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What is the area of the plates of a 2 farad parallel plate air capacitor, given that the separation between the plates is 0.5 cm? |
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Answer» Here, `A = ? C = 2F, d = 0.5m = 5xx10^(-3) m`, As `C = (in_(0) A)/(d) :. A = (Cd)/(in_(0)) = (2xx5xx10^(-3))/(8.85xx10^(-12)) = 1.13xx10^(9) m^(2)`, which is too large. That is why ordinary capacitance are in the range of `muF` or less. However, in electrolytic, d is too small. Therefore, their capacitance is much larger `(= 0.1F)` |
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