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]

IRStuff gave an excellent answer to cbarn's question about moving the plates apart in an air capacitor (not connected to any external circuit). As he proves, the capacitance havles, voltage doulbes, and stored energy doubles.

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.

 

[IRstuff]

The converse is that once you have separated plates and allow them to get closer together, you get the energy back.

This is ultimately where bulk of the energy goes.

TTFN

 

[electricpete]

At first it seems a little non-intuitive that voltage can increase simply by pulling plates apart. But a simple way to see it is to remember that volts = joules per coulomb. If I put in work (joules) without changing the charge (coulombs), I will increase the voltage.

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

 

[cbarn24050]

Hi, i didn't think it was a good answer, he just restated the question. What I want to know is how the charged plate theory causes this effect.

 

[electricpete]

The increase in energy should be explained by the force applied. I worked it through and it appears we will have to expend an energy equal to twice the increase in potential energy in order to move the plates apart from d0 to 2xd0.

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?

 

[electricpete]

Ok, I'd like to take a stab at the reason for the factor of 2 error in my last problem.

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.

 

[electricpete]

"...0.5*C0V0^2 = 0.5*W0" should be changed to "...0.5*C0V0^2 = 1.0*W0"

(you knew what I meant)

 

[WhMcC]

Hello all!

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

 

[IRstuff]

"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? "

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

 

[cbarn24050]

Hi, you are still not explaining HOW the plates change their charge, there is no external circuit, so how do the charges get from 1 plate to the other?

 

[electricpete]

whmcc -
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.
&quot;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?&quot;
Yes - no charge, no voltage.

&quot;If so, would it be unreasonable to then say in similar fashion that there would be no effect on the charged device?&quot;
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).

&quot;In the uncharged device the amount of charge that would be stored is a function of the voltage to be applied.&quot;
Maybe I don't understand what you mean by uncharged device. If there is no charge then voltage is zero, very uninteresting case.

&quot;In the charged device the voltage difference between the plates is a function of the charge now stored&quot;
... 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 &quot;0&quot; (initial) of &quot;f&quot; (final) postscript because it doesn't change if there is no connection to an external circuit.

 

[WhMcC]

electricpete,

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

 

[IRstuff]

Work in physics is defined as the integral of force*distance. Therefore, when there is no charge, there is no attractive force between the plates, so as far as the ideal world of capacitors are concerned, if there is no charge, there is no work involved.

In the real world, there's friction and there's gravity, but they're outside of the capacitance and charge question.

TTFN

 

[cbarn24050]

Hi elecricpete, thats exactly my point. Your claim is that the enrgy in a charged cap is held on the plates, you also agree that moving the plates apart increases the energy, which must by your definition be stored on the plates. Now you are saying that the charge on the plates dosen't change.

 

[IRstuff]

The charge is on the plates. The energy is stored in the positioning of the charge.

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

 

[electricpete]

cbarn - I hope you don't mind me responding very specifically to each statement. I think maybe we're getting down to only a difference in terminology.

&quot;Your claim is that the enrgy in a charged cap is held on the plates&quot;
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.

&quot;you also agree that moving the plates apart increases the energy&quot;
yes.

&quot;which must by your definition be stored on the plates.&quot;
No, the energy is stored in the field.

&quot;Now you are saying that the charge on the plates dosen't change.&quot;
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.

 

[electricpete]

WHMCC - In my mind the uncharged case is where q=0. That means v=0. There is no electric field and no force. Moving the plates against zero force does not require any energy. It seems like a very uninteresting problem. Have I understood correctly your definition of &quot;uncharged case&quot;?

 

[cbarn24050]

Hi, what happens if we now slide another metal plate between the plates of our charged cap? So now we have 2 caps in series or do we? It we assume we do then is the center plate charged? If so, how did the charge get there?

 

[electricpete]

That's easy. To set the whole problem go back to the beginning.

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.

 

[cbarn24050]

Hi, ok now take 1 of the outer plates away, does the center plate now have a charge or not, if it does then where did the charge come from, if it doesn't then we are left with a cap with 1 plate charged and 1 plate not charged. I'm away for 2 weeks so i'll pick up this then.