Parallel Plate Capacitor

[MattNotrick]

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.

 

[berkeman]

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.

 

[MattNotrick]

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

 

[MattNotrick]

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.

 

[DeeNos]

Hello Rick,

I can provide some guidance and clarification on your questions regarding the parallel plate capacitor.

Firstly, your approach of using the equation c=εoεrA/d is correct. However, there are a few things to keep in mind when using this equation.

1. The value of εo (permittivity of free space) is 8.85x10^-12 F/m. This value is always used when calculating capacitance in air or vacuum. In this case, since there is a dielectric sheet present, you will need to use the value of the relative permittivity (εr) of the dielectric material.

2. Your conversion of 2 microns to 0.000002 meters is correct.

3. The equation ε=εo * εr is correct. This is known as the permittivity constant.

Now, let's plug in the values in the equation:

c= (8.85x10^-12 F/m)(6)(4.5238m^2) / 0.000002m

= 1.69x10^-8 F

Therefore, the capacitance of the parallel plate capacitor is 1.69x10^-8 Farads.

I hope this helps to clarify your doubts. It is always important to double-check your calculations and make sure you are using the correct values for the variables. Keep up the good work in your scientific endeavors!