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!

 

[gjones33]

Hi Guys,

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

 

[electricpete]

I see that Whmcc and cbarn have figured out the vanishing energy question. Part of the answer is that when you connect a charged and uncharged cap, you will get a big boom which will dissipate energy. (similar to shorting the terminals of a single capacitor). There is always some small resistance in the circuit and as that resistance appraoches zero, instantaneous current at moment of shorting approaches infinity. Integral of power dissipated in the resistor over time will explain the "lost" energy.

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 &quot;bound&quot; 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 &quot;paired&quot; 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.

 

[electricpete]

By the way I'd like to take a moment to complement everyone on a good discussion.

I hope it's ok to disagree as long as we do it in a professional manner.

More comments to cbarn:

&quot;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&quot;
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.

&quot;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.&quot;
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.

 

[IRstuff]

I was going to write a big long thing, but I found pretty good web site with information about the question of how charge is distributed. This site is specifically about semiconducting p-n junctions, but the basic idea applies, which is that the drift current imposed by the electric field is countered by the diffusion current caused by the high concentration of charge (essentially Fick's Law applied to charged carriers).

http://ece-

http://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

 

[gjones33]

Hi All,

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 &quot;ether&quot;, 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

 

[electricpete]

From my perspective, there is no mystery in the energy loss and no reason as to why the energy dissipated would always be exactly 50% if some small resistance is included in the simple model, along with the two capacitors.

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.

 

[electricpete]

No criticism was intended by my comments about making the subject too complicated. There has been some good discussion by all. Hopefully there will be more.

 

[electricpete]

Minor correction to my writeup above. Substitute the following under
&quot;Total energy dissipated is the integral of power from t=0 to infinity&quot;:

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.

 

[electricpete]

you know what I mean...
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.

 

[IRstuff]

It's an interesting calculation, but seems somewhat unnecessary. The assumption is that the a charged capacitor C with a charge Q has 1/2Q^2/C energy stored. By doubling the C, but keeping the charge the same, the stored energy is 1/4Q^2/C, starting from scratch. This is equivalent to putting Q on a capacitor C and then adding C in parallel.

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

 

[electricpete]

IRStuff - You have pointed out that potential energy considerations lead us to conclude 1/2 the initial energy is lost. Specifically , when we double the effective C and keep Q the same, we come up with Ef=0.5*Q^2/(Cf) = 0.5*Q^2/(2C0) = 0.5*[0.5*Q^2/C0] = 0.5*E0.

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 &quot;lost&quot; 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 &quot;lost&quot; 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 &quot;lost&quot; from potential energy considerations (E0/2).

So, from my perspective, you have restated the obvious. Am I missing something?

 

[IRstuff]

It's the same result, but the point is that the resistance heating loss as you indicate can be set to 0 with zero series resistance, so the energy is not necessarily lost in heating the resistor, since your conclusion was:

&quot;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. &quot;

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

 

[electricpete]

Well, I'm not much of a superconducting particle physics guy. If you say it's so, I won't argue.

I liked it better when we were going to get free energy.

 

[cbarn24050]

Hi all, it seems i leave you for 2 days and you go beserk. Lets take the charge on plate problem, if its on the plate, surface or bulk, then where do all these electrons go? we know there is no room for them because if there were the conductors would act like a large sponge, and the speed of conduction would be very slow.If we have a vacuum capacitor and we charge it up to some voltage, then we move the plates apart so that the distance beween them is now double, the voltage on between them is now double, how can you explain this with your charge on plate theory? Lets take a capacitor with a solid dielectric can charge it up, remove the plates and short them together, wash the in the ocean whatever, when we replace them on the dielctric the capacitor is still charged, how come if the charge was on the plate all the time. Can an electron be strung out in space? yes it can, remember that an electron is a wave as well as a particle. Where does the lost energy go? all circuits have inductance so our 2 cap circuit will ocsilate, with each cycle some energy is lost to conductor and dielectric resistances but most goes in electromagnetic radiation.By the way these are easy experiments you can do at home, so you don't need to take my word for it.

 

[IRstuff]

V=Q/C

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

 

[IRstuff]

Actually, this example is an excellent corollary to electricpete's poser. If you then calculate the stored energy, it's doubled as well.

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

 

[electricpete]

bezerk…. who us? Well, maybe a little ;-)

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 &quot;the charge is on the plate&quot; 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.

 

[electricpete]

... and also for cbarn a star for keeping me in line. That's not always easy to do.

Thanks for all y'all's patience.

 

[electricpete]

More elaboration on my comments regarding the two types of charge:

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.

 

[electricpete]

A minor/obvious correction:
&quot;So if er = 4, then Qb = (3-4)/4 [/b]*Qf[/b] = 75% * [/b]Qf&quot;