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 realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

#### Solution

Capacitance of a parallel capacitor, V = 2 F

Distance between the two plates, d = 0.5 cm = 0.5 × 10^{−2} m

Capacitance of a parallel plate capacitor is given by the relation,

C = `(in_0"A")/"d"`

A = `("Cd")/in_0`

Where,

`in_0` = Permittivity of free space = 8.85 × 10^{−12 }C^{2 }N^{−1} m^{−2}

∴ A = `(2 xx 0.5 xx 10^-2)/(8.85 xx 10^-12)` = 1130 km^{2}

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.