Capacitor Charging- Where are all the electrons? 11

[WhMcC]

Greetings,

I am hoping that someone may have some insight here:

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

Thanks for your thoughts on this one!

 

[IRstuff]

Just to satisfy WhMcC's request to get to 100 posts ;-)

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

 

[gjones33]

Hi IRstuff,

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

 

[electricpete]

gjones - so now you are ruling out both resistive heat losses and radiative emissions (and sparks) as energy dissipation mechanisms?

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?

 

[peebee]

I'd then suspect the energy is being converted to mass and would suggest weighing the circuit after running the experiment.

 

[gjones33]

Hi all,

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

 

[cbarn24050]

Hi electricpete, these charges will have a great attraction for each other, the metal will allow them to meet and recombine, wont it? gjones if you build this cicuit in space from superconducting materials the circuit will oscillate at its resonant frequency.As time passes and energy is radiated away the amplitude of the oscillation decreases, this sinewave is centered on V/2, so the wave decays to V/2.

 

[electricpete]

gjones - you wrote "Indeed it would be nothing short of a miracle (an inexplicable phenomenon) if the losses always added to up to 50%"

Why would it be a miracle if energy is conserved? Why would we ever expect anything else?

 

[electricpete]

cbarn - in the problem we were talking about we had a plate on bottom with net charge -Q, and a plate in the middle with net charge zero. You asked how the middle plate could maintain +Q on its bottom and -Q on its top (referring to bottom and top faces of the middle plate). i.e. why don't the +Q and -Q recombine.

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.

 

[electricpete]

cbarn - Maybe there is one more way to explain it. Consider what would happen if the charges on the perfect conductor in presence of external electrostatic field did not split up around the outer edges.... then you would have an electrostatic field inside the conductor... which has to lead to current flow... which will violate the assumption of electrostatics and cause the charges to move/change over time. The charges will not stop moving until there is no field within the conductor. The only way to get no field within the conductor in presence of an external applied field is for the conductor to provide charges around it's outer edges.

 

[cbarn24050]

Hi electricpete, the force of attraction between oposite charges is proportional to the inverse of the distance between them, therfore a charge some distance away (on another plate) would exert much less force trying to seperate charges from each other that the force trying to recombine them. There is no voltage difference between any surface of the conductor and therefore no electric feild within it to seperate charges either.

 

[electricpete]

greetings, cbarn

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.

 

[cbarn24050]

hi electicpete, it seems youv'e discovered a new law.

 

[electricpete]

What I have written is just normal textbook electrostatics. Nothing new, creative, or controversial to the best of my knowledge.

 

[cbarn24050]

hi, try and find the bit about force not changing with distance.

 

[gjones33]

Hi all,

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

 

[electricpete]

cbarn -
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.

 

[IRstuff]

Nature indeed has an inverse-square law phenomenon right below your feet.

F=Gm1m2/r^2

TTFN

 

[electricpete]

I should clarify in my above example that d does not have to be separation distance between two plates. It is any small distance from a plane charge where we want to determine the field (which is proportional to force on a charge at that location).

So d could also represent separation distance for charges drawn to opposite edges of the middle plate.

 

[kamenges]

Hi electricpete-
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.

 

[electricpete]

Hi kamenges
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.