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!

 

[electricpete]

cbarn -
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.

 

[WhMcC]

IRstuff, thanks for the clarification on work in this instance, that's where I was somewhat unclear.

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

 

[cbarn24050]

Hi electricpete, lets look at your middle paragraph. You state that the middle plate has q+ on 1 side and q- on the other, bearing in mind that this is a conductor and no current is flowing, how can this be?

 

[xnuke]

It's a charge distribution induced by the charges on Plates 1 and 3. Notice he said the middle plate is still electrically neutral (no net charge).

 

[peebee]

Back to the E/2 question, yes, the energy is burnt up in resistance, and the ball & hill analogy is a very good one.

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).

 

[peebee]

So far as the removed plate question goes, I agree completely with electricpete. One small clarification in terminology to keep IR off your back, rather than +/- Q, we're really talking about +/- delta-Q. The "uncharged" plate actually already had a bunch of charge on it, as it already had a bunch of electrons. They were just offset with protons so the net charge was zero. It can still have a delta in charge with respect to the other plate.

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.

 

[electricpete]

cbarn - glad to see your back and ready to get things lively again. I agree with xnuke's and peebee's explanation.

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.

 

[electricpete]

..... so you may ask: "How can both answers be right? Doesn't the energy have to go to one place or the other?"

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.

 

[cbarn24050]

Hi, has the middle plate got no charge? or has it got q+ on 1 side and q- on the other? please make up your mind. Peebee PLEASE lets try and keep the clouds out of this. As far as the lost energy problem is concerned its very easy to show experimentaly that the loss is mostly down to em radiation in a low resistance circuit.

 

[electricpete]

Midde plate has q+ on 1 side and q- on the other. Net (total) charge on the middleplate is therefore zero.

 

[electricpete]

I guess I can envision a few experiments to test how much lost energy goes into resistive heat or em radiation. Put the whole thing in a calorimeter to find heat? Try to measure the released em radiation? All well beyond my capabilities. Have you done any experiments or seen results of such experiments?

 

[cbarn24050]

Hi, ok so now answer my first question. As far as the experiment goes here's what you do. Get two large quality non electrolitic caps and a low resistance mosfet to act as a switch.Hook up the circuit, charge up 1 cap, then turn on the fet, monitor the result and do the calcs. The fet will aborb nearly all of the non em losses and they are easy to measure whats missing has been radiated.You can prove its radiation by changing the lenght of the hook up wire and observing the effect it has, a nearby am radio off tune will pick it up as well.

 

[gjones33]

Hi Cbarn24050,

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

 

[electricpete]

cbarn/gjones - OK, I don't disagree that energy can be dissipated by electromagnetic radiation. I am also sure that you can pick up something on an AM radio during the experiment that will demonstrate that there is em. I still am not 100% convinced these experiements have provided the means to demonstrate quantitatively how much is dissipated by each means. One question: what voltage are you measuring? How do you know you have captured all of the resistance between the terminals of your measurement. I agree a FET looks like much bigger resistance than perhaps a cap lead if we are talking about small currents. But as the current increases the FET voltage drop stays constant and the FET effective resistance goes down. So in the first instance of initial very high current (which I am also slightly skeptical that brief peak can be accurately measured) the effective resistance of the FET may be small in comparsion to other reistances in the circuit.

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?

 

[cbarn24050]

Hi electricpete, yes you didn't answer my first question 27 sept.

 

[electricpete]

Your question:
"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.

 

[gjones33]

Hi electricpete,

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

 

[cbarn24050]

Hi electricpete, no not that question the 1 before. How can you have q+ on 1 side and q- on the otherside of a conductor if there is no current flowing?

 

[electricpete]

gjones - you tell me the rate of discharge has no bearing on where the missing energy goes? I cannot accept that.

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.

 

[peebee]

If you want to measure what's being radiated, it seems to me you could run the experiment twice in a calorimeter, once as described above and the second time in a Faraday cage, just wrap the whole thing in aluminum foil. The EM should get converted to heat in the foil. The difference in measured heat should equal the EM radiation.