What is the Capacitance of a Parallel Plate Capacitor with a Dielectric Sheet?

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Discussion Overview

The discussion revolves around calculating the capacitance of a parallel plate capacitor that includes a dielectric sheet. Participants explore the appropriate equations and unit conversions necessary for the calculation, addressing both theoretical and practical aspects of the problem.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the initial parameters of the capacitor, including the radius, thickness of the dielectric, relative permittivity, and potential difference, and seeks clarification on the capacitance formula.
  • There is uncertainty about whether to include the permittivity of free space (εo) in the capacitance calculation when a dielectric is present.
  • Participants discuss unit conversions, with one participant noting a mistake in converting the area from cm² to m².
  • Another participant provides a corrected area value and asks for clarification on the term "mks" and the validity of the equation used for capacitance.
  • One participant calculates capacitance using two different approaches, questioning which result is correct and seeking verification from others.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correct application of formulas and unit conversions. There is no consensus on the correct capacitance value, and multiple approaches to the calculation are presented without resolution.

Contextual Notes

Participants have not fully resolved the implications of using squared area in the capacitance formula, nor have they clarified the correct interpretation of "mks" units in this context.

MattNotrick
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A parallel plate capacitor is defined of two circular plates with a radius of 12cm, it is separated by a dielectric sheet with a thickness of 2 microns (um) and relative permittivity of 6. It is charged with a potential difference of 110V.

The area of one plate I found is A=pie*R^2
= 452.389cm^2
= 4.5238m^2 is the area of each plate.

I am trying to work out the Capacitance, but I am not 100% sure on which equation route to take.

c=εoεrA
------
d

right? I am not sure whether to include the εo as there is it separated by a dielectric sheet, not just air space, or where to go from here.

Are my conversions correct? 2 microns = 0.000002 meters

Also ε = εo * εr, yes / no?

I tried doing

εA
___ =
d

(1.2x10^-5)(4.5238)^2
____________________
0.000002

= 122.7885986

I feel this is wrong, or I am going down the wrong route, I've tried several different "plate capacitor calculators" on the internet, all giving me a different answer each time, confusing me even further, I think I may be using the wrong equation, any help would be much appreciated,

thanks Rick.
 
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MattNotrick said:
A parallel plate capacitor is defined of two circular plates with a radius of 12cm, it is separated by a dielectric sheet with a thickness of 2 microns (um) and relative permittivity of 6. It is charged with a potential difference of 110V.

The area of one plate I found is A=pie*R^2
= 452.389cm^2
= 4.5238m^2 is the area of each plate.

I am trying to work out the Capacitance, but I am not 100% sure on which equation route to take.

c=εoεrA
------
d

right? I am not sure whether to include the εo as there is it separated by a dielectric sheet, not just air space, or where to go from here.

Are my conversions correct? 2 microns = 0.000002 meters

Also ε = εo * εr, yes / no?

I tried doing

εA
___ =
d

(1.2x10^-5)(4.5238)^2
____________________
0.000002

= 122.7885986

I feel this is wrong, or I am going down the wrong route, I've tried several different "plate capacitor calculators" on the internet, all giving me a different answer each time, confusing me even further, I think I may be using the wrong equation, any help would be much appreciated,

thanks Rick.

Welcome to the PF!

Convert to mks units in the beginning, and use them in all of your calculations. Your calculation of the disk area is wrong because of the way you converted units from cm^2 to m^2.
 
Hey, and thanks for the fast response, thanks for pointing out my error, needed to convert the squared number thanks, its 0.0452m^2.

What are you referring to when you say convert to "mks"? Thanks, is the equation on the right lines, or I am on the wrong track? I am fairly new to the course.

thanks, rick
 
Ok so I have changed all units to mks and using the equation

(relative permittivity)x(permitivity of free space)x(area of plate)^2
______________________________________________________
(separation between plates)

=
εo x εr x A
_________
d

=

6 x (8.85x10^-12)x(0.0452)^2
_________________________
(2x10^-6)

=

5.4267288 x 10^-20

I then did the same calculation, but i did not put the area as squared, instead I left it as just the single value, this gave me

= 1.2006024 x 10^-18

Please can someone verify which answer is correct?

Thanks, rick

thanks.
 
Last edited:

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