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.

  • #61
gracy said:
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?
 
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  • #62
gracy said:
And the direction of given capacitor plates is always considered to be horizontal. Right?
Yes.
 
  • #63
gracy said:
given capacitor plates
The plate which are at extreme positions i.e the left and right most.
 
  • #64
gracy said:
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.
 
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  • #65
cnh1995 said:
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?
 
  • #66
gracy said:
Any exceptions?
None. But the partitions should be continuous from one end to the other.
 
Last edited:
  • #67
cnh1995 said:
But the partions should be continuous from one end to the other.
You mean there should not be empty space in between ?
 
  • #68
I meant something like this..
%2F%2Fwww.physicsforums.com%2Fproxy.php%3Fimage%3Dhttp%253A%252F%252Fi.stack.imgur.com%252FeWBHL.jpg
 
  • #69
The two plates of parallel capacitors should have same area . Is that correct?
 
  • #70
gracy said:
The two plates of parallel capacitors should have same area . Is that correct?
What do you mean by "two plates"? Which two plates?
 
  • #71
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?
 
Last edited by a moderator:
  • #72
gracy said:
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.
 
Last edited by a moderator:
  • #73
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"
 
  • #74
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.
 
  • #75
Ok. What will be the formula of capacitance for the plate 2 here

kdi.png
 
  • #76
What are those blue portions? Dielectrics? What are the red lines joining them? Wires?
 
  • #77
cnh1995 said:
What are those blue portions? Dielectrics? What are the red lines joining them? Wires?
Yes and yes.
 
  • #78
Given the area, separation and permitivitty, its capacitance would be C=Aε/d..
 
  • #79
A is area of ? And why there isn't "K" present?
 
  • #80
gracy said:
A is area of ? And why there isn't "K" present?
What's the need of k when there is 'ε' which is itself kεo? A is the area of each plate(which is same throughout). I don't understand why you are analyzing this diagram.
 
  • #81
Ok. Fine. Let's end this thread here as it is going off topic.
 

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