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 C2 N−1 m−2
∴ A = `(2 xx 0.5 xx 10^-2)/(8.85 xx 10^-12)` = 1130 km2
Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.
