Three dielectric slabs in a capacitor

[NascentOxygen]

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.

 

[haruspex]

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.

 

[sophiecentaur]

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.

 

[sophiecentaur]

Do you see that each boundary between the dielectric slabs is an equipotential surface?

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.

 

[gracy]

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")

 

[SammyS]

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.

 

[gracy]

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.

 

[SammyS]

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.

 

[haruspex]

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.

 

[gracy]

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?

 

[haruspex]

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.

 

[sophiecentaur]

I think it has something to do with induced charges in metal plate

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.

 

[sophiecentaur]

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.

 

[Gordianus]

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.

 

[SammyS]

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.)

 

[xareu]

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.

 

[sophiecentaur]

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.

 

[gracy]

My textbook says this makes three pairs of parallel capacitors.

(In my OP)
In this manner?

First pair of capacitor

second pair

And the third pair

 

[cnh1995]

(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.

 

[gracy]

You have shown 6 plates here.

No , there are only 4 plates in total.

 

[cnh1995]

No , there are only 4 plates in total.

Ok. But the middle pair looks shorter than the other two. You need to put the plates at the boundaries of the dielectrics in order to distinguish the three capacitors.

 

[sophiecentaur]

"My textbook says this makes three pairs of parallel capacitors."
I'm sure this is a typo on copying out what the book says. There are 'Parallel Plate Capacitors' and there are 'Parallel Connected Capacitors' and Capacitors Connected in Parallel. We are talking about parallel plate capacitors which have symmetry and which have which have plane shaped equipotential surfaces.

There is no problem in considering charges on the surfaces of adjacent dielectric plates. They can be touching OR infinitesimally far apart OR separated by a thick metal sheet OR pairs of metal sheets, connected by ideal wires. There will be no difference in the field situations within the dielectrics. This conversation seems to be going round in circles about this point. Why? If anyone could be bothered to draw a good diagram of each case, the thing would be obvious and would avoid all the miscommunications in the thread. I reckon that we all know what's really going on, electrically.

 

[SammyS]

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.

No , there are only 4 plates in total.

Let's look at a much earlier post.

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.

When you split these, keep them connected with a conducing wire. You now have the six plates in total.

 

[sophiecentaur]

Ok. But the middle pair looks shorter than the other two. You need to put the plates at the boundaries of the dielectrics in order to distinguish the three capacitors.

Yes - the picture could have been drawn better and be less open to confusion.
Most of the diagrams that people want to draw on PF can be done on PowerPoint or equivalent - available to most (?) people and easy to learn. A lot easier than Physics, at any rate.

 

[cnh1995]

When you split these, keep them connected with a conducing wire.

Right! I forgot to mention that key part in every post...

 

[gneill]

It should be pointed out that the "inserted" plates can be arbitrarily thin and do not change the amount of dielectric involved, nor effect the total size of the regions where the electric field is. The conductive plates assume the potential of the equipotential line where they are inserted and have the same potential on both surfaces. Charge separation ensures that the field is "transmitted" from one face to the other, and it's as though no plate were there as far as the field is concerned on either side of the plate.

You can split such a plate in two and move the halves apart, maintaining a conductive connection with a wire. The potential of the two must remain the same due to being connected, and the field is still "transmitted" from one outer face to the other.

It might be instructive to look at a simpler setup first where there are two capacitors in series and no dielectrics to complicate matters. Both have the same plate area but one has separation d1, the other d2. Use the capacitor equation for parallel plate capacitors and the series connection formula to find the net capacitance. Show that this capacitance is the same as a single capacitor with the same plate area but a separation of d1+d2. This should lend support to the idea that inserting parallel plates and forming series connected capacitors is legitimate.

 

[gracy]

You mean if the following two are equivalent

Here also conductive plates have been inserted and inserted plates are split and separated by conductive wires in (1) but it is [picture (1)] is equivalent to picture (2) that means this kind of assumptions are legitimate.

 

[cnh1995]

Here also conductive plates have been inserted and inserted plates are split and separated by conductive wires in (1) but it is [picture (1)] is equivalent to picture (2) that means this kind of assumptions are legitimate.

Yes. If you combine the middle plates in picture 1, you'll end up with picture 2.

 

[gracy]

I have come to a conclusion

If dielectrics partition the space horizontally , they are in series and if dielectrics partition the space vertically then they are parallel. Correct?

 

[NascentOxygen]

I have come to a conclusion

If dielectrics partition the space horizontally , they are in series and if dielectrics partition the space vertically then they are parallel. Correct?

Correct, provided that the parallel plates are vertical...and spaced horizontally apart as in the figures in this thread.