Capacitor plate current skin effect

  • #1
FusionJim
42
11
Is it true that for a charged capacitor with a DC potential across it, the actual surface charge depth is very shallow, just few angstroms or few atomic layers so to speak, because if the plates of a parallel plate capacitor are made from a metal like copper where the free electron number is large these electrons can effectively screen the opposite charge with minimal thickness?


If this is so, then is it true that the surface charge depth/thickness would be actually deeper for a capacitor that is attached to an AC current source and is continually charged and discharged?

I think, I'm asking whether a changing current causes the E field to penetrate deeper into the capacitor plate surface than a static E field would with a steady potential across the capacitor?


Finally, is this also the case for a capacitor that is being physically dynamically changed, like a capacitor where one plate is much larger than the other and the smaller plate is being dragged around the bigger plate at some high velocity , would this cause continual charge redistribution on the larger plate and effectively cause similar effects to the surface charge depth - to increase the depth deeper than if the plates were stationary?




PS. To those that will answer this thread, could you also please look into my other thread in the classical physics category under electromagnetism - the topic is similar but I don't get any answers there :)
 
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  • #2
Each surface area of a conductor has capacitance to every other nearby conductor. Capacitance is a geometric and material parameter that relates charge imbalance, to differential voltage. C = Q / V.

FusionJim said:
Is it true that for a charged capacitor with a DC potential across it, the actual surface charge depth is very shallow, just few angstroms or few atomic layers so to speak, ...
Yes, A relatively negative conductor voltage causes electrons to protrude above the surface. While electrons on all relatively positive conductors nearby, are pushed down between the nuclei, exposing the positive nuclear charge.

FusionJim said:
If this is so, then is it true that the surface charge depth/thickness would be actually deeper for a capacitor that is attached to an AC current source and is continually charged and discharged?
No. The depth of electron displacement is related to voltage and capacitance. As a conductor voltage begins to change, a wave of electron displacement crosses the surface, establishing a new electron surface, across the entire capacitor. Charge flows a very short distance parallel with the surface. The total charge that flows is called the displacement current. There is a one-to-one relationship between the changing AC voltage, and the static DC voltage differences, so they involve exactly the same electron surface depth for the same capacitance.

FusionJim said:
I think, I'm asking whether a changing current causes the E field to penetrate deeper into the capacitor plate surface than a static E field would with a steady potential across the capacitor?
The initial E field propagates across, and does not penetrate the surface, you should consider the magnetic field. The displacement current propagates across the surface, not deep within the conductor. A perpendicular magnetic field, induced by the surface current, creates an equal and opposite current, deeper within the surface of the conductor, that cancels any current flow deeper within the conductor. The surface of a perfect conductor is therefore a perfect mirror, that prevents deeper penetration of changes on the surface.

FusionJim said:
Finally, is this also the case for a capacitor that is being physically dynamically changed, ...
If the geometrical or material capacitance parameters change, the ratio of Q / V will change, the electron surface displacement will change, so a surface current will flow to redistribute the charge. Changes in capacitance, that redistribute energy storage, may involve physical forces and the movement of materials. Energy conservation holds.
Dynamic changes in surface capacitance does not involve deeper effects. It is the surface geometry and material that is important.

FusionJim said:
PS. To those that will answer this thread, could you also please look into my other thread in the classical physics category under electromagnetism - the topic is similar but I don't get any answers there :)
If it was important, you would provide a quote or a link.
 
  • #4
Baluncore said:
No. The depth of electron displacement is related to voltage and capacitance. As a conductor voltage begins to change, a wave of electron displacement crosses the surface, establishing a new electron surface, across the entire capacitor. Charge flows a very short distance parallel with the surface. The total charge that flows is called the displacement current. There is a one-to-one relationship between the changing AC voltage, and the static DC voltage differences, so they involve exactly the same electron surface depth for the same capacitance.

So are you saying that an AC current flows differently within a parallel plate capacitor VS a regular conductor? Because in a regular conductor the skin depth is dependent on frequency. A 60 Hz AC current would actually penetrate at least 7 mm into the conductor. Let me guess before you answer, the difference is because within a current carrying conductor the E field drives the current along the conductor (assume a closed circuit) and even though the E field changes travel on the surface of the conductor but because the current flows it then has time to penetrate and go deeper into the conductor, meanwhile within a capacitor the current only flows until the charge has redistributed and once that happens the field doesn't penetrate deeper because the surface charge screens the field, correct?



Baluncore said:
The initial E field propagates across, and does not penetrate the surface, you should consider the magnetic field. The displacement current propagates across the surface, not deep within the conductor. A perpendicular magnetic field, induced by the surface current, creates an equal and opposite current, deeper within the surface of the conductor, that cancels any current flow deeper within the conductor. The surface of a perfect conductor is therefore a perfect mirror, that prevents deeper penetration of changes on the surface.
This is the same as with a changing B field within the core of a transformer right? The same reason why they can't use a solid core but need thin steel plates with laminated surfaces so that the field can travel fast (almost instantaneously) across the core and penetrate it but the penetration happens slower correct?
So what if I had a parallel plate capacitor where the plates where made from a very fine interconnected wire mesh, would there the field penetrate deeper with less "resistance" from the conductive surface?




Baluncore said:
If the geometrical or material capacitance parameters change, the ratio of Q / V will change, the electron surface displacement will change, so a surface current will flow to redistribute the charge. Changes in capacitance, that redistribute energy storage, may involve physical forces and the movement of materials. Energy conservation holds.
Dynamic changes in surface capacitance does not involve deeper effects. It is the surface geometry and material that is important.

I forgot to say that the smaller plate always keeps the same distance between the larger plate , just moves its position around.

You mention displacement current, which consists of the surface charge of the plate as I assume. Let me ask this question in the form of an analogy.
Can I think of the displacement current and the E field that drives it as a ship sailing the sea, the ship is the field and just like with a real ship the water directly underneath it flows a short distance with it until it mixes with the water further away from the ship and loses its kinetic energy.
So is it true that whether it is a stationary capacitor with a changing E field , or a small plate moved across a larger plate, the change in E field causes the surface charge to redistribute and this causes each charge (electron for example) to move in the direction of the field a little (by its mean free path?) So , for example, electrons don't just "jump up and down" from the sea of free electrons to match the surface charge of the opposite plate but they also travel a small distance each towards the changing E field , or the physically moving plate (if viewed from the position of the larger plate) ? This electron travel is parallel to plate surface ?


One more question then, electron random kinetic motion in metals is rather fast, but the bulk net movement in a specific direction due to E field is rather slow. What about the surface charge on the larger capacitor plate when the small plate is physically moved across it? Assume the smaller plate moves really fast, like 0.2c , is the short drift each electron experiences parallel to the surface happening at the same speed as the smaller plate moves?
 
  • #5
Your style is intractable. You tell a fragmented and cluttered fairy story, then ask if it is, "correct?". That fairy story is stuck in your mind, and cannot be migrated, or evolved, into the current theory. You must start from scratch to build a solid foundation.

You now bring up "skin effect", but that is not important when considering ideal capacitor plates as being perfect conductors, where C = Q / V operates between all opposed surface patches. Skin effect is important in transmission line conductors that are used to transport energy.

A displacement current flows in the leads of the capacitor, to distribute charge over all the plate area patches, proportional to the part they play in the total capacitance.

You are correct, transformer laminations provide magnetic access to the volume of the core. The laminations have a thickness of about one skin depth, in the lamination material, at the intended frequency of operation. The orientation of the laminations is arranged to reduce eddy currents.
 
  • #6
Baluncore said:
Your style is intractable.
Please excuse my style, this topic is complex enough.



1) Is the displacement current that you said the same in both an AC connected capacitor as well as in the example of a smaller plate that moves across a larger one? In the latter one the current would run across the larger plate as the smaller has already charge across its full surface area.


2) Specifically with regards to the example of a smaller plate being moved across a larger one , do the charges (electrons) on the large plate have a short run (mean free path before scattering?) velocity that is in the direction of the movement of the smaller plate - so a velocity parallel to the larger plate surface?

In other words if I was an observer from the stationary frame (same frame as the larger plate) and I could ignore the charge on the smaller moving plate, would the displacament current on the larger plate seem to me like an actual current similar to the one that runs in current carrying conductors?


3) And lastly, is the speed of actual electrons within the surface charge displacement current the same as the net drift velocity in current carrying conductor or is it different, higher?


Thank you, I hope I made the questions understandable and "tractable"
 
  • #7
1737999951686.png
Hello! I think this illustration will help you figure out your questions. The figure shows the electric field lines between the plate and the wire. The wire in this case can be considered a "smaller plate". As you can see, the field lines are evenly distributed between the larger and smaller plates. If the smaller plate is moved (dragged) relative to the larger plate without changing the distance between them, the system parameters will not change.
 
  • #8
@Ivan Nikiforov I believe this is only the case if the "conductor" or second plate is a considerable distance away from the first plate.

If the separation distance is small (much less than a milimeter) and the plate surface area is also small compared to the larger plate, then I believe this image doesn't hold , because the E field is largely contained between the overlapping surfaces.

Maybe other can comment too.
 
  • #9
FusionJim said:
@Ivan Nikiforov I believe this is only the case if the "conductor" or second plate is a considerable distance away from the first plate.

If the separation distance is small (much less than a milimeter) and the plate surface area is also small compared to the larger plate, then I believe this image doesn't hold , because the E field is largely contained between the overlapping surfaces.

Maybe other can comment too.
1738003813365.png

It seems to me that the pattern of electric field lines would look something like this, right? It is fundamentally important that the large plate, being the lining of the capacitor, has a charge over the entire surface.
 
  • #10
Ivan Nikiforov said:
View attachment 356435
It seems to me that the pattern of electric field lines would look something like this, right? It is fundamentally important that the large plate, being the lining of the capacitor, has a charge over the entire surface.
I haven't been following this thread, but that diagram does not look correct to me. The charge and field lines will be concentrated on the large plate in the vicinity of the small plate, I believe. The charge distribution on the large plate would not be uniform. Think of an infinitely large plate -- there is no way the charge could be uniform.
 
  • #11
berkeman said:
I haven't been following this thread, but that diagram does not look correct to me. The charge and field lines will be concentrated on the large plate in the vicinity of the small plate, I believe. The charge distribution on the large plate would not be uniform. Think of an infinitely large plate -- there is no way the charge could be uniform.
You're right. The second figure #9 shows that the density of the lines of force between the plates is higher than outside them. The main idea that I wanted to outline is that a large plate has a charge over its entire surface. If you move the smaller plate in Figure #9 to the left, the lines of force will enter the space between the plates on the left and will exit this space on the right. At the same time, the system parameters won't change, right?
 
  • #12
Ivan Nikiforov said:
View attachment 356435
It seems to me that the pattern of electric field lines would look something like this, right? It is fundamentally important that the large plate, being the lining of the capacitor, has a charge over the entire surface.
@Ivan Nikiforov That is easy to disprove, add another large plate on top of the small plate so that there are now two equally large plates with the small plate between them, in this case the E field can truly only mostly exist in the overlapping areas. I can't see how the very thin sides of the small plate (assume some 0.1mm copper plate) could extend an electric field to the sides because that would require for there to be a very high charge density on the very small surface on the plate sides.


Also is there any law that requires for there to be charge outside the overlapping region on the larger plate? I am not aware of any such law or rule.
PS. I read that the electric field falls off with distance, so in a sense you are right, but introducing another plate on top greatly reduces this effect so in that case the field would be mostly contained in the overlapping region
 
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  • #13
FusionJim said:
@Ivan Nikiforov That is easy to disprove, add another large plate on top of the small plate so that there are now two equally large plates with the small plate between them, in this case the E field can truly only mostly exist in the overlapping areas. I can't see how the very thin sides of the small plate (assume some 0.1mm copper plate) could extend an electric field to the sides because that would require for there to be a very high charge density on the very small surface on the plate sides.


Also is there any law that requires for there to be charge outside the overlapping region on the larger plate? I am not aware of any such law or rule.
1738007436660.png
In this book it is written as follows: "The electric field lines always end at the bodies forming the capacitor and approach the surface of these bodies perpendicular to it." A very high charge density on a very small surface on the sides of the plate does not depend on the size of the large plate, but on the magnitude of the voltage between the plates. As an argument that a large plate has a charge over its entire surface, I'll give you an example. There are thousands of miles of direct current transmission lines. If such a wire is connected to the "+" terminal of the generator, then it is obvious that the wire has a charge along its entire length. This charge can be measured at any point on the wire. Let's say at some point we bring a smaller plate closer to this wire and connect it to the "-" generator. Will this cause the "+" to disappear along the entire length of the wire?
 
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  • #14
If the smaller plate is shifted relative to the larger plate, the number of charges entering and exiting in the space between the plates will be equal. Thus, the area of high tension will simply move along the larger plate. In this case, there will be no electric current flow. The capacity and voltage remain the same.
 
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  • #15
FusionJim said:
Also is there any law that requires for there to be charge outside the overlapping region on the larger plate? I am not aware of any such law or rule.
Yes. It's probably easiest to think in terms of equipotential lines. Those equipotential lines are at the surfaces of the two conducting plates, and encircle the smaller plate. The E-field lines must cross the equipotential lines at right angles, and those E-field lines must start and end on charges.

1738010597900.png

https://www.researchgate.net/figure...of-microstrip-line-is-shown-in_fig6_291339518
 
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  • #16
Ivan Nikiforov said:
As an argument that a large plate has a charge over its entire surface, I'll give you an example. There are thousands of miles of direct current transmission lines. If such a wire is connected to the "+" terminal of the generator, then it is obvious that the wire has a charge along its entire length. This charge can be measured at any point on the wire. Let's say at some point we bring a smaller plate closer to this wire and connect it to the "-" generator. Will this cause the "+" to disappear along the entire length of the wire?
@Ivan Nikiforov I think the example of the HVDC wire is not the best one because there as you already said , the + is the long wire and the - is the small plate - in this situation ofcourse all the charge will reside between the two bodies with the small plate having an enormous charge density and the long + wire having charges across its length , although even here the charge density would be higher directly next to the small plate and decrease further away from it.

Rather why won't you look at the example I gave, two very long rectangular plates with a small rectangular plate between them, the width is the same for all 3 plates for simplicity but the length is not. Would you say that there is almost equal charge density across the two long plates (both long plates assume +, small plate -)?
It makes no sense because the E field is strongest in the area of overlap, so the further you go from the small plate the E field should get weaker and charge density decrease , is this not true?




@berkeman yes I agree the picture you provided with just 2 plates where the bigger one is not that much bigger than the smaller one seems true, but consider the scenario I provided with a short plate between two very long ones, I believe the situation would be much different.


My argument is this - with two long plates and one small, being very close together with a very thin dielectric the E field would naturally want to exist in the overlap region, for there to be charges further along the long plates it would mean the E field would need to extend from the small plate side through a very long layer of dielectric (the direction being parallel to the plate surface) in order to reach further down the long plates, from my understanding the E field would drop in value rather quickly therefore I can't see how there would be any considerable charge density further away from the small plate.
 
  • #17
FusionJim said:
I can't see how there would be any considerable charge density further away from the small plate.
It certainly falls off quickly, but is not zero. :smile:
 
  • #18
The key thing about a conductive capacitor plate, is that it is an equipotential, but the charge density distribution at the surface, will depend on the presence of other potentials nearby, that will repel or attract electrons, in or out between the more positive nuclei of the capacitor plate.

A small positive patch, close to a large negative plate, will draw electrons out of the large plate surface. Since C = Q / V, the charge drawn out will be Q = C⋅V ; That is, for a fixed voltage, charge is proportional to the capacitance.

If the positive patch is then moved around on the plate, the mound of electrons on the plate surface will be dragged about, following the moving patch. A momentary surface current of electrons will flow, where the edge of the small moving patch, covers or uncovers the larger plate.
 
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  • #19
Ivan Nikiforov said:
If the smaller plate is shifted relative to the larger plate, the number of charges entering and exiting in the space between the plates will be equal. Thus, the area of high tension will simply move along the larger plate. In this case, there will be no electric current flow. The capacity and voltage remain the same.
I agree , given neither the separation distance nor voltage nor total capacitance is changed, the number of charges will be the same. But because electrons travel a very small distance before scattering, as the small plate is moved I think it will still cause there to be a local current, because as it moves the field moves and it causes new electrons to be "drawn up and along" in the direction of movement. Even though each electrons net drift distance is small I believe it is still real and should be similar to a current, only on the surface. Would you agree?


berkeman said:
It certainly falls off quickly, but is not zero. :smile:
Yes it seems that way to me too. There is a gradient of the E field that falls off fast (although how fast is depending on the plate size ad geometry) and then continues to fall off slower until it becomes zero outside of the plates.



Baluncore said:
If the positive patch is then moved around on the plate, the mound of electrons on the plate surface will be dragged about, following the moving patch. A momentary surface current of electrons will flow, where the edge of the small moving patch, covers or uncovers the larger plate.
Great, just like I thought. And this differs from a non moving stationary capacitor where surface charge just "comes up" but doesn't move parallel to plate surface , correct?

In the stationary capacitor the surface charge is very shallow, is it true that for the moving plate scenario due to the spatially changing E field the depth to which charge is affected increases and becomes equal with the skin depth normally observed in conductors per given frequency. The frequency in this case is the speed with which the small plate is moved across.?


Thank you.
 
  • #20
FusionJim said:
Great, just like I thought. And this differs from a non moving stationary capacitor where surface charge just "comes up" but doesn't move parallel to plate surface , correct?
No.
 
  • #21
Baluncore said:
No.
Given your short answer I have to guess. I think the reason you said "No" is because unlike in the text book examples where a parallel plate capacitor is shown as two identical plates with each plate having a wire at the exact middle, in regular capacitors the wires attach to one side or both sides, so upon connecting a PD to the capacitor the E field would not appear uniformly across the capacitor plate overlap area but instead "move" from the side where the wires are connected and then with some delay reach the furthest side where there are no wires connected. Upon this movement there would be some charge movement that is parallel to the plate surface. Am I correct in my assessment ?
 
  • #22
FusionJim said:
but instead "move" at the speed of light in the dielectric medium
(fixed that for you) :wink:

You are talking about transient transmission line effects...
 
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  • #23
berkeman said:
Yes. It's probably easiest to think in terms of equipotential lines. Those equipotential lines are at the surfaces of the two conducting plates, and encircle the smaller plate. The E-field lines must cross the equipotential lines at right angles, and those E-field lines must start and end on charges.

View attachment 356442
https://www.researchgate.net/figure...of-microstrip-line-is-shown-in_fig6_291339518
When the large plate is infinite and earthed we can apply mirror image method to get electric field E, i.e. that of a small plates pair with double distance. The infinite plate holds same amount of negative charge with the small plate.

( OP question in another thread may be staed in this infinite ideal condition that when the infinite Earthed plate moves horizontally with close to light speed does E change or not ? )
 
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  • #24
FusionJim said:
Am I correct in my assessment ?
No.
Do you remember what I wrote about your intractable style.

The external circuit, and the capacitor leads, are irrelevant. When you move a small charged patch, over a larger plate, no displacement current flows in the capacitor leads.

You continue to extrapolate from your previous guesses, in an attempt to prove that, some irrelevant assumption, made earlier, must be true in all situations.
 
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  • #25
Baluncore said:
No.
Do you remember what I wrote about your intractable style.

The external circuit, and the capacitor leads, are irrelevant. When you move a small charged patch, over a larger plate, no displacement current flows in the capacitor leads.

You continue to extrapolate from your previous guesses, in an attempt to prove that, some irrelevant assumption, made earlier, must be true in all situations.
Look @Baluncore I am sorry my writing is intractable. I am trying my best but maybe you are too demanding for my ability. It takes two to tango...


I understood, I did not mean a moving capacitor in my last post.
I agree with you actually.
1) The charged capacitor where one plate is smaller than the other and is moved across the larger plate , there should be a surface current within the larger plate directly beneath the smaller plate, just like you said. I think we both think the same here.
I believe the same effect could be done electronically - by having a bunch of small segmented plates that are switched in sequence and discharged in sequence , ON/OFF, this should simulate a moving E field and result in a surface current on the plate beneath the segmented plates.


2) In my last post I was asking about a stationary capacitor, like a regular parallel plate capacitor and whether you also have a surface current within a stationary parallel plate capacitor.
My guess is that you have because the E field spreads out over the plate area from the point where the incoming wire is attached to the plate. So if the wire is attached at one plate end, then the E field starts there until it moves across the plates (which happens fast) but still I think there should be some electron velocity parallel to the surface of the plates , correct?
 
  • #26
1738167216542.gif
Perhaps this animation demonstrates the process you're talking about? If this is the case, then as the smaller plate moves, the electrons simply oscillate up and down. Obviously, there is no surface current in this case, since the electrons do not move in the direction of the plate movement.
 
  • #27
Ivan Nikiforov said:
View attachment 356545 Perhaps this animation demonstrates the process you're talking about? If this is the case, then as the smaller plate moves, the electrons simply oscillate up and down. Obviously, there is no surface current in this case, since the electrons do not move in the direction of the plate movement.
I am not confident that this simulation shows the full picture. I believe there is a perpendicular to surface current present just like in your animation, but I believe there also has to be a parallel to surface current present.

I think of a positive test charge being placed above the long plate and then moved around, the E field is moving so it interacts with multiple electrons at any given time, so as the field moves it must be that the electrons can't just move up and down , because the field interacting with those electrons has a parallel to plate direction therefore the electrons must also be deflected and move in the direction of the field by their mean free path just like they do in a current carrying wire.

Sure there is not a net current because both positive and negative charges are equal and move at the same speed but if we ignore the upper plate or charge, then there I think is a real net movement of electrons on the lower larger plate.

This is my understanding.
 
  • #28
FusionJim said:
... because the field interacting with those electrons has a parallel to plate direction therefore the electrons must also be deflected and move in the direction of the field by their mean free path just like they do in a current carrying wire.
"The electric field lines always end at the bodies forming the capacitor and approach the surface of these bodies perpendicular to it"
 
  • #29
Ivan Nikiforov said:
"The electric field lines always end at the bodies forming the capacitor and approach the surface of these bodies perpendicular to it"
I believe this is true only in the static case. If there is a moving E field (whether because of physical motion of charged object or electronic switching) then there should be two components, one perpendicular and one parallel to surface or at an angle, because it takes time for the charge to redistribute to the new position and as it is redistributing/aligning the field is not perpendicular which I believe is what drives that redistribution, otherwise there would be no need for charge to redistribute.

Please correct me if I am wrong. I believe @Baluncore said the same thing.
Baluncore said:
If the positive patch is then moved around on the plate, the mound of electrons on the plate surface will be dragged about, following the moving patch. A momentary surface current of electrons will flow, where the edge of the small moving patch, covers or uncovers the larger plate.
 
  • #30
FusionJim said:
Please correct me if I am wrong.
You are wrong.

The mound of electrons under the patch is moved, very slowly, by electrostatic attraction from above. The entire capacitor plate is an equipotential, so there is no differential voltage across the plate. C = Q / V, has fixed voltage, so Q is decided by the changing distribution of C.
 
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  • #31
Baluncore said:
A small positive patch, close to a large negative plate, will draw electrons out of the large plate surface. Since C = Q / V, the charge drawn out will be Q = C⋅V ; That is, for a fixed voltage, charge is proportional to the capacitance.

If the positive patch is then moved around on the plate, the mound of electrons on the plate surface will be dragged about, following the moving patch. A momentary surface current of electrons will flow, where the edge of the small moving patch, covers or uncovers the larger plate.
Well , is the surface current that you talk about parallel to the plate surface or perpendicular? Because @Ivan Nikiforov claims and shows even with an animation that there is only perpendicular movement of electrons due to the change in plate location. You left out that distinction but that is an important one, which causes some misunderstanding to me.
 
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