A capacitor as an open wire that may induce current

Thread Starter

pigpen

Joined Jan 26, 2016
23
What is this "LIEC" thing?
Oh sorry it is Lessons in Electric Circuits, the book that was the origin of the AAC book / this website (unless I'm wrong). I learned it is called that from the "Read before posting" message for new members.
 

Thread Starter

pigpen

Joined Jan 26, 2016
23

The thinking is often the other way around: an open wire as a capacitor.
Thank you.


But what you won't learn is just how a field can act over a distance, with no medium in between. How exactly does a magnet work, or gravity? You have to study to an advance level to begin to grasp how a simple magnet works, beyond precisely describing what anyone can see.
What would be a reasonable way to describe the purpose of the insulator between the two metal plates, if we don't know enough to say no significant current passes between them? Or even, as other members state, current does literally flow between them.

I understand there will always be leakage current, which is negligible unless the capacitor is broken, and of course fields are an intractable concept, but what for you would be a "good enough" description of what an insulator does in a capacitor?
 

wayneh

Joined Sep 9, 2010
17,496
What would be a reasonable way to describe the purpose of the insulator between the two metal plates...
In theory you don't need the insulator. A vacuum would be a superior insulator for stopping any leakage.

In reality, it's not practical to have a perfect vacuum inside a capacitor, and it would be mechanical design challenge to ensure the plates don't crash into each other. So, they use an insulator. No vacuum, and a big boost in mechanical strength keeping the plates in place.
 

RBR1317

Joined Nov 13, 2010
713
Perhaps this is a good time to consider just what is meant by the term 'capacitor'.

First consider the network model of a capacitor. It is a two-terminal device whose behavior is governed by a differential equation. Current flowing through the network model of a capacitor is I=C*dV/dt, and follows the standard network laws for voltage and current in a circuit. There is no physical description of the network model of a capacitor - it is just two terminals and a differential equation.

Then there is a physical approximation of the network model of a capacitor. It also has two terminals, but those terminals are connected to plates separated by a dielectric. This gives rise to physical analysis of electromagnetic fields in the dielectric, Maxwell's equations, div, grad, curl, and all that, in order to explain how it can approximate the behavior of its network model namesake. It appears that the discussion in this thread is primarily concerned with the physical capacitor at a technical focus somewhat below the level of Maxwell's equations, e.g. see "The Maxwell Displacement Current" https://www.phy.duke.edu/~rgb/Class/Electrodynamics/Electrodynamics/node28.html

But there is another approximation of the network model of a capacitor. This is based on the use of a gyrator with inductor. Although it is more common to use a gyrator with a capacitor to simulate an inductor, a gyrated inductor in a circuit will behave as a capacitor.

So if you are just interested in circuit design & analysis, then all you need to know is the network model of the capacitor. But if you want to really understand how a physical capacitor works, then Good Luck! And if you are at all familiar with gyrators, then you are probably an integrated filter designer.
 

Thread Starter

pigpen

Joined Jan 26, 2016
23
Perhaps this is a good time to consider just what is meant by the term 'capacitor'....
Thank you for widening the perspective, I will definitely need this stuff for further studies. For the purposes of reaching a working description or even a provisional description of what the insulator is for, do you disagree that the insulator or dielectric is there to force charges to accumulate in the metal plates, and to carry the EM field?
 
Last edited:

RBR1317

Joined Nov 13, 2010
713
...a provisional description of what the insulator is for...
The purpose of the dielectric material is to increase the relative permittivity of the space between the plates, thereby reducing the field strength for a given charge and increasing the capacitance. A good dielectric will have many electric dipoles that rotate to absorb the electric field. The permittivity is a characteristic of space, and the relative permittivity or "dielectric constant" is a way to characterize the reduction in effective field because of the polarization of the dielectric. Since more charge will then be necessary to achieve the same field strength as in free space, and C=Q/V, capacitance is increased by the presence of a dielectric material.
 

anhnha

Joined Apr 19, 2012
905
With dielectric material in between the total electric field is reduced. Is there an explanation how this would affect charging and discharging of capacitor physically?
I'd like to know how this happen in physics not by using time constant RC.
 

anhnha

Joined Apr 19, 2012
905
Have you considered how would charging and discharging be affected by different values of capacitance?
I know this by using time constant RC but I would like to understand how this happens in physics.
Let's consider the case there are two capacitors with different dielectric material and so different capacitance C1, C2 and C1>C2.
With C1 the electric field is reduced much more than C2 and does the reduced electric field affect charging time?
 

RBR1317

Joined Nov 13, 2010
713
The electric field strength is determined by the applied voltage and the distance between the plates. Are we really talking about a reduced electric field in the charged capacitor?
 

anhnha

Joined Apr 19, 2012
905
From your post:
The purpose of the dielectric material is to increase the relative permittivity of the space between the plates, thereby reducing the field strength for a given charge and increasing the capacitance.
So here is what I infer:
Increase relative permittivity -> reduce electric field strength and increase capacitance -> increase time constant.
So is there a physics explanation for the increasing of time constant by increasing relative permittivity?
Is it possible to explain why increasing relative permittivity will increase time to charge capacitor without using the concept/definition of capacitance?
 

RBR1317

Joined Nov 13, 2010
713
Increasing permittivity in a capacitor means the increased capacitance will hold the same charge (coulombs) with a reduced field strength, or hold more charge at the same field strength (voltage). Either way you look at it, C=Q/V.
 

MrAl

Joined Jun 17, 2014
11,389
Thank you.
-------

2.


I wonder why I have never seen a comparison made between transformers and capacitors: transformers require a changing field for voltage to be induced, hence they don't work for DC, can only work for AC. Capacitors require a changing field for current to be induced, hence they "block" DC and "pass" AC.


3.


I think everyone agrees with this stuff. The problem is more in the basic description... I'm sure there are other newbies out there too who are wondering how current can flow THROUGH the metal plates when there is an insulator between them. Why doesn't the basic description of capacitors say immediately that this insulator prevents current flow BUT is designed to carry a changing field that helps to induce current.... etc. Why do basic definitions of capacitors not focus on the changing field in the insulator?

Why is it that saying

---that current doesn't flow between the metal plates but rather current is induced via a change in the field inside the insulator/dielectric--.

why is that so controversial? I mean, it is in LIEC. Why are people fighting over "current flow" and why do so many refuse to accept the term "displacement current" or "induced current" when it is a better descriptor of what capacitors do?

Hi again,

Some good points.

What the real question is, is it that the current is 'induced' or is the current 'produced' from the changing field?
In a transformer the current is said to be 'induced' but really it is the field that pushes the electrons that forms this new current. So a better way to look at it is that the current is formed because of the force produced by the field. The current in this case is completely circular in that it forms a loop through whatever load there is and then back again to the other terminal of the transformer. So the electrons that were in the wire at the time this action first started at still there, and are the same electrons that were always there, and there are no new ones introduced.

In the vacuum capacitor however, we cant seem to be sure if there are new electrons introduced or not. In other words, did any electrons from terminal A ever reach terminal B as there is in a transformer secondary circuit loop? It seems that the electrons for the current in terminal B might come from the plate itself, and they may have come from that plate as the result of the field forcing them out of that plate, at least temporarily. This view is interesting to me because there are physical limits to how long we can actually do this and that is matched by the fact that there must be a limited amount of electrons in one plate, no matter how large. These two limiting factors are in agreement, although yes it is not direct proof.
But the only other possibility is that the electrons are able to change form from an electron to a field 'photon' and then back again on the other plate.

There is another theory in physics that states that there is no real 'identity' for a single particle. That means that at that level the particles loose any real enumeration of any kind so if particle 1 enters node A and particle 2 leaves node B, we have no way of knowing nor do we care if particle 2 was really particle 1 or a new particle. If this holds here too, then there would be no circumstance that could exist that would ever make a difference to us whether the particle went through the cap or went around the cap somehow, or changed form first.
This may change in the future i guess, as we dig deeper into the workings of nature.

The other 'controversy' is the difference between the network view of the capacitor and the physical view, which someone else here brought up too. The network version would have us believe that current flows right through the capacitor, without any proof, because that works for what it is intended to work for: network analysis. It appears from the EXTERNAL view that current flows right through the cap, even though the INTERNAL view might be different. The external view works for all cases of network analysis so we use it anyway. It might not work for a pure physical view but that doesnt matter there.
 

Thread Starter

pigpen

Joined Jan 26, 2016
23
Hi again,

Some good points.

What the real question is, is it that the current is 'induced' or is the current 'produced' from the changing field?

In the vacuum capacitor however, we cant seem to be sure if there are new electrons introduced or not. In other words, did any electrons from terminal A ever reach terminal B as there is in a transformer secondary circuit loop? It seems that the electrons for the current in terminal B might come from the plate itself, and they may have come from that plate as the result of the field forcing them out of that plate, at least temporarily. This view is interesting to me because there are physical limits to how long we can actually do this and that is matched by the fact that there must be a limited amount of electrons in one plate, no matter how large. These two limiting factors are in agreement, although yes it is not direct proof.

But the only other possibility is that the electrons are able to change form from an electron to a field 'photon' and then back again on the other plate.

There is another theory in physics that states that there is no real 'identity' for a single particle.
So, would my simplistic explanation- that the presence of the dielectric forces charges to accumulate on either plate, and this accumulation produces a field in the dielectric -- would this be acceptable as a provisional description, considering all the unknowns?
 

MrAl

Joined Jun 17, 2014
11,389
So, would my simplistic explanation- that the presence of the dielectric forces charges to accumulate on either plate, and this accumulation produces a field in the dielectric -- would this be acceptable as a provisional description, considering all the unknowns?
Hi again,

Well if i understand you right, i dont think there was ever a question of whether or not that was true, as it was always held to be true i believe.
Electrons accumulate, and that causes a field. I dont think it requires an actual physical dielectric material however, as even a vacuum can cause this to happen, so even a break in the wire would cause this although the capacitance might be small.

What else might be interesting to look at though might be why some actual dielectrics produce a capacitance that is greater than other dielectrics. If we build up a voltage on the one with more capacitance (and 'better' dielectric) we see more energy stored than with the other. But then again that voltage would have been 'harder' to build up on the one with the better dielectric, requiring more electron movement. This would be analogous to a hard piece of rubber versus a soft piece of rubber. The hard piece would be harder to compress and thus store more energy while the soft piece would be easier to compress and thus store less energy. Perhaps the electrons undergo a stress from their natural motions that force them to change position long enough to see the overall capacitance effect. There are so many involved that this could be true, although some calculations might help here.
I think it is best to deal with the vacuum capacitor first though in order to prevent the complexities of an actual dielectric from just complicating the whole picture and thus making it more difficult to resolve. The dielectric capacitor could then be treated as a separate case where we try to understand the action of the dielectric itself as it is placed inside an electric field or a field is built up around it.
 

Thread Starter

pigpen

Joined Jan 26, 2016
23
Hi, thanks so much, I understand that a vacuum or even air between two conductors will cause chargers to accumulate. I think that's what happens in metal arc welding? I was just confused by some messages here that seemed to imply that my view was totally wrong and that a current besides leakage current does indeed pass between the metal plates -- so I was interested to get a perspective on that... but the guys behind that idea seem to have left the thread now. In any case you've offered the clearest explanation here and I really appreciate your help.
 

anhnha

Joined Apr 19, 2012
905
Hi again
...
What else might be interesting to look at though might be why some actual dielectrics produce a capacitance that is greater than other dielectrics. If we build up a voltage on the one with more capacitance (and 'better' dielectric) we see more energy stored than with the other. But then again that voltage would have been 'harder' to build up on the one with the better dielectric, requiring more electron movement. This would be analogous to a hard piece of rubber versus a soft piece of rubber. The hard piece would be harder to compress and thus store more energy while the soft piece would be easier to compress and thus store less energy. Perhaps the electrons undergo a stress from their natural motions that force them to change position long enough to see the overall capacitance effect. There are so many involved that this could be true, although some calculations might help here.
...
Interesting explanation! However, I think you answer based on what is observed from reality not that rigorous proof.
Is there one that explain why some actual dielectrics produce a capacitance that is greater than other dielectrics in physics?
BTW, Happy Lunar New Year to all!
 
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