Three dielectric slabs in a capacitor

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    Capacitor Dielectric
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Discussion Overview

The discussion revolves around the configuration of three dielectric slabs in a capacitor and whether they can be treated as three pairs of parallel capacitors. Participants explore the necessary conditions for forming these capacitors, including the role of metal plates and the concept of equipotential surfaces.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • Some participants assert that three pairs of parallel capacitors require at least four plates, while others suggest that inserting metal plates at the boundaries of the dielectrics can create the necessary configuration.
  • There is a proposal that splitting existing plates into two can yield six plates, thus forming three capacitors.
  • Some participants argue that even without inserting additional plates, the configuration can still be viewed as three capacitors in series.
  • Questions arise regarding the relevance of equipotential surfaces and whether the boundaries between dielectrics can be treated as metal plates.
  • Concerns are expressed about the understanding of basic concepts, with some participants suggesting that a deeper engagement with the material is necessary for clarity.

Areas of Agreement / Disagreement

Participants generally disagree on the interpretation of the capacitor configuration and the necessity of additional plates. Multiple competing views remain regarding the treatment of dielectric boundaries and the overall classification of the capacitors.

Contextual Notes

Participants express uncertainty about the definitions of equipotential surfaces and the implications of the dielectric arrangement. The discussion highlights a lack of consensus on the fundamental principles involved.

gracy
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di.jpg


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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