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2.24 What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of \mu F or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

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Given that,

The capacitance of the parallel-plate capacitor C=2F

Separation between plated d=0.5cm

Now, as we know

C=\frac{\epsilon _0A}{d}

A=\frac{Cd}{\epsilon _0}=\frac{2*5*10^{-3}}{8.85*10^{-12}}=1.13*10^9m^2

A=\1.13*10^3km^2=1130km^2

Hence, to get capacitance in farads, the area of the plate should be of the order of a kilometre, which is not good practice, and so that is why ordinary capacitors are of range \mu F

Posted by

Pankaj Sanodiya

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