Calculating Capacitance of Two Metal Plates with 1μF Capacity

[Mark Martinello]

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]

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]

The answer I get is 2*10 to the power -6. I don't think that is the right answer.

 

[SpeedBird]

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]

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]

No worries