Can electrons "fly" between capacitor plates?

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SUMMARY

The discussion clarifies that electrons do not "fly" between capacitor plates; instead, they move through the circuit connected to the capacitor. When a voltage is applied, charge builds on the plates, creating an electric field that influences charge movement in the circuit without crossing the dielectric gap. This phenomenon is explained through the concept of "displacement current," which accounts for the apparent flow of current in a closed loop involving capacitors. The energy is stored in the dielectric material rather than on the plates themselves, and excessive voltage can lead to dielectric breakdown, causing capacitor failure.

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x86
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How exactly do the electrons move from the capacitor to the rest of the circuit? Do they fly between the capacitors plates, or can they only travel where there is a wire connection? (I.e., if a small chunk of our circuit looks like this: A-------| |-------B can an electron start from A, go to our plate, then fly across the plate and go to B? Or can the electron only move backwards from the plate back to A?)
 
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From what I understand, if you were to track an electrons journey, it would never cross the gap (provided a perfect dielectric between the plates). If A were connected to the positive side of a battery, and B to the negative side, a negative charge would build on plate B, repelling the electrons from plate A resulting in a net positive charge on plate A.
 
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Electrons don't cross the gap. If they did it would represent dielectric breakdown and a resistive path, and the capacitor would not be able to store charge for very long.

The plates in a capacitor are so close together that any build-up of charge on one plate causes field lines to reach across and influence an equal number of charges on the other plate. So if Q electrons are added to one plate, they will repel that same number from the other plate, sending them into the circuit that second plate is connected to. To an outside observer it may therefore seem that current is flowing "through" the capacitor, but the non-conducting dielectric in the path means it is not.
 
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The above description is based on the effect called "electrical influence".
More than that, in case the capacitor is part of a closed current loop we observe a current within the loop:
* a decaying current if a dc voltage is switched-on, or
* a continuous ac current in case of an ac voltage source.

Because this observation needs an explanation in the "current domain", we have invited the term "displacement current" (through the capacitor).
 
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Well, they can fly between capacitor plates, but it is not recommended they do that too much... It is way much better they just dance on plates.
 
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NascentOxygen said:
Electrons don't cross the gap. If they did it would represent dielectric breakdown and a resistive path, and the capacitor would not be able to store charge for very long.

The plates in a capacitor are so close together that any build-up of charge on one plate causes field lines to reach across and influence an equal number of charges on the other plate. So if Q electrons are added to one plate, they will repel that same number from the other plate, sending them into the circuit that second plate is connected to. To an outside observer it may therefore seem that current is flowing "through" the capacitor, but the non-conducting dielectric in the path means it is not.

But then isn't the circuit not closed? For instance, if we have a simple capacitor connected to a battery, then electrons wouldn't be able to flow across the capacitor. (Although this now negative plate of the capacitor will repel electrons from the opposite plate, making it positive). What happens when eventually tons of electrons are crammed at the negative plate, and there are no more electrons to repel on the opposite plate?

Like, eventually there will be so many electrons crammed at one of the capacitor plates, and there will be no more electrons to repel on the opposite plate?? Therefore it seems like there will be no more current in the circuit, even though it is hooked up to a battery?
 
The secret to understanding capacitors is to realize that the energy is stored not on the plates but in the dielectric.

Good dielectrics have polar molecules that the electric field 'twists' out of their preferred alignment.

diel.gif

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html

Water for instance is a very polar molecule. That's why water has a dielectric constant 80X that of free space.
0016.jpg


Why free space has a dielectric constant i do not know - there's nothing there to deform.
old jim
 
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x86 said:
But then isn't the circuit not closed? For instance, if we have a simple capacitor connected to a battery, then electrons wouldn't be able to flow across the capacitor.

that's correct

(Although this now negative plate of the capacitor will repel electrons from the opposite plate, making it positive). What happens when eventually tons of electrons are crammed at the negative plate, and there are no more electrons to repel on the opposite plate?

then the capacitor fully charged equal to the voltage potential across it. Increasing the voltage applied to the plates will increase the charge
Eventually, depending on the construction of the capacitor, the increasing voltage will get to a point where the electric field between the plates will cause a
breakdown of the dielectric and it will cause a current to flow. At that point, the capacitor has failed.

Dave
 
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  • #10
x86 said:
But then isn't the circuit not closed? For instance, if we have a simple capacitor connected to a battery, then electrons wouldn't be able to flow across the capacitor.
How do you define "closed"? What do you observe in the above described case?
s there a flow of charges (not "electrons")? Yes, there is.

We have a so called charging current (or loading curent) until the capacitor is charged up and its voltage is equal to the battery voltage.
This charging process assumes (at least!) a certain wiring resistance and/or a finite source resistance (internal to the battery), otherwise the connecting of the capacitor to the battery would cause something like a "Dirac impuls" which is unrealistic.

As mentioned in my reply#4 during this charging process we define a so-called "depletion current" to fill the "optical gap" caused by the capacitor.
Please note that, therefore, we consider a circuit as "closed" if such a depletion current can flow. This is true for all ac waveforms as well as for dc circuits during the switch-on period (charging period).
 
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  • #11
Here is yet another example where electrons just add complication. The fact that they don't actually flow across the gap in a Capacitor is no more relevant to the electrical situation than the fact that they may not even reach the end of the wire between the switch being turned on and then off.
What have electrons actually got to do with electricity? (A Joke but not totally a joke)
 
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  • #12
sophiecentaur said:
Here is yet another example where electrons just add complication. The fact that they don't actually flow across the gap in a Capacitor is no more relevant to the electrical situation than the fact that they may not even reach the end of the wire between the switch being turned on and then off.

Yes - this illustrates the fact that "current" is NOT identical to "flow of electrons".
Instead, current is "movement of charges" - caused by electrons, but the "body" of the electrons is moving much slower than the charges.
 
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  • #13
LvW said:
Yes - this illustrates the fact that "current" is NOT identical to "flow of electrons".
Instead, current is "movement of charges" - caused by electrons, but the "body" of the electrons is moving much slower than the charges.

I'm sorry but I don't understand this point at all. How can current not be identical to the flow of electrons when current = coulombs/second? Yes it is a flow of charge but that charge can only flow with the flow of electrons, which not carry, but ARE that charge. One coulomb has a defined number of electrons associated with it. So if charge flows, so do electrons.
 
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  • #14
It is really necessary to distinguish between the "drift velocity" of electrons (see here: http://en.wikipedia.org/wiki/Drift_velocity) and the "charge velocity" which is app. identical to the speed of light. This is because the charge of an electron is a property of an electron (and not the electron itself).
 
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  • #15
I think you're confusing charge of charge carriers and the propagation of voltage across a circuit, which is what travels near to the speed of light. Even if charge is a property of an electron, you can't transfer that charge any faster than the movement of the actual electrons in a circuit.
 
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  • #16
No - I am not mixing things, but you are right: I have expressed myself rather unclear.
Let me try agian: The drift velocity of one single electron is rather low.
However, the thing we call "current" consists of charge movement (charge per time unit) - and because we have a huge number of electrons in a wire, we have a rather large amount of charges crossing a certain point within or at the end of the wire.
 
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  • #17
x86 said:
What happens when eventually tons of electrons are crammed at the negative plate, and there are no more electrons to repel on the opposite plate?
I think you needn't worry about running out of free electrons, the free electron density in metals is so high. http://hyperphysics.phy-astr.gsu.edu/hbase/tables/fermi.html#c2
 
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  • #18
LvW said:
No - I am not mixing things, but you are right: I have expressed myself rather unclear.
Let me try agian: The drift velocity of one single electron is rather low.
However, the thing we call "current" consists of charge movement (charge per time unit) - and because we have a huge number of electrons in a wire, we have a rather large amount of charges crossing a certain point within or at the end of the wire.

I suppose we were probably chasing each others tails and going round in circles. Glad to have cleared that up.
 
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  • #19
x86 said:
But then isn't the circuit not closed? For instance, if we have a simple capacitor connected to a battery, then electrons wouldn't be able to flow across the capacitor. (Although this now negative plate of the capacitor will repel electrons from the opposite plate, making it positive). What happens when eventually tons of electrons are crammed at the negative plate, and there are no more electrons to repel on the opposite plate?

Like, eventually there will be so many electrons crammed at one of the capacitor plates, and there will be no more electrons to repel on the opposite plate?? Therefore it seems like there will be no more current in the circuit, even though it is hooked up to a battery?

Maybe this will help with the confusion.
sefton.pdf
 
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  • #20
nsaspook said:
Maybe this will help with the confusion.
sefton.pdf

This article looks amazing. I'm reading it now. Thank you.
 
  • #21
nsaspook said:
Maybe this will help with the confusion.
sefton.pdf

As good as that article is, it doesn't actually explain how the energy is transferred into the light bulb filament and converted into light (and heat), unless I missed it.
 
  • #22
FeynmanFtw said:
As good as that article is, it doesn't actually explain how the energy is transferred into the light bulb filament and converted into light (and heat), unless I missed it.

No it doesn't but how a battery actually works is not vitally important to the concept to EM energy transfer past it's terminals or into the terminals of the light bulb filament. Once you understand the basic principles of the article it's much easier to visualize those processes in electrical terms that are built from a solid foundation.
 
  • #23
When I was young and had no sense, I sent electrons flying between capacitor plates any number of times. Just cringe a bit as you up the voltage... Till an event happens.
Sigh... no more capacitor...
 
  • #24
nsaspook said:
No it doesn't but how a battery actually works is not vitally important to the concept to EM energy transfer past it's terminals or into the terminals of the light bulb filament. Once you understand the basic principles of the article it's much easier to visualize those processes in electrical terms that are built from a solid foundation.

I have to say though, that pdf has given me more questions than answers. I have never ever heard of a non-uniform surface charge on a straight conducting wire being the cause of current flow. Surely such an effect would be measurable and hence mentioned at least somewhere?
 
  • #25
Thanks Mr 'spook

that sefton paper answered a question I've carried around for fifty years (really).
Namely - i once saw a statement that 'energy doesn't flow in the wires but in the fields surrounding them.'
That really confused me because i thought of volts as analogous to enthalpy of steam, and the fields surrounding a wire carrying power just a curious side effect.

loooks like now i'll have to figure out what is a Poynting vector...

Drat now you've got me wondering is there a parallel in Thermodynamics. There's some reason everybody loves a steam engine...

old jim
 
  • #26
FeynmanFtw said:
I have to say though, that pdf has given me more questions than answers. I have never ever heard of a non-uniform surface charge on a straight conducting wire being the cause of current flow. Surely such an effect would be measurable and hence mentioned at least somewhere?

It is but usually in the context of high-voltage where the effects are easily measurable.
http://adsabs.harvard.edu/abs/1996AmJPh..64..855J
 

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  • #27
jim hardy said:
Thanks Mr 'spook

that sefton paper answered a question I've carried around for fifty years (really).
Namely - i once saw a statement that 'energy doesn't flow in the wires but in the fields surrounding them.'
That really confused me because i thought of volts as analogous to enthalpy of steam, and the fields surrounding a wire carrying power just a curious side effect.

In the case of low frequency and DC circuits it's the system that makes it work. The wire with mobile charges or the fields in isolation don't completely explain it. It's only when you look at them as a whole working together that it really clicks. Beyond basic theory the Poynting vector is not really useful for power engineering. It's a cool thing to understand but has limited use in the field.

http://ieeexplore.ieee.org/xpl/login.jsp?tp=&arnumber=1564217&url=http%3A%2F%2Fieeexplore.ieee.org%2Fxpls%2Fabs_all.jsp%3Farnumber%3D1564217

http://www.lsczar.info/wp-content/uploads/2014/07/97-Poynting-Vector.pdf
 
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  • #28
nsaspook said:
It is but usually in the context of high-voltage where the effects are easily measurable.
http://adsabs.harvard.edu/abs/1996AmJPh..64..855J

Thanks for the link, that was an interesting read. But unfortunately, it doesn't answer my question. :(
 
  • #29
FeynmanFtw said:
Thanks for the link, that was an interesting read. But unfortunately, it doesn't answer my question. :(

If the question is about light bulbs then maybe a Google search on 'the physics of a light bulb' would help.
 
  • #30
nsaspook said:
If the question is about light bulbs then maybe a Google search on 'the physics of a light bulb' would help.

No, not about the light bulb, but about the non uniform surface charge. That paper simply goes on to show how we can see A surface charge. It doesn't explain anything about how it varies across the wire's length.
 

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