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

AI Thread Summary
The discussion focuses on calculating the capacitance of a parallel plate capacitor with a dielectric sheet. The capacitor has circular plates with a radius of 12 cm, separated by a dielectric of 2 microns and a relative permittivity of 6. The correct formula for capacitance is C = ε₀εᵣA/d, where ε₀ is the permittivity of free space, εᵣ is the relative permittivity, A is the area of the plates, and d is the separation distance. There was confusion regarding unit conversions and the area calculation, which should be in square meters. After correcting the area and using the proper formula, the user seeks verification of their final capacitance calculations.
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
______________________________________________________
(seperation 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.
 
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