wires and plates) and in addition also molecules in dielectric force electrons from plate with excess to plate with lack of them, a current flows for a moment, until both plates have same potential (until electrons and holes are evenly distributed) and molecules in dielectric are again randomly oriented: capacitor is discharged.
I think this is a little closer to real answer, but still far from ideal.
Best regards to WhMcC, IRstuff and others, Jmarko

 

[cbarn24050]

Hi irstuff, still having problems reading, you do not need a material to hold charges in order to support an electric field, if you did we wouldn't be able to see the sun. If your theory held, then if you remove the plates from a charged capacitor then the charge would stay on them, but of course it doesn't, but it does reappear when you replace them.

 

[IRstuff]

Of course the charges go with the plates. How would they get off the plates? Particularly with a vacuum capacitor.

What physical mechanism could you possibly invoke for getting the charges off the plates and have them stay put in space?

This is no different than with a charged comb or glass rod, You move the rod, the charge follows the rod.

TTFN

 

[jmarko]

Hi Cbarn24050!
I think that what you have in mind is not a capacitor phenomenon, but body, charged by triboelectricity or radiation or some other way. Let me explain: capacitor always has two plates with dielectric material (or vacuum) inbetween and if we charge it, and then we remove plates (regardless or one or both or one by one), almost all charge stays on plates, not on dielectric. For dielectrics special materials are used (like ceramic, polyester, polypropilen etc.), and they usually not accept easily surplus or lack of electrons and holes.
But some other materials do, like: glass, styrene, leather, PVC , metals etc. and some of them are also used for capacitors as dielectric material.
The point is that we can charge body made of isolator or conductor, and if we touch it (or bring it close enough, so the spark appears) by other body, charge will distribute on both bodies (usual experience of this is if we walk on carpet and then touch metal part, small spark appears). While this phenomenon is described by the same electrotechnical laws and equations, it is not called capacitor.

Hi, WhMcC!
Some additional information : special case of capacitors are electrolytic capacitors, and my message from sept. 2 2002 close briefly describes only that type of capacitors.

Best regards to all, Jmarko

 

[cbarn24050]

Hi irstuff, your nearly getting there with the charged rod! lots of charges and electric fields, but NO plates.

 

[IRstuff]

Jmarko,
thanks for the help!

As far is body electrostatics are concerned, it is still treated as a capacitance problem, where your body, being a a conductive material is one plate of the capacitance with the other plate being the earth. In most electrostatic discharge problems, the body is treated as a 100-150 pF capacitor, with a large series resistance.

TTFN

 

[IRstuff]

Cbarn,

A plate is simply the physical manifestation of making real-world and practical/useful capacitors. The exact same equations and physics work with the metal shaped as a rod, a sphere, or a key for that matter. The word "plate" as used in physics texts is simply for convenience, because most students understand that it's a generic label for physical plates, balls, spheres, etc. In fact, Halliday and Resnick's chapter on capacitance has a figures showing a capacitor composed of two separated sheres, concentric spheres and a metal ROD with a concentric shell.

Jmarko describes triboelectric effects that result in electrical charges on your body; my usual approach is to use by house key to discharge myself. During this discharge, the key is part of the plate of the capacitance of my body to the earth.

 

[WhMcC]

Hello all,

Thanks for the information and clarifications. IRstuff & Jmarko, thanks for sharing the specific references and the very detailed descriptions of this subject matter. I have learned from your responses to further research topics such as the Free Electron Theory and electric dipoles to provide myself with greater depth of understanding in the matter - While I have yet to try to locate the specific books cited I already have about half a dozen papers printed that discuss such topics in one form or another that I am reviewing.

Best regards,

WhMcC

 

[cbarn24050]

Hi Jmarko, no i am not geting confused, i am refering to the capacitor problem. Let me state it again for you all. The energy in a capacitor is stored in the electic field between the plates, it is not stored in the metal plates.

 

[IRstuff]

I finally see where you're going. Halliday and Resnick use the same terminology in their chapter on capacitors, but I think they were too glib and failed to reconcile their remarks with their discussion about single charges in a uniform electric field.

Since a single charge can experience energy changes in a uniform electric field, where is the energy stored? Not the field, since it's uniform. It's stored in the distortion of the field around the charge, due to the work incurred in moving the charge against the field.

Similarly, the field in a capacitor is a property of the negative charges separated from the positive charges and the concentration of like charges into a limited volume. Since the field is concentrated through the dielectric between the plates, and the field that the isolated charges would have is distorted and the energy is stored in the strain of the field. It's a semantical thing, but the glossing of this in Halliday and Resnick leads to confusion.

The electric field in the capacitor is a direct function of having moved the charges onto the plates, so without the charges, there would be no field and no energy stored. Consider moving the plates of a charged capacitor 10 ft apart. There is still an electric field around each plate, which can be easily measured by a field meter. Moreover, using Gauss's law, the net flux leaving a boundary box enclosing the plate is the same no matter where the plates are. And, there is still energy stored on the plates, since you can still do useful work or non-useful zapping of people or things. So, where is the energy stored now? It's actually in the plate, since the electric field of each individual excess charge is distorted by the Coulombic repulsion. This distortion is only evident within the plate, so at least some of the energy is stored within the plate!

TTFN

 

[electricpete]

Wow, there is a lot going on here. I hesitate to step into the mire. Lot of arguing and I haven't payed close enough attention to really see the differences in opinions.

I'll sum up my views. The free charge (that which can exchange with the external circuits) is on the plates. The energy associated with the charge on those plates is stored in the field, not on the plates.

jmarko brought out very important point that there are bound charges within the dielectric (due to dipole alignment) which play an important role. Net effect of all of those dipoles realigning is as if there was a surface charge on the the surface of the dielectric adjacent to each plate. That bound surface charge draws a matching opposite charge on the adjacent plate. The amount of free charge stored on the plates (per applied voltage) resulting from interaction with this bound charge will be much larger than it ohterwise would (with only air between the plates).

Bottom line. Everybody is right. I think jmarko's insight was under-appreciated. Good comments from IRstuff and cbarn as well.

Here's an old brain teaser. Let's say you have two capacitors of equal capacitance C0. Charge one of them up to voltage V0.

The amount of charge on positive plate is Q0=C*V0.

The amount of energy stored in the charged cap is E0 = 1/2C*V0^2.

The amount of energy in the uncharged cap is zero.

Total energy in both caps is E0=1/2C*V0^2.

Now connect the uncharged cap (same capacitance) in parallel with the charged cap and allow the charges to equalize onto both caps.

Each cap will now have half the original charge or Qf=1/2*Q0=[1/2*C*V0]

The voltage on each cap will be Vf=[Qf]/C = [1/2*C*V0]/C = V0/2. Makes sense intuitively so far.

The energy associated with either cap alone is now Ef = 1/2*C*Vf^2 = 1/2*C*(V0/2)^2 = 1/2*C*(V0^2)/4 = E0/4.

The total energy from both caps is twice this or 2*Ef = 2*E0/4 = E0/2.

What happened to the remaining E0/2 of energy? We have conserved charge but not energy. How is this possible?

 

[IRstuff]

we apparently forgot that we paid the electricity bill with it...

TTFN

 

[cbarn24050]

Hi electricpete, i woundered when you would turn up, I take it that you already know the answer to your little poser, if not i'll tell you in a week or 2 if someone else doesn't. As for the other problem, if you remove the plates from a charged capacitor you will find that they have no charge and can be shorted together without anything hapening at all so there cannot be any storage capacity in the plates. Also the formula for capacitance does not have any terms related to the plate material, thickness, volume etc. which you would expect if the plates were the energy stogage medium instead it only contains the field area, field thickness and dielectic permiability. Another point is the duality of formulas between capacitive circuits and inductive circuits, nobody doubts that energy stored in an inductor is stored in its magnetic field.

 

[WhMcC]

Hi electricpete,

My understanding is that the energy is expended as heat during charging due to inherent resistance of the connections between the capacitors and that one-half of all energy is lost in this manner when charging a capacitor regardless of the value of the (series) resistance.

Otherwise, thanks for the comments. I'm still mulling all this over as I am also wondering along with cbarn24050 how the type of material or third spatial dimension of the material comprising the plate has no bearing on the capacitance. I'm reading more on electric flux and Gaussian surfaces to try to get a handle on those aspects but may not have read enough yet...

IRstuff, thanks for the first offering of humor in the recent posts!

Best regards,

WhMcC

 

[IRstuff]

WhMcC,

Our canonical capacitor has metal, conductive plates and once you get the charge on them, you'll need a conductive path and a potential difference to move them.

TTFN

 

[Bung]

Isn't the electric field between the plates the result of the charge difference between the plates? (ie the potential difference). Arguing about "where" the energy sits seems pretty pointless.
Bung
Life is non-linear...

 

[cbarn24050]

Hi Bung, no it's the result of the voltage difference between the plates, Charge is quite a different matter.

 

[IRstuff]

Last time:

Q=C*V

Q=0, V=0 no charge, no voltage

TTFN