Capacitor Charging- Where are all the electrons?
Capacitor Charging- Where are all the electrons?
(OP)
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!
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!





RE: Capacitor Charging- Where are all the electrons?
http://www.amasci.com/emotor/cap1.html
RE: Capacitor Charging- Where are all the electrons?
TTFN
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
- Capacitors store energy.
- Energy is stored in the electrostatic field created by a charge imbalance between the plates of the capacitor.
- The charge imbalance is a result of more electrons in one plate versus fewer electrons in the other.
My ongoing confusion is still this: in the plate with more electrons where are they held? In the water sphere analogy it is simple to see that when charged one side has a greater volume than the other even though the overall volume water in the sphere is unchanged and further that energy is held in the displaced rubber plate. It is simple also to say that the volume of electrons has increased on one side of the capacitor but it seems unlikely to say that the dielectric has been bent in a physical manner similar to the rubber plate in the water sphere analogy. If (thinking aloud now) electrons have mass and no space can be occupied by more than one electron at any given instant the electrons on a plate in excess of those normally associated with the material constructing the plate must be somewhere in or around the plate? Again, where? In the normally vacant orbitals of the valence shell of the atoms of material comprising the plate? Randomly across the surface area of the plate as dictated by maximum charge separation based on the repulsion of like charges given the strength of the electrostatic field? Or elsewhere?
Thanks again for any follow-up thoughts!
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
In metals, the electrons are free to move about, which is where the electrical conductivity comes from. When the metal is charged with additional electrons, they are simply shared amongst the atoms that are closest to the other plate. Anywhere else, there would be an internal electric field that would cause a current flow.
The internal distribution is driven externally by the dielectric field attracting electrons to the inner surface of the plate, but limited by the Coulombic repulsion between electrons in the cloud.
TTFN
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
There is no electric field without charges.
TTFN
RE: Capacitor Charging- Where are all the electrons?
It is apparent now that this question leans more towards the underlying physics rather than the electronics applications and I appreciate all the responses as they have helped shine light in many of the dim corners of the topic.
I have a next question for anyone who might respond to a true physics follow-up based on this first query discussion. It would not be appropriate for this forum at that level so I would invite individuals to contact me at whksmcc@aol.com if interested in helping tackle the next question.
Otherwise, I want to pass along a link to a great site I found very recently while searching for information. The site covers a lot of ground with regard to electricity, magnetism, and physics that one might find useful and does so in a very user-friendly and informative manner:
http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html
RE: Capacitor Charging- Where are all the electrons?
TTFN
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
See also Chapter 4.6 "Electrical Conductitivty of Metals" in "Electricity and Magnetism" by Purcell, McGraw-Hill, (c) 1965.
See also, Chapter 15-B, "The Free-Electron Theory of Metals" in "Electronic Structure and the Properties of Solids" by Harrison, W.H. Freeman, (c) 1980.
As to the capacitor itself, see any physics text such as "Fundamentals of Physics" Halliday and Resnick, J. Wiley & Sons, (c) 1974, Chap 26, "Capacitors and Dielectrics"
If there is no excess charge, how does the displacement current arise from discharging a capacitor and how does the integral of the displacement current conveniently equals Q=V*C?
TTFN
RE: Capacitor Charging- Where are all the electrons?
I think that key of how capacitor works is in material between metal plates : in dielectric .
There is where really takes place a distribution of charge carriers (electrons or holes-positive charged atoms or molecules).
And really, in solid dielectric, some atoms (or molecules) loses electrons, and some atoms (or molecules) gains electrons, and the charge is then evenly distributed in solid material between conductive plates : geometrical middle is a virtual border between positive and negative charge .
Special case is gas or vacuum or liquid dielectric capacitor. Here are all excess electrons (and holes - negative electrons) distributed on surface of capacitor conductive plates : in space between plates there are no charge carriers (true only when transition effects are bygone - DC voltage only).
So, partially is everbody right.
Best regards, Jmarko
RE: Capacitor Charging- Where are all the electrons?
TTFN
RE: Capacitor Charging- Where are all the electrons?
You are right, equations for capacitance are the same regardless the dielectric. But original question was: where are the electrons; and physical picture is not the same. In solid dielectric, charge carriers are distributed evenly in the material between plates - atoms and molecules are hold still, while in gas dielectric all charge carriers are atracted to plates.
And also, there is no surplus of electrons -regardless charged or discharged- capacitor (in whole capacitor sum of negative and positive charge carriers is always zero - DC conditions);in charged capacitor electrons are only distributed from one half of material or plate to another with help of external voltage (or electric field).
Therefore, maybe better word is polarised instead of charged, which is probably confusing.
Best regards, Jmarko
RE: Capacitor Charging- Where are all the electrons?
Great dialog folks- thanks!
JMarko, am I to understand that it is the atoms/molecules of the dielectric that lose or gain electrons based on being located on one side or the other of the geometric middle of the dielectric material as a whole? The conductive plates neither gain nor lose electrons in the process but rather provide a conduction path for the migration of the electrons to and from the dielectric?
Just to review, if an atom gains an electron it then occupies a space within the orbital of the valence shell of the atom, with the total numer of electrons to be gained being limited by the number of vacancies in that valence orbital.
Best regards,
WhMcC.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
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.
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
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?
RE: Capacitor Charging- Where are all the electrons?
TTFN
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
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
RE: Capacitor Charging- Where are all the electrons?
Bung
Life is non-linear...
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
Q=C*V
Q=0, V=0 no charge, no voltage
TTFN
RE: Capacitor Charging- Where are all the electrons?
What an interesting an instructive series of posts leading to Electricpete's vanishing energy. If the two capacitors have the same values then, within measuring tolerences, exactly half the energy is "lost" each time a capacitor is charged and connected to the second capacitor, no matter what values or voltages are used. Now if energy can be destroyed then surely it can also be created.
Inductors have opposite properties to capacitors so come on you guys, how do I connect my inductors to double the energy and save the planet?
Cheers
G
RE: Capacitor Charging- Where are all the electrons?
cbarn - I agree that the ENERGY is stored in the field and not the plates. But that doesn't mean that the free CHARGE is not stored on the plates (it is). Once again I will mention there are two types of charge: 1 - free charge which resides on the plates and participates in circuit current flow; and 2 - bound charge which is bound within the dielectric and only helps to draw more free charge onto the plates for given applied voltage. Bound charge cannot leave the dielectric.
Side Note on the role of free and bound charge - The free charge will be associated with the field D. The free plus bound charge is associated with the field E. Since bound charge dipoles oppose the free charge, we find E < D. (Contrast to magnetic where dipoles assocaited with "bound" current line up with free current magnetism to increase field due to total current B beyond field due to free circuit current only - H). Voltage is associated with E. The presence of dielectric means we have lower E (voltage) for a given D (charge). That should coincide with our intuitive understanding that high dielectric constant gives rise to high capacitance (C=Q/V).
Back to the main point, the free charge cannot leave theplates. As IRstuff has pointed out... two plates in a vacuum form a capacitor (although capacitance is lower than if we had a high-dielectric medium between plates). Surely you're not saying that the electrons are hanging out somewhere in the vacuum?
The question has been raised: if charge storage on plates plays a role in capacitance, then why doesn't play thickness show up in the formula for capacitance. The answer is that the charge is stored on the surface of the plate. Each unit of charge is "paired" with an opposite unit of charge on the opposite plate (vacuum capcitor). The attractive force between the opposite charges is what draws the charge to the surface. You can make the plate as thick as you want, it won't matter because the charge is stored on the surface.
Another way of looking at the same thing: There is an E field in the space between the plates. But there can be no E-field in the plates if they are assumed perfect conductor. How can E field change instantaneously as we enter the metal? Only when there is surface charge. In an electrostatic system, surface charge must reside at the boundary of the metal, not the interior of the metal (otherwise there would be potential gradient).
I believe it can be shown in general that for a static (non-time-varying) system including perfect conductor, the charge will NEVER reside inside the bulk volume conductor. Always on the surface. If there were charge within the conductor then there would be potential gradient within the conductor and current would flow to eliminate that gradient.
RE: Capacitor Charging- Where are all the electrons?
I hope it's ok to disagree as long as we do it in a professional manner.
More comments to cbarn:
"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"
I agree with IRStuff on this. If the plates are in a vacuum and not connected to any circuit then there is simply nowhere for the charge to go. When you pull the plates apart the charge will follow the plates. I wonder if you are referring to the fact that the total charge on a cap connected to an external circuit must be zero... and trying to apply that to the single plate? If so, that reasnoing doesn't apply if we consider the plate not connected to an external circuit.
"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."
Energy in both types of circuits is stored in the fields (magnetic field or electric field). Current for magnetism plays the similar role as charge for electrostatics. The free current in a magnetic system resides within the conductor (coil) portion of the circuit in a similar manner to how the free charge for an electrostatic system resides within the conductor (plate) portion of the circuit.
RE: Capacitor Charging- Where are all the electrons?
http://ece-www.colorado.edu/~bart/book/book/chapter4/ch4_2.htm
The reason that semiconducting junctions, particularly the lightly doped region, has spatially distributed charge is that the free carriers are limited by the doping concentration, which is not the case for metals, but what is analogous is that a spatially distibuted charge could exist in the metal, and drift currents in would be balanced in thermal equilibrium by an opposing diffusion current coming out of the region.
TTFN
RE: Capacitor Charging- Where are all the electrons?
First of all let me correct the erroneous explanation given in text books about the vanishing energy. I have conducted this experiment with very large value capacitors and high voltages with automated set ups to try and find the heat that is supposed to be produced and I have not found it. When using two capacitors banks of equal values even if each capacitor is made up of several others in parallel or series, the result is always the same, within limits of measurement exactly half the energy is lost. If anyone still believes the energy is lost as heat then where is it?. Why are the losses always exactly 50%.
Ok now for my explanation which originated from research into magnetic fields. There is a medium between the plates even in a material vacuum. J.C.Maxwell called this substance the "ether", I will call it anti-matter. I have studied and published several papers on magnetic anti-matter until academia threaten to cancel their subscriptions
and write nasty letters to the editors involved.
Charging a capacitor is like blowing up a balloon except electric antimatter does not seem to be affected by heat and therefore the mechanical pressure on the plates is proportional to the square of the distance between them unlike the balloon analogy where the thermal properties of the gas equalise the pressure throughout the enclosed space.
I would value your comments friendly or unfriendly.
Cheers,
G
RE: Capacitor Charging- Where are all the electrons?
Let v1(t) represent the voltage on the initially charged cap and v2(t) represent the voltage to ground on the initially uncharged cap. Define current i(t)in direction from V1 to V2. Superscript prime (') indicates d/dt.
C v1' = -i (eq 1)
C v2' = +i (eq 2)
(v1-v2)/R = i (eq 3)
Define vd (d for difference) vd = v1-v2.
eq1 - eq 2 gives:
C vd' = -2i (eq 4)
eq 3 gives
vd/R = i (eq 5)
Substitute equation 5 for i into equation 4:
C vd = -2vd/R (eq 6)
Rearrange eq 6 as
C vd' + 2 vd/R = 0 (eq 7).
Equation 7 has the following solution:
vd = V0* exp(-2t/RC)
where V0 = initial value vd = same as initial value of V1
From this it can easily be shown that
v1 = V0/2 + V0/2*exp(-2t/rc)
v2 = V0/2 - V0/2*exp(-2t/rc)
But these are not required for the solution….
All we need is I:
i = vd/R = V0/R * exp(-2t/RC)
Energy dissipation power P is i^2*r
P = =I^2*R = V0^2/R*exp(-4t/RC)
Total energy dissipated is the integral of power from t=0 to infinity:
E = Int(P from 0 to Infinity)
= [V0^2]/[-4/(RC)*R*exp(-4t/RC)] from 0 to infinity
= {[C*V0^2]/[-4*exp(-4t/RC)]}from 0 to infinity
= [C*V0^2]/[-4*exp(-4t/RC)] at infinity - [C*V0^2]/[-4*exp(-4t/RC)] at zero
= 0 - -CV0^2/4
= cv0^2/4
= E0/2.
Energy is conserved. Note that the result is independent of the R. You could take the limit as R approaches zero and the relation will still hold true. The rate of power dissipation approaches infinity, the time for discharge approaches zero, the integral remains the same. In a real-world scenario, there may be other complex phenomena such as arcing and current oscillations due to circuit inductance (normally low inducatance associated with leads etc becomes more significant during high-frequency transient). No matter how complex you make the analyis, I can pretty well guarantee you that energy will be conserved.
A philosophical question might be: why do I need to include resistance in the model? Why can’t we just move the charge without changing the energy? My answer is that moving the charge from one cap to two caps involves doing negative work. An analogous situation is gravitational force. We can’t just move a weight up a hill without doing work on it. We can’t move a weight down a hill without it releasing work/energy (the potential energy has to go somewhere). Likewise we can’t redistrbute the charge to a lower-potential energy configuration without it releasing work/energy.
Back to the subject of where are the electrons located. I believe that perhaps it has been made too complicated. From my simplistic way of thinking, the charge distribution is predicted by electrostatics (no time variation need be considered). Directly from electrostatics we can derive the formula for c associated with the relation q=c*v. For time-varying circuits we use this electrostatic result, take the derivative and apply the result i = cdv/dt. I don't think that currents need to be considered in this simple discussion. So electron clouds, drift velocity, valence shells, anti-matter (!)... it seems more than is necessary for the simple question. If you have opposite charges on the two plates, those charges will attract each other and therefore move to the inside edge of each plate. Maybe there's a lot more that needs to be considered, but for me that tells the whole story. (my opinion).
With that said, I'll admit I am very eager to hear what magnetic anti-matter has to do with capacitors. Please elabortate.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
"Total energy dissipated is the integral of power from t=0 to infinity":
E = Int(P from 0 to Infinity)
= [V0^2/R]*[-RC/4]*[R*exp(-4t/RC)] from 0 to infinity
= {[C*V0^2/4]*[-exp(-4t/RC)]}from 0 to infinity
= [C*V0^2]*[-exp(-4t/RC)] at infinity - [C*V0^2]/[-4*exp(-4t/RC)] at zero
= 0 - -CV0^2/4
= cv0^2/4
= E0/2.
RE: Capacitor Charging- Where are all the electrons?
E = Int(P from 0 to Infinity)
= [V0^2/R]*[-RC/4]*[R*exp(-4t/RC)] from 0 to infinity
= {[C*V0^2/4]*[-exp(-4t/RC)]}from 0 to infinity
= [C*V0^2/4]*[-exp(-4t/RC)] at infinity - [C*V0^2/4]*[exp(-4t/RC)] at zero
= 0 - -CV0^2/4
= cv0^2/4
= E0/2.
RE: Capacitor Charging- Where are all the electrons?
As in the case of rolling things downhill, the initial stored energy is consumed through kinetic energy of moving the charges around. A series resistance does not affect the total energy expenditure, since the starting and ending points don't change, just as in the case of rolling a ball downhill with rolling friction. The starting and ending potential energies are as predicted by 1/2mgh, the only difference is how fast the it happens.
Thus, for a series resistance case, while it takes longer for the charge to equalize, the stored energy in the end is still simply 1/2 of the initial energy.
TTFN
RE: Capacitor Charging- Where are all the electrons?
I obviously have no disagreement whatsoever. If you read my initial question of 9/6, you will see that I predicted that the energy changed to 1/2 it's initial value based on potential energy calculations equivalent to yours. My originally-posed question was: where did the energy go? Potential energy considerations alone do not answer that question (although they identify the exact amount of the "lost" energy).
On 9/9/02 I gave my short answer - the energy must be dissipated in resistance. There is no question that the amount of energy I am referring to is the "lost" potential energy E0/2.
On 9/10 gjones challenged my short explanation and asked to explain why the energy dissipated in such a resistance would always be precisely E0/2, regardless of the value of resistance. My answer earlier today responds to this gjones' query by proving that the amount of energy dissipated (E0/2) in the resistance is exactly equal to the amount of energy "lost" from potential energy considerations (E0/2).
So, from my perspective, you have restated the obvious. Am I missing something?
RE: Capacitor Charging- Where are all the electrons?
"Note that the result is independent of the R. You could take the limit as R approaches zero and the relation will still hold true. "
This means the while the energy COULD be lost through heating, it doesn't have to be. The energy is consumed as kinetic energy by moving the charges to the other plate.
TTFN
RE: Capacitor Charging- Where are all the electrons?
I liked it better when we were going to get free energy.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
No magic:
2V=Q/(C/2) Halve the capacitance double the voltage. It's the opposite form of electricpete's example, double the capacitance, halve the voltage. Even electricpete would agree with that, I think...
TTFN
RE: Capacitor Charging- Where are all the electrons?
Moreover, given the same conditions and moving the plates closer together, we double the capacitance and halve the voltage, which means the energy stored is likewise halved. In this case, no wires, no mirrors and poof, half the energy is gone!
Cool!!
TTFN
RE: Capacitor Charging- Where are all the electrons?
I think I follow your discussion cbarn. It's a very good point. And I will plead guilty to sloppy use of terminology.
In my first message I made a careful distinction of the difference between free charge and bound charge. In my subsequent messages after my first, I have been talking about vacuum or air capacitors (only free charge). No dielectric. I revert to that simple example because it is easier for me to discuss and illustrates the points I was trying to make. But I have not been careful to identify it as such. I apologize.
Bascially, the free charge is what is on the plates and available to return to the circuit in the event of changing voltage. The bound charge is created from polariztion of the dielectric (net result is a layer of bound charge on each of the outer edges of the dielectric), and is contained within the dielectric. The presence of bound charge in the dielectric draws more free charge (per applied voltage) onto the plates, enhancing the capacitance.
When we say c=q/v we are talking about the free charge q (on the plates). There is some formula that allows us to calculate the bound charge (in the dielectric) based on free charge q and based on relative permittiviy. I forget the forumla.
So... back to your dielectric. There is charge in there. Bound charge. I have never done the experiment but from your description it sounds like it has some memory....ie it apparently does not depolarize (at least for a period of time) unless acted upon by an outside field.
So, if anyone (including myself) has ever said "the charge is on the plate" without clarifying whether they're talking about free charge or restricting the discussion to a vacuum or air capacitor, then I agree that statement is not 100% correct. There are two different types of charge... one on the plates and one on the dielectric.
As far as the lost energy dissipated as electromagnetic radiation, I will accept that as a fair comment. In general we grow accustomed to the lumped-circuit model which is valid when the frequency is low and the circuits are small. All circuits have some wave tendencies which will become more important as size of the circuit grows in comparsion to the wavelength (higher frequencies). It is ceratinly possible that this scenario may give rise to wave significant wave behavior that dissipates energy, particularly if the leads are long with low resistance.
So, in summary, I have to offer my apologies and respect to cbarn. Those are good points.
Likewise, good discussion by IRstuff.
RE: Capacitor Charging- Where are all the electrons?
Thanks for all y'all's patience.
RE: Capacitor Charging- Where are all the electrons?
The relationship between bound charge Qb and free charge Qf can be found as follows:
E is electric field considering all charges, free or bound. (E is associated with voltage)
D = er*e0*E where er is relative permitivity, e0 is permitivity of free space
D =Int{Qf}d_A = Qf*A is (1/e0 times) the field associated with free charge only
Let’s say we have cap with free charge Qf, unknown bound charge Qb, and permitivity ef on area A
#1 – Find the field Ein terms of free charge only
Q = Qf/A
E = [1/(er*e0)]*D = [1/(er*e0)]* Qf/A
#2 – Find the same field in terms of free and bound charge
it is equivalent to the field E’ which would occur if we had free charge Qf’ = Qf-Qb and er’=1
in that case
D’ = Qf’/A = (Qf-Qb)/A
E’ = [1/(er’*e0)]* D’ =[1/(1*e0)]*(Qf-Qb)/A
Set E#1 = E#2
E = E’
[1/(er*e0)]* Qf/A = [1/e0]*(Qf-Qb)/A
Qf = erQf – erQb
Qb = Qf*(er-1)/er
So if er = 4, then Qb = (3-4)/4 = 75%
That means that for every 4 units of free charge, 3 units are “canceled out” by bound charge, leaving only one unit net to contribute to E and V. We can carry 4 units of free charge for the same applied voltage as would be required to sustain one unit of charge in an air capacitor. The capacitance is 4 times that which would occur if er=1.
As er increases Qb gets to be a higher fraction of Qf, but never gets to 100%
If Qb was equal to Qf, then infinite charge Qf could be sustained without any applied voltage and capacitance would be infinite.
So bottom line, there is a known quantitative relationship between Qb and Qf. Qb resides on the dielectric and Qf resides on the plate. The one on the plate (Qf) is the one we talk about when we say C = Q/V.
RE: Capacitor Charging- Where are all the electrons?
"So if er = 4, then Qb = (3-4)/4 *Qf = 75% * Qf"
RE: Capacitor Charging- Where are all the electrons?
One thing that is probably obvious but I will state it anyway: you had to put energy into the system to push apart the two plates that were electrostatically-attracting each other. That is where the energy comes from.
RE: Capacitor Charging- Where are all the electrons?
This is ultimately where bulk of the energy goes.
TTFN
RE: Capacitor Charging- Where are all the electrons?
So, if I take those two plates let's say 100in^2 area with 100 volts at distance of 0.05" and pull them apart to 0.10, then my voltage will jump to approx 200 volts. That is a true statement.
So here's another mini brainteaser. If I now take one of one of those same plates and drive it 1000 miles without discharging, does the voltage between the plates go to a hundred giga-volt range?
200*volt*1000*mile/(0.1*inch)*12*inch/foot*5280*foot/mile = 126,720,000,000*volt
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
Postscript 0 means initial conditions
Postscript f means final conditions
First find increase in potential energy, repeating the simple equations that predict doubling of voltage and doubling of energy
C0 = (e*A)/d0 where e is permitivity of free space…
Cf = (e*A)(2*d0) = C0/2
V0 = Q/C0
Vf = Q/Cf = Q/(C0/2) = 2Q/C0 = 2V0
W0 = 0.5*C0*V0^2
Wf = 0.5*Cf*Vf^2 = 0.5*(C0/2)*(2V0)^2 = 2*E0
(where W means energy or work)
Now find a physical explanation for the increase in V based on electrostatic force.
There is An electric field E = Q/(e*A)
This produces a force on each plate F = Q*E = Q^2/(e*A)
By separating the plates from distance d0 to distance 2*d0, I have to move them against that force F through a distance d0. The energy W is
W = Integral (F) with respect to d
Assume that A >> (2d0)^2 => end-effects of the field can be neglected, field E remains constant and the force remains constant.
(This is the assumption that will be violated when we move the plates 1000 miles apart.)
W = F* d0 = Q^2/(e*A) * d0 = Q^2*d0/(e*A) = Q^2/C0 = (C0V0)^2/C0 = C0V0^2
It appears the energy we have to expend is a factor of 2 higher than the increase in potential energy (which is 0.5*C0V0^2)
I'm not sure whether I have made a factor of 2 error in my calculation or perhaps… similar to the "lost energy" problem, if a fuller model were used then an energy loss mechanism would become apparent, similar to the resistance in the other problem.
cbarn, Irstuff, etc – what do you guys think?
RE: Capacitor Charging- Where are all the electrons?
When I calculated the field E (let's say on the positive plate), I did this based on the field from charge on the other plate (negative plate), ignoring the effect of the positive plate's charge upon the field E.
There may be good reasons for this. I invite comments. But I am going to use the opposite approach.
I am going to include the effect of the positive plate charge upon E (correct me if I'm wrong). Now we see that the field is E = Q/(e*A) on one side of the charge (toward the other plate) and the field is zero on the other side (toward outside of the cap).
So if Q lies on the boundary between two different fields, E= Q/(e*A) and E=0, what E do we use when calculating force F=Q*E. I propose that E=0.5Q/(e*A) is the appropriate charge. This of course would take care of the factor of 2 error in previous message.
Two justifications for this approach, a simple one and a more rigorous one.
The simple justification…. the charge lies on the boundary. If we imagine the charge to be a little sphere, it is half in the field and half out of the field. Use the average field of 0.5Q/(e*A)
A little more rigorous. Move the charge in small elements dq (perhaps stored on separate elements d_A).
That way it seems clear that dq itself has no effect on the overall field (very small element of charge), but all of the remaining charge should be considered.
When I move the first element dq from d0 to 2d0…. it immediately enters an area of zero field (the rest of the positive charge lies between it and the negative charge). The field that the first element sees is zero.
When I move the last element dq from 0 to 2d0…. it is between the rest of the charge where it sees the full field E=Q/(e*A)
So the field that is seen as each charge is moved will vary linearly from 0 (for the first differential element dq) to E=Q/(e*A) (for the last element). The average field seen is halfway between these: E=0.5Q/(e*A)
If I substitute this field E=0.5Q/(e*A) in my previous message, the mechanical energy F*d is equal to the change in potential energy.
Start here:
There is An electric field E = 0.5 Q/(e*A)
This produces a force on each plate F = Q*E = 0.5Q^2/(e*A)
By separating the plates from distance d0 to distance 2*d0, I have to move them against that force F through a distance d0. The energy W is
W = Integral (F) with respect to d
W = F* d0 = 0.5*Q^2/(e*A) * d0 = 0.5*Q^2*d0/(e*A) = 0.5*Q^2/C0 = 0.5*(C0V0)^2/C0 = 0.5*C0V0^2 = 0.5*W0.
I have to put an energy W0 into the system to move it from d0 to 2*d0. This increases stored energy from W0 to 2*W0.
RE: Capacitor Charging- Where are all the electrons?
(you knew what I meant)
RE: Capacitor Charging- Where are all the electrons?
With great possiblity of being entirely off-base in this line of reasoning I humbly submit the following for your consideration:
Would there be a general agreement that separating the plates of our capacitor when it is uncharged would have no effect on the voltage difference between the plates- it would unchanged? If so, would it be unreasonable to then say in similar fashion that there would be no effect on the charged device? In the uncharged device the amount of charge that would be stored is a function of the voltage to be applied. In the charged device the voltage difference between the plates is a function of the charge now stored - two different situations entirely. In the context of our discussions once a device of certain construction is charged the subsequent, appropriately qualified deconstruction can not alter the quantity of charge stored. Yes, the capacitance of the device has been altered and one would not again be able to store the same amount of charge in the altered state as one did in the unaltered state by applying the same voltage but that wouldn't be news, would it?
WhMcC
RE: Capacitor Charging- Where are all the electrons?
No, because the V=Q/C
so if Q=0 then V=0 regardless of capacitance
But if Q<>0, then V is inversely proportional to C, so if you separate the plates, C decreases and V increase, while if you move the plates closer together, C increases and V decreases.
It's all described by the same equation.
TTFN
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
I'm having a hard time understanding your point. Let me try to respond to some of those (similar to IRStuff's response) and see if we can come closer to understanding each other.
"Would there be a general agreement that separating the plates of our capacitor when it is uncharged would have no effect on the voltage difference between the plates- it would unchanged?"
Yes - no charge, no voltage.
"If so, would it be unreasonable to then say in similar fashion that there would be no effect on the charged device?"
No - in the charged device, you have to do work to move the plates. That adds energy (joules). Cap is open-circuited so charge (coulombs) is constant. Increasing joules with same coulombs means increasing volts. (volts ~ joules/coulombs).
"In the uncharged device the amount of charge that would be stored is a function of the voltage to be applied."
Maybe I don't understand what you mean by uncharged device. If there is no charge then voltage is zero, very uninteresting case.
"In the charged device the voltage difference between the plates is a function of the charge now stored"
... and the separation between the plates.
cbarn - in the entire discussion of moving the plates apart, there was never any change in the charge stored on the plates. I always called in Q. No "0" (initial) of "f" (final) postscript because it doesn't change if there is no connection to an external circuit.
RE: Capacitor Charging- Where are all the electrons?
I'm impossibly slow so I ask for your indulgence though all of this...
It seems that work is required to separate the plates regardless of the amount of charge on the plates, given that the greater the charge the more work required. Why is uncharged a special case where the work adds nothing to the picture?
Regards,
WhMcC
RE: Capacitor Charging- Where are all the electrons?
In the real world, there's friction and there's gravity, but they're outside of the capacitance and charge question.
TTFN
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
The same situation exists for gravitational potential energy. When you leap out a 10 story window, you convert the stored potential energy into kinetic energy. Neither you nor the Earth has any change to your gravitational attraction.
TTFN
RE: Capacitor Charging- Where are all the electrons?
"Your claim is that the enrgy in a charged cap is held on the plates"
That is not what I claim. The free charge is on the plates. The energy is stored in the field. The amount of energy in the field (for a given charge) depends upon distance between the plates.
"you also agree that moving the plates apart increases the energy"
yes.
"which must by your definition be stored on the plates."
No, the energy is stored in the field.
"Now you are saying that the charge on the plates dosen't change."
That was an assumption of the problem. Charge doesn't change. Distance is changed, causing a change in capacitance, energy, and voltage. Charge is assumed constant.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
Initially we had one cap C0 with plates at distance d0 and energy W0, and voltage V0.
We moved plates to distance d0 (required adding energy W0), resulting in new separation d=2d0, new voltage V=2V0, new energy W=2W0, same charge Q.
Now you want to take a conducting plate and insert it from the side in between the two plates. That requires no energy because there is no charge on the new plate, so total energy remains W=2W0. We now have the equivalent of two series caps. Each of these new caps is equivalent of one of the original caps in the following respects:
Each has a separation distance d0
Each has an energy of W0 (total energy 2W0)
Each has a voltage V0 for total voltage V0 across the two caps.
Each has a capacitance C0 for a total capacitance of C0/2 across the series combination (that's the way series caps add).
Each has a charge magnitude Q on the outer plate. The center plate might be viewed as having +Q belonging to one cap and -Q belonging to the other cap... but those cancel out and the center plate has no charge. It really doesn't alter the problem at all (how could it if it doesn't have any charge). The terminal relationships associated with the series combination remain the same.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
OK, so we had the outer two plates Plate 1 and Plate 3 at distance 2*d0. Then we installed a new perfectly-conducting infinitely thin plate (Plate 2) between them. This resulted in two capacitors, each with capacitance c0, voltage V0, energy W0. Charge on outer plates (1,3) is +Q and –Q. Middle plate (2) has +Q on one side and –Q on the other. But the middle plate has not net charge and really had no effect on the problem (remember it didn’t change the total energy stored, terminal voltage or capacitance of the series combination). Since the middle plate (2) is charge-neutral, it does not change the field distribution (except within plate #2 itself).
Now we remove let’s say the positive outer plate (Plate 3) to an infinite distance . Energy will be required to move plate #3 away from plate #1 (The same amount of energy as would be required even if plate #2 were not present). The field external to plate #1 will change to E0/2 on both sides of plate #1 (it used by be E0 only toward inside of capacitor). Middle plate (2) will keep charge +Q on it’s side toward plate #1 and –Q toward the other side. There is a net neutral charge on plate #2 => the field on the opposite side of plate #2 (away from #1) remains E0/2 (within short distances where plane-field approximation holds).
If we were to measure the voltage between plate #1 and plate #2 we would see voltage E0/2*d0 = V0/2. But there is NO energy stored in the interaction between plate #1 and plate #2. Only in the field associated with plate #1. This may sound surprising given that we have a voltage between these two adjacent plates, but consider the following:
1. There is no force on plate #2 since it is charge-neutral. We can move plate #2 around without expending any energy against an electric field. We can remove plate #2 altogether without adding any energy and without changing the field from plate #1 and energy stored in that field.
2. Note that we could also place our voltage probe at a distance d0 on the opposite side of plate #1 and we would still measure V0/2 volts difference from plate #1.
RE: Capacitor Charging- Where are all the electrons?
electricpete, yes, you have understood my uncharged case. Thanks for the additional information. Again, I was somewhat unclear as to why the work would not add to the system in that case.
This has been a great discussion. During my education in electronics this topic was covered from an electrical perspective but without any substantial foundation with regard to the physics behind the equations. You all have helped me greatly - thank you!
Think we can get to 100 posts on this one? Who would have imagined it?
WhMcC
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
If you want to idealize the question and assume no resistance, then obviously no energy is burnt off, so where does it go? To kinetic energy, you'd have an oscillator with perpetual energy transfer between potential charges and inductive current flow. The mechanical analogy would be a bouncing ball that continued bouncing forever, or a rolling ball that perpetually rolled back and forth between two hills.
Even with a resistance, some oscillation will occur. Same thing with the ball, again the analogy is exact. You'll have a slowly decaying oscillation between the caps, or ringing, which would translate to a ball which bounced a little lower each time or made it a little lower up the hills each time.
No matter what the resistance on the ball is (except for zero, which has already been discussed), the final state of the ball will be in the valley between the hills E=0. The final energy stored in the caps will be E/2. The analogy breaks down here due to the difference in final potential states: the ball is in the valley, whereas with the caps the charge is split between to caps (sooner or later internal resistance will drop the charge to zero and once again the analogy works).
RE: Capacitor Charging- Where are all the electrons?
This phenomenon happens all the time with one of the bigger capacitors we deal with in daily life, the atmosphere-earth capacitor. The earth is defined as zero volts, or "ground." No charge, right? But the clouds constantly develop a difference in potential, that is a difference in charge, with the earth, which causes lightning.
RE: Capacitor Charging- Where are all the electrons?
One more comment on the disappearing energy and resistance problem.
#1 - Notice the expression for energy dissipated by resistance does not depend on R. That means that I am perfectly allowed to set the limit as R->0. I feel satisfied to say that the energy remains constant in limit as R->0. A really loose way of looking at it:
As R->0:
* I increase in proportion to 1/R
* I^2 increases in proportion to (1/R)^2
* I^2*R increases in proportion to (1/R)
* Power =I^2*R increases in proportion to (1/R)
* Time interval ~ Tau ~ RC dependent upon R
* Energy ~ Power * Time ~ (1/R)*(R) is constant.
To me this gives some physical intuition to what is gong on. Yes R is going to zero which implies zero energy dissipation but I^2 is going toward infinity twice as fast which more than compensates. The limit as R->0 seems fine to me for this model.
#2 - I agree with cbarn that energy dissipation by em radiation due to high frequency oscillation is also a valid energy dissipation mechanism. This involves a more complex model.
Both answers are right IMHO. They are just based on different math models of the physical process. It should be recognized that any mathetmatical model will never fully represent all the complexities of real life.
RE: Capacitor Charging- Where are all the electrons?
My answer is: It will depend on the non-idealities in your attempt to implement this ideal circuit. The resistive heat dissipation depends upon some infinitessimally small element of resistance. The electromagnetic energy radiation for simplicity will depend upon some infinitessimally small element of inductance (after all there is no oscillation without inductance). A real circuit will have some of both. The relative proportions of remaining resistance and inductance will in some complex manner determine which proportion of energy is dissipated by which mechanism.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
Why use non-polarised capacitors?
The radiated energy can be measured easily and accurately with a good quality oscilloscope.
An experiment with 2 x 10,000 mfd capacitors charged to 150 volts will show a loss of about 6oj and that is easily detected. Try discharing the capacitors with a resistance of the same value as that used to link the capacitors together you can compare the energy lost with the energy retained.
In another thread I stated that this problem has its reciprocal equivalent in inductive circuits where the energy stored in two inductively coupled inductors is greater than the sum of the energy stored in each inductor.
Energy stored = 1/2La Ia^2 + 1/2 Lb Ib^2 + MIaIb
Cheers,
G
RE: Capacitor Charging- Where are all the electrons?
I'm not disagreeing, just still not 100% convinced. If you can provide more details of an actual experiment that has been done with the test setup and results accurately recorded I'd be interested.
cbarn - was there a previous question on the middle plate or something that I didn't answer?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
"Hi, has the middle plate got no charge? or has it got q+ on 1 side and q- on the other?"
Both. The middle plate has no NET (total) charge. It has got q+ on 1 side and q- on the other. The total of +q and -q is zero.
RE: Capacitor Charging- Where are all the electrons?
In the capacitor experiment it is not compulsory to use a switch or low value resisitor. Modern capacitors have very little leakage over relatively long periods of time and therefore a high value resistor can be used to limit the maximum current and radiated energy to levels that can be monitored and measured very accurately. The rate of charge and discharge has no bearing on the 'missing energy'.
Cheers,
G
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
I have already given a lumped circuit model with no inductance which shows that all of the energy goes to resistive heat dissipation in that model. Surely that model becomes more and more valid as we increase resistance which decreases rate of change and decreases e-m wave behavior.
Going the other direction, as resistance decreases, current rate of change and high-frequency content increases, inductive /reactance increases and becomes more and more important to the problem, more wave behavior is seen.
cbarn - The reason that we can have q+ on 1 side and q- on the otherside of a conductor if there is no current flowing... it is induced charge. We had an adjacent charged plate with net charge which drew opposite charge of the middle (uncharged plate) toward it. The middle plate ends up being polarized. Once again for a perfect conductor there can be no internal field so the conductor's charge will redistribute around it's edges in a manner that cancels the external field to provide zero field inside the conductor. If that conductor started as charge neutral and was not connected to any circuit, then it will remain charge neutral (total charge). I think maybe IRstuff gave some good explanations of this before also.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
There are configurations of the experiment that would require near-zero resistance and inductance. This can be accomplished with a moving-vane variable capacitor. If you believe that the equations that describe behavior of capacitors to be correct, then the same experiment of changing capacitance and stored energy must work equally well on a moving-vane capacitor. In that case, both resistance and inductance should be negligible and the energy behave according to the equations. Since this experiment is reversible and the time span is adjustable, it should be easy to show that the energy is not dissipated solely through heating or emission.
The experiment is no different than placing a ball on a seesaw. It gains and loses potential energy as it moves up and down with the seesaw. You can move the seesaw slow enough to not incur drag and other physical effects, but the potential energy must still change.
TTFN
RE: Capacitor Charging- Where are all the electrons?
I disagree. In potential / kinetic energy conversions as in a pendulum, for example, both energies can be measured at every stage from zero to 100% less losses which, in a well designed experiemnt such as a clock can be very small indeed. Not so with the capacitor experiment over 50% of the energy is always lost no matter how slow or how fast you conduct the experiment and that energy cannot be accounted for by measurements no matter how large the capacitors.
Heat losses, sparks, radiated emissions and all the other fairy tales used to explain the phenomenon do not account for the practical reality that when a charged capacitor is connected to another discharged capacitor having the same value over half the energy is lost and unaccounted for.
Cheers,
G
RE: Capacitor Charging- Where are all the electrons?
Please give us your analyis of what energy conversion is going on.
By your term "unaccounted for", am I correct in interpretting that you are suggesting that energy is not conserved?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
Oh dear how good our education system is at indoctrination. Of course I am not saying energy is not dissipated in sparks, radio emissions or heat. What I am saying is that in the first instance these losses can be minimised to almost negligible values as in other experiments, secondly the losses can be measured and finally, having measured and/or calculated all the known losses they do not account for the energy lost.
Indeed it would be nothing short of a miracle (an inexplicable phenomenon) if the losses always added to up to 50%. As far as capacitors are concerned charge is conserved Q = CV and this can easily be proved by practical experiments and does not involve mysterious losses. Energy stored = 1/2CV^2 and it is impossible mathematically and practically to conserve both and therefore in this experiment it would appear energy is not conserved.
Cheers,
G
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
Why would it be a miracle if energy is conserved? Why would we ever expect anything else?
RE: Capacitor Charging- Where are all the electrons?
The +Q is attracted to th bottom of the middle plate by the -Q on the bottom plate.
Imagine that the middle plate has 10 layers of charge-neutral atoms. Electrons are repelled away from the bottom (10th) layer of the middle plate by the -Q charge on the bottom plate. That electron from 10th layer moves to 9th layer and displaces and electron which will move to 8th layer and so-on until you have extra electron on the top side of the first layer. Certainly not an exact description, but just a way to imagine it.
Or if you prefer, think of the metal plate as a bunch of dipoles. When all the dipoles align in presence of external field, you will have one charge on bottom and opposite charge on top. Also not an exact descritpion, just another way to imagine what is going on.
Why don't they recombine? Because of the external field. That's what separated the charges and that's what keeps them apart.
A perfect conductor in presence of an external electrostatic field will always develop opposite charges at some points along it's outer edges.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
In response to your first sentence:
The electric field varies inversely as the distance for a point charge. But for a charge distributed on a plane such as parallel plate capacitor, the force does not change with distance (as long as you don't go too far away from plate that plane-behavior starts changing to point behavior).
In response to your second sentence:
The field which causes separation of the charges is external. There can be no internal field within a conductor during electrostatic conditions.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
I suggest it is physically impossible for any naturaly occuring mechanisim to exert a force of attraction with inverse square law characteristics.No one has ever succeeded in constructing a mechanical model with these characteristics.
Cheers
G
RE: Capacitor Charging- Where are all the electrons?
The assumption for a plane geometery (such as parallel plate capacitor) is
A >> d where A is area of the chargeand D is separation distance.
If you draw the field lines they are very nearly parallel to each other and perpendicular to the plates in close vicinity to the plates. Let's say you double the distance from d to 2d but still A>>2d. Then the field is still the same. Same field means same force per charge at 2d as at d
Nothing new there. College freshman have been learning that for years.
RE: Capacitor Charging- Where are all the electrons?
F=Gm1m2/r^2
TTFN
RE: Capacitor Charging- Where are all the electrons?
So d could also represent separation distance for charges drawn to opposite edges of the middle plate.
RE: Capacitor Charging- Where are all the electrons?
I must admit you make a compelling case for the lost energy in the charge transfer brainteaser. Good math back-up on it, too. I just don't know if I can buy it.
The reason for my skepicism is the two base equations you start with:
Q = C * V and E = 0.5 * C * V^2
have no means built into them to account for any of the losses discussed so far and yet they correctly predict the result. None of the other energy conversion models correctly anticipate the final result of a test. If, for example, I roll a car down a hill, I should be able to predict is final speed using basic, Newtonian equations. But I can't because none of those equations account for friction or other losses. So the car will not go as far as I predict it would. How can we assume that the idealised capacitance equations, which do not account for losses, correctly predict the final outcome and also attribute 1/2 the total energy distribution in the system to losses?
Looking at it a different way, if I simply reduced the voltage applied from an infinite energy source (which would replace any lost energy) to 1/2 it's original value, the single original cap would store 1/4 of the original energy, just like it would if we discharged it into another equivalent cap. This in spite of the fact that it is connected to an infinite energy source that will replenish any losses.
I don't claim to have the answer. I'm just not sure any of the explanations proposed so far are correct either, given the prediction of the ideal equation.
RE: Capacitor Charging- Where are all the electrons?
I must admit you make a compelling case for the lost energy in the charge transfer brainteaser. Good math back-up on it, too. I just don't know if I can buy it.
The reason for my skepicism is the two base equations you start with:
Q = C * V and E = 0.5 * C * V^2
have no means built into them to account for any of the losses discussed so far and yet they correctly predict the result.
Yes, conservation of energy using these equations can be used to correctly predict the energy dissipation without knowing the mechanism for energy dissipation. I see no problem with that. Regardless of where the energy goes… I know that if potential energy goes down by E0/2 than energy E0/2 has been transferred somewhere. That is the principle of conservation of energy.
None of the other energy conversion models correctly anticipate the final result of a test. If, for example, I roll a car down a hill, I should be able to predict is final speed using basic, Newtonian equations. But I can't because none of those equations account for friction or other losses. So the car will not go as far as I predict it would. How can we assume that the idealised capacitance equations, which do not account for losses, correctly predict the final outcome and also attribute 1/2 the total energy distribution in the system to losses?
Let’s say the car starts on top of a hill and then is allowed to roll into a valley between two hills where it will oscillate and finally come to rest. I don’t have to know the exact energy dissipation mechanism (air friciton, rolling friction, etc) to know that the total energy dissipated is equal to the gravitational potential energy (m*g*h).
Knowing the initial capacitance C and voltage V is equivalent to knowing the initial height of the car… it allows me to calculate initial potential energy.
Looking at it a different way, if I simply reduced the voltage applied from an infinite energy source (which would replace any lost energy) to 1/2 it's original value, the single original cap would store 1/4 of the original energy, just like it would if we discharged it into another equivalent cap. This in spite of the fact that it is connected to an infinite energy source that will replenish any losses.
I agree with that. No matter how you change the voltage, the potential energy formula PE=0.5*C*V^2 will still hold. I’m not sure what your the dilemma or contradiction is.
Maybe here is a different way to look at it. PE=0.5*C*V^2. Substituing V=Q/C we have PE = 0.5*C*(Q/C)^2 = 0.5*Q^2/C. The basic behavior which gave rise to our dlima is that potential energy does NOT vary linearly with Q (it actually varies with Q^2). So there is NOT a fixed amount of PE per charge regardless of voltage. So I can NOT expect in general to be able to transfer charges from one cap to another without changing the total energy.
Why shouldn’t each unit of charge carry an equal unit of PE? deltaPE = V*deltaq….where V itself is a function of total stored charge Q (V=Q/C). As a charge up the cap my V is increasing so each incremental unit of charge requires more and more energy to place it on the cap. Total PE is integral V*dq from q = 0 to Q = integral q/Cdq from 0 to Q = [0.5q^2/c] evaluated at Q minus [0.5q^2/c] evaluated at zero = 0.5*Q^2/C. Possibly drawing that integral graphically as a right triangle area shows the intuition that each unit of charge added (or removed) from a cap does not require the same amount of energy.
Once again... modeling resisitve energy dissipation was my first choice based on math ease in modeling and relative simplicity to understand. I don't rule out other possiblities like electromagnetic which will may also dissipate significant energy depending on the experimental setups. No matter what, we can rest assured that energy conservation will hold.
RE: Capacitor Charging- Where are all the electrons?
RE: Capacitor Charging- Where are all the electrons?
We are talking about forces between plates in a capacitor and forces between charges on opposite side of a given plate. I have repeatedly identified the constant field assumption applies for plane geometry A>>d^2 which clearly applies in this context. Nothing hidden here. You have free choice to disbelieve the constant-field assumption if you are so-inclined but then of course you must also discard the formula C=epsilon*A/d which is based upon constant field assumption.
RE: Capacitor Charging- Where are all the electrons?
I need to stop posting stuff so late. After reading them again my comments were pretty disjointed.
As you state, the energy required to incrementally add charge to a cap effectively increases with the square of the charge, due to the voltage increase in the cap. That part makes perfect sense so I will leave that alone.
But let's consider your car between two hills analogy. You correctly state that, intuitively, the car would eventually come to rest in the valley. However, if we used only the two energy equations that would seem to directly apply:
PE = mgh and KE = 0.5mv^2
and set the system in motion the car would never stop. Those two equations would swap energy perfectly indefinitely because there are no terms in the 'ideal' equations that produce any losses. What you would expect is that in a real test if you measure the speed of the car in the valley, the value of v would NOT be what you predict based on KE = PE = mgh. The difference would be the system losses.
In the capacitor question the voltages go to exactly what we expect them to. So where is the error from the 'ideal' equations that we would expect to see?
I think I may have just seen the error of my ways. We have the same discrepancy here as we see in mechanical systems with conservation of momentum. If you have two bodies of equal mass, one stationary and the other moving at velocity V and you model a collision between them with both bodies elastically deforming perfectly, conservation of momentum says you will end up with the two bodies moving at 0.5 * V. Momentum was conserved but the energy level of the system is cut in half. It's the same question; where's the energy lost to? So the thing I missed, as you state above, it the two base equations define fundamentally different quantities. You can't expect them to be equivalent.