What is the area of the plates of a 2F 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 μF or less. However, electorlytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductor.]

Solution

Given,
Capacitance of parallel plate capacitor, C=2F
Seperation between the parallel plate capacitor, d =0.5 cm =5 x 10^–3 m

Now, using the formula for capacitance of a parallel plate capacitor,

C = ε_{0} \frac{A}{d} A = \frac{Cd}{ε_{0}} (ε_{0} = 8.85 \times 10^{- 12} {Fm}^{- 1})

= \frac{2 \times 5 \times 10^{- 3}}{8.85 \times 10^{- 12}}

A = 1.13 \times 10^{9} m^{2} = 1.13 \times 10^{3} {km}^{2} = 1130 {km}^{2}

Therfore, inorder to get the capacitance in Farads, the area of the plates should be in kilometres which, practically cause inconvenience. And, that is why, the ordinary capacitors are in the range of

μF

.

.

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