Three dielectric slabs in a capacitor

[gracy]

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

 

[cnh1995]

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.

 

[gracy]

I think we need only four plates to form three parallel plate capacitors.

 

[cnh1995]

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.

 

[gracy]

So we do need to insert metal plates there otherwise it is not possible to have 3 pairs of capacitors.

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.

 

[cnh1995]

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.

 

[gracy]

How?That is all my OP about.

 

[SammyS]

How?That is all my OP about.

How _____what_____ ?

 

[gracy]

To make three pairs of parallel plates capacitors we at least need four plates.

 

[cnh1995]

https://www.physicsforums.com/posts/5317154/
This older thread of yours might help you..

 

[SammyS]

To make three pairs of parallel plates capacitors we at least need four plates.

See post #2.

What is he not clear about.

 

[gracy]

You can insert metal plates at the boundaries of two dielectrics

He is also talking about inserting metal plates .

 

[cnh1995]

He is also talking about inserting metal plates .

See mentor NascentOxygen's post #14 in that thread.

 

[gracy]

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.

 

[SammyS]

https://www.physicsforums.com/posts/5317154/
This older thread of yours might help you..

gracy,

Also consult the old thread(s) in which @gneill worked with you regarding "nodes".

 

[cnh1995]

See mentor NascentOxygen's post #14 in that thread.

What is not clear about that post?

 

[gracy]

https://www.physicsforums.com/threads/capacitor-with-dielectric.847950/#post-5317154

I haven't marked this thread as solved yet. I was not able to understand there but I left that question as well as thread. That's why still confused.

 

[gracy]

I don't understand how is it relevant to being "equipotential"?

 

[SammyS]

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?

 

[gracy]

No. Is it perpendicular to electric field lines ?

 

[cnh1995]

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!

 

[gracy]

You have exams ? Then what are you doing here ? Go sleep well . Good luck for the test !

 

[SammyS]

No. Is it perpendicular to electric field lines ?

The end plates are conductors and therefore each is equipotential ...

 

[gracy]

each boundary between the dielectric slabs

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 .

 

[SammyS]

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:

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 ...

 

[gracy]

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?

 

[SammyS]

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.

 

[gracy]

I still don't understand even if these surfaces are equipotential , how can we treat the dielectric boundaries/surfaces as metal plates?

 

[Vanadium 50]

I posted this in another thread:

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.

 

[NascentOxygen]

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.

 

[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")