View Full Version : Capacitance Question
Mark Martinello
Nov17-04, 04:53 AM
Hi Guys,
Here is a question that I cannot get my mind on. Bad day I guess. I'm trying to figure out the answer. Here comes the question:
How far away from each other would two metal plates, 2 square meters in area each, have to be in order to create a capacitance of 1 micro Farad? Assume that the plates are separted by air.
SpeedBird
Nov17-04, 06:19 AM
Capacitance = (Permitivity of air * Area of overlap of the plates)/d between plates
you have all the variables apart from d
rearrange the formula with "d" on one side of the equals sign and everything else on the other.
Permitivity of air = 1.0006 (in case you didnt know)
Mark Martinello
Nov17-04, 09:51 AM
The answer I get is 2*10 to the power -6. I don't think that is the right answer.
SpeedBird
Nov17-04, 01:49 PM
i made an error there sorry
1.0006 is the relative permittivity of air
you multiply this by the permittivity of free space.
permittivity of free space is 8.85*10^-12
permittivity of air is (8.85*10^-12)*1.0006 = 8.85531*10^-12
d = dist between plates = unknown
A = overlapping area of plates = 2 metres square
E = permittivity of air = 8.85531*10^-12
C = capacitance of the parallel plate capacitor
your formula was C = (E*A) / d
you rearrange it to d = (E*A) / C
plug in your values for E, A and C
then see what you get.
Mark Martinello
Nov18-04, 05:30 AM
Thanks for helping out SpeedBird. I forgot completely about the permittivity of free space. Answer correct after I got the full menu!!
Thanks again, Mark
SpeedBird
Nov18-04, 12:14 PM
No worries :biggrin:
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