A 0.20-f capacitor is desired. what area must the plates have if they are to be separated by a 3.2-mm air gap?

Air gap means that the dielectric is air.

So ε0 = 8.85 x 10^-12 [F/m] ... permitivity of free space

Lets use the equation

C = (ε0x A) / d

Where A is the area of the plate

And d the distance between the plates

d = 3.2-mm = 3.2 E-3 m

so ... > A = C * d / ε0 = 0.20 F * 3.2 E-3 m / 8.85 x 10^-12 [F/m]

A = 7.23 E 7 [m2]

Capacity = 0.20 F Distance d = 3.2 x 10^-3 m We know the vacuum permittivity e0 = 8.854Ă-10^-12 We have the equation, C = e0A / d = > A = Cd / e0 A = (0.20 x 3.2 x 10^-3) / 8.854Ă-10^-12 = 0.64 x 10^9 / 8.854 = 0.07228 x 10^9 Area of the plates A = 7.228 x 10^7