Three dielectric slabs in a capacitor

AI Thread Summary
The discussion centers on the confusion regarding how three dielectric slabs in a capacitor can be interpreted as three pairs of parallel capacitors, despite having only two physical plates. Participants explain that inserting metal plates at the boundaries of the dielectrics creates equipotential surfaces, allowing for the conceptualization of six plates, which can be treated as three capacitors. There is debate over whether the configuration actually represents three parallel capacitors or if it is more accurately described as three capacitors in series. The importance of understanding electric fields and equipotential surfaces in this context is emphasized, with some suggesting that relying on equivalent circuits may hinder deeper comprehension. Ultimately, the conversation highlights the complexities of analyzing capacitors with dielectrics and the need for a solid grasp of underlying principles.
gracy
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My textbook says this makes three pairs of parallel capacitors.
I don't understand how?To make three pairs of parallel plates capacitors we at least need four plates but here we have only two
plates.png
 
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gracy said:
View attachment 95770

My textbook says this makes three pairs of parallel capacitors.
I don't understand how?To make three pairs of parallel plates capacitors we at least need four plates but here we have only two
View attachment 95771
You can insert metal plates at the boundaries of two dielectrics. They will be equipotential. Then splitting each plate into two plates in series, you can have 6 plates i.e. 3 capacitors.
 
I think we need only four plates to form three parallel plate capacitors.
 
gracy said:
I think we need only four plates to form three parallel plate capacitors.
Yes. But to distinguish three separate capacitors, you can make six plates from them.
 
So we do need to insert metal plates there otherwise it is not possible to have 3 pairs of capacitors.

pair.png

Here I have shown green and orange metal plates placed at boundaries of dielectric slab . The left most and the right most plates were already there.
 
gracy said:
So we do need to insert metal plates there otherwise it is not possible to have 3 pairs of capacitors.

View attachment 95772
Here I have shown green and orange metal plates placed at boundaries of dielectric slab . The left most and the right most plates were already there.
Even if you don't insert plates, still it is a 3 capacitors in series combination.
 
How?That is all my OP about.
 
gracy said:
How?That is all my OP about.
How _____what_____ ?
 
To make three pairs of parallel plates capacitors we at least need four plates.
 
  • #11
gracy said:
To make three pairs of parallel plates capacitors we at least need four plates.
See post #2.

What is he not clear about.
 
  • #12
cnh1995 said:
You can insert metal plates at the boundaries of two dielectrics
He is also talking about inserting metal plates .
 
  • #13
gracy said:
He is also talking about inserting metal plates .
See mentor NascentOxygen's post #14 in that thread.
 
  • #14
cnh1995 said:
This older thread of yours might help you..
At that time also I was having same doubt. But I tried to focus on different aspect of that problem.
 
  • #15
  • #16
cnh1995 said:
See mentor NascentOxygen's post #14 in that thread.
What is not clear about that post?
 
  • #18
I don't understand how is it relevant to being "equipotential"?
 
  • #19
gracy said:
I don't understand how is it relevant to being "equipotential"?
Do you see that each boundary between the dielectric slabs is an equipotential surface?
 
  • #20
No. Is it perpendicular to electric field lines ?
 
  • #21
Well, I really wish to continue here but I MUST get up early tomorrow for my exam..It is midnight here, almost 12:00am. Got to go:frown:!:sleep::sleep:
 
  • #22
You have exams ? Then what are you doing here ? Go sleep well . Good luck for the test !:smile:
 
  • #23
gracy said:
No. Is it perpendicular to electric field lines ?
The end plates are conductors and therefore each is equipotential ...
 
  • #24
SammyS said:
each boundary between the dielectric slabs

SammyS said:
The end plates are conductors and therefore each is equipotential
Which ones are equipotential? The end plates (right most and left most plates) or boundaries between the dielectric i.e shown in green and orange color in my post #5? Should be end plates as these are metal plates .
 
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  • #25
gracy said:
Which ones are equipotential? The end plates (right most and left most plates) or boundaries of dielectric i.e shown in green and orange color in my post #5? Should be end plates as these are metal plates .
I had said the following:
SammyS said:
The end plates are conductors and therefore each is equipotential ...
- hoping that would lead you to further realize -

... therefore, the E-field lines are horizontal (in the picture) so any vertical surface is equipotential, in particular the boundaries between dielectric slabs are equipotential surfaces ...
 
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  • #26
SammyS said:
so any vertical surface is equipotential, in particular the boundaries between dielectric slabs are equipotential surfaces ...
Boundaries between dielectrics are denoted by green and orange color in my post #5 ,Right?
 
  • #27
gracy said:
Boundaries between dielectrics are denoted by green and orange color in my post #5 ,Right?
It looks to me as if that's the case.

If that's where you placed them then, yes, that's correct.
 
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  • #28
I still don't understand even if these surfaces are equipotential , how can we treat the dielectric boundaries/surfaces as metal plates?
 
  • #29
I posted this in another thread:

Vanadium 50 said:
Gracy, I don't think PhysicsForums is helping you. You've asked us about vectors in the past, and now it's clear you haven't learned them.

The problem is that you immediately jump to asking a question here without having put much work into it, and when you are guided by someone towards the answer, you don't try and work it out for yourself, but instead ask for another hint. And another. And another. Eventually, you have been hinted all the way to the answer. Well, you've gotten the answer, but you haven't really learned.

You're going to have to decide if you want to learn or not. If you want to learn, you are going to have to spend more time thinking and working on your own. In this case, your starting point for basic vectors shouldn't be asking for more hints - it should be going back to your past materials and see if you can solve this using what you have already been told.

While the topic is different, the point is the same.

At the time I wrote this, the average time between a message and Gracy's reply is 3 minutes, 35 seconds. This includes the time to compose and post the message. The median time is 2 minutes and the most probable time is 1 minute. The reason that Gracy-threads turn into marathons is not because she spends a lot of time seriously thinking about what was said, and still doesn't get it. It's because immediately after getting a hint she wants another one. And another one and another one. And, as I have said above this isn't helping her. Sure, eventually she gets the answer, but it has left her completely unable to solve a similar problem. Examples abound.

Gracy, as I said before, if you want to learn, you are going to have to spend more time thinking and working on your own. Asking for more hints immediately after getting one is not working.
 
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  • #30
Because of the medium's uniformity and shape, any vertical line you care to draw on your figure will happen to be a line of equipotential, i.e., it joins all points having the same potential because these all have the same path distance to the battery. You can add in a conductor joining points of identical potential and nothing changes---no current flows in a conductor where no potential differences exist.

You started this thread asking how does the figure represent 3 parallel capacitors. It doesn't. Your figure represents 3 capacitors in series.

To show 3 capacitors in parallel, you would arrange the dielectric blocks so every dielectric is in contact with both end plates.
 
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  • #31
gracy said:
No. Is it perpendicular to electric field lines ?
Electric field lines are drawn as originating on + charge and terminating on - charge. These electric field lines are perpendicular to lines joining points of equal potential.

The pattern that you see when electric field lines and equipotential lines are superimposed, for any general distribution, are called curvilinear squares. Their general shape is they resemble squares or rectangles but with bowed sides.
 
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  • #32
gracy said:
I still don't understand even if these surfaces are equipotential , how can we treat the dielectric boundaries/surfaces as metal plates?
In any situation, we can invent equal and opposite pairs of charges at the same point in space. That won't change anything. If we were to insert infinitesimally thin conducting plates at these boundaries, that is exactly what would happen.
 
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  • #33
gracy said:
To make three pairs of parallel plates capacitors we at least need four plates.
The equivalent capacitors are, of course, in series. You can imagine infinitely thin metal sheets between the layers and then you can treat it in a very straightforward way.
 
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  • #34
SammyS said:
Do you see that each boundary between the dielectric slabs is an equipotential surface?
gracy said:
No. Is it perpendicular to electric field lines ?
Equipotential surfaces are parallel with the plates and the surfaces between the layers of dielectric, ignoring any edge effects.
 
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  • #35
NascentOxygen said:
You started this thread asking how does the figure represent 3 parallel capacitors. It doesn't. Your figure represents 3 capacitors in series.
By three parallel plates I mean 3 pairs of parallel plate capacitors (a specific type of capacitor called "parallel plate capacitor")
 
  • #36
gracy said:
By three parallel plates I mean 3 pairs of parallel plate capacitors (a specific type of capacitor called "parallel plate capacitor")
The plates are parallel. The 3 resulting capacitors are in series.
 
  • #37
How inserting infinitesimally thin conducting plates at the boundaries of dielectric is same as placing equal and opposite pairs of charges at the same point in space?
I think it has something to do with induced charges in metal plate.
 
  • #38
The main idea with this is to find a set of capacitors which are easy to analyze and which are equivalent electrically to the capacitor you are given in the problem.

Inserting the plates at the boundaries of the dielectrics will not alter the electric fields in the dielectric materials.
 
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  • #39
gracy said:
How inserting infinitesimally thin conducting plates at the boundaries of dielectric is same as placing equal and opposite pairs of charges at the same point in space?
I think it has something to do with induced charges in metal plate.
Because the field is perpendicular to the boundaries, yes.
 
  • #40
And even if it does not alter electric fields in dielectric medium , how we can assume metal plates to be at boundary even though they are not present there?
 
  • #41
gracy said:
And even if it does not alter electric fields in dielectric medium , how we can assume metal plates to be at boundary even though they are not present there?
We can compare two situations - one with metal sheets and one without. If we can see that the electric fields will be the same then we know that the results for the situation with metal sheets will carry over to the case without.
 
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  • #42
gracy said:
I think it has something to do with induced charges in metal plate
gracy said:
how we can assume metal plates to be at boundary even though they are not present there?
The metal plates have no effect on the situation but they are just a convenient way to think about it. It is a common ploy to analyse an electrical situation in terms of an 'equivalent circuit'. That's why the virtual plates are introduced; it reduces the situation to one that is more familiar.
PS You could put a metal plate half way through one of the dielectric layers if you wanted to and not alter the situation either.
 
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  • #43
Equivalent Circuit: In 1962 (prehistoric sixth form stage!) someone showed me a pamphlet about transistors and it had a very complicated Equivalent Circuit which allowed them to derive some equations to describe it (totally beyond me at the time - and again, at this time!). I was convinced that you could actually build a transistor, using the components on that Circuit. Later, the penny dropped and I became able to use equivalent circuits for my own work.
 
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  • #44
This way of analyzing the problem as three capacitors in series is, in my opinion, a poor way of dodging the boundary conditions in a dielectric interface.
Students "love" these magic tricks and, instead of focusing in the relationship between the fields in the different dielectrics, they struggle memorizing these rules.
Although these "recipes" give the right result, I'm not convinced they lead to good comprehension of the problem.
 
  • #45
Gordianus said:
This way of analyzing the problem as three capacitors in series is, in my opinion, a poor way of dodging the boundary conditions in a dielectric interface.
Students "love" these magic tricks and, instead of focusing in the relationship between the fields in the different dielectrics, they struggle memorizing these rules.
Although these "recipes" give the right result, I'm not convinced they lead to good comprehension of the problem.
Your reference to "magic tricks" in this context is not helpful in my opinion.

I would not consider it a "trick" to split up this capacitor into an equivalent combination of three capacitors in series. That's a fairly common way to find this capacitance in a physics course at the likely level of sophistication of the OP.

(Any extensive discussion of what might or might not constitute such "magic tricks" should probably be taken up in a separate thread. Let's not take this thread on that tangent.)
 
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  • #46
Yes gracy, if fringe effects in the extremes (up and down faces in your drawing) are neglected, each boundary between plates can be regarded as an equipotential surface, with the electric field lines normal to it.There is no need for an additional metal electrode.
 
  • #47
Gordianus said:
This way of analyzing the problem as three capacitors in series is, in my opinion, a poor way of dodging the boundary conditions in a dielectric interface.
Students "love" these magic tricks and, instead of focusing in the relationship between the fields in the different dielectrics, they struggle memorizing these rules.
Although these "recipes" give the right result, I'm not convinced they lead to good comprehension of the problem.
You may be right but only to some extent. There are always multiple ways of approaching problems and they are often complementary - which they are, in this case. The concept of inserting metal sheets is not just a magic trick. It acts as a reminder that the surfaces are equipotential and it also links to basic circuit theory - series connected capacitors are a familiar idea so why not use it?
This approach is not one that anyone would take in the case of 'optical' dielectrics and your "boundary conditions" consideration is, perhaps, more appropriate in that situation.
 
  • #48
gracy said:
My textbook says this makes three pairs of parallel capacitors.
(In my OP)
In this manner?

First pair of capacitor

1.png

second pair
2.png


And the third pair
3.png
 
  • #49
gracy said:
(In my OP)
In this manner?

First pair of capacitor

View attachment 95893
second pair
View attachment 95895

And the third pair
View attachment 95896
You have shown 6 plates here. Plates will have induced charges. I believe you should insert two plates between the two dipole boundaries and draw the induced charges on each plate according to the direction of net electric field. Then you can make 6 plates out of these 4 by splitting each middle plate into two, each carrying equal and opposite charge.
 
  • #50
cnh1995 said:
You have shown 6 plates here.
No , there are only 4 plates in total.
 

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