Three dielectric slabs in a capacitor

[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.

 

[gracy]

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

And the direction of given capacitor plates is always considered to be horizontal. Right?

 

[cnh1995]

And the direction of given capacitor plates is always considered to be horizontal. Right?

Yes.

 

[gracy]

given capacitor plates

The plate which are at extreme positions i.e the left and right most.

 

[cnh1995]

The plate which are at extreme positions i.e the left and right most.

If the partition is parallel to the plates, they're in series and if the partition is perpendicular to the plates, they're in parallel.

 

[gracy]

If the partition is parallel to the plates, they're in series and if the partition is perpendicular to the plates, they're in parallel.

Any exceptions?

 

[cnh1995]

Any exceptions?

None. But the partitions should be continuous from one end to the other.

 

[gracy]

But the partions should be continuous from one end to the other.

You mean there should not be empty space in between ?

 

[cnh1995]

I meant something like this..

 

[gracy]

The two plates of parallel capacitors should have same area . Is that correct?

 

[cnh1995]

The two plates of parallel capacitors should have same area . Is that correct?

What do you mean by "two plates"? Which two plates?

 

[gracy]

http://astarmathsandphysics.com/university-physics/electricity-and-magnetism/parallel-plate-capacitors-html-9fd61e5.gif

Here the two plates have same area "A" ,I want to ask what if they had different areas , will they still form parallel plate capacitors together?

 

[cnh1995]

http://astarmathsandphysics.com/university-physics/electricity-and-magnetism/parallel-plate-capacitors-html-9fd61e5.gif

Here the two plates have same area "A" ,I want to ask what if they had different areas , will they still form parallel plate capacitors together?

Yes. But the capacitance will be different.

 

[gracy]

What will we write in place of A in the below formula ? A or B?
$\frac{Aε0}{d}$ If one plate has area "A" and the other "B"

 

[cnh1995]

The charges have to be uniformly distributed on both the plates and the plates have different areas. I believe the capacitance calculation will be a little complicated.

 

[gracy]

Ok. What will be the formula of capacitance for the plate 2 here

 

[cnh1995]

What are those blue portions? Dielectrics? What are the red lines joining them? Wires?

 

[gracy]

What are those blue portions? Dielectrics? What are the red lines joining them? Wires?

Yes and yes.

 

[cnh1995]

Given the area, separation and permitivitty, its capacitance would be C=Aε/d..

 

[gracy]

A is area of ? And why there isn't "K" present?

 

[cnh1995]

A is area of ? And why there isn't "K" present?

What's the need of k when there is 'ε' which is itself kε_o? A is the area of each plate(which is same throughout). I don't understand why you are analyzing this diagram.

 

[gracy]

Ok. Fine. Let's end this thread here as it is going off topic.