Parallel Plate Capacitor and distance

AI Thread Summary
To find the separation distance in a parallel-plate capacitor with a given capacitance, the correct formula is d = (E₀A)/C. The user initially struggled with rearranging the equation, leading to incorrect calculations. It was clarified that the area does not need to be doubled and that careful algebraic manipulation is essential. The dielectric assumption of air allows for a calculated distance of approximately 2.656 mm. Proper dimensional analysis can help verify the correctness of the final equation.
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I'm going through my book and one of the Example exercises has the following problem:
Two plates, each of area 3X10^-4 m^2 are used to construc a parallel-plate capacitor with capacitance 1 pF. (a) Find the necessary separation distance.

Seems pretty obvious right? C = Eo(A/d)
So, I figure I'm finding d = C/(Eo)(A)
but it doesn't come out right, so I must be missing something either in my math (not surprising) or the area. Do I need to do a calculation with the area? Like 2 X A? That doesn't work either, and it doesn't seem right to do it that way. Help, please??
 
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Its just your manipulation which is letting you down. It should be;

C = \frac{E_{0}A}{d}
C \cdot d = E_{0}\cdot {A}
d = \frac{E_{0}A}{C}

-Hoot:smile:
 
Hootenanny said:
Its just your manipulation which is letting you down.


You are awesome! Thank you!

:approve:
 
If you are assume the dielectric is air (k ~ 1) and \epsilon_o = 8.854 x 10^{-12} F/m, you may get d = 2.656 mm. A more general formula for this type question may be found here.
In agreement with Hoot, i noticed the same fault in rearranging your equation. It is a good idea to do a couple of things to double check maths. One is to do the algebra step by step, the other is to do a quick dimensional analysis of the final equation. In your case, you have A left in the denominator. The only way that would work and have answer come out in meters is to have m^3 in the numerator, which you don't have.
 
Wow. Thank you, that is most helpful. The link is terrific. Again, thank you!
 
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