Question

Two aluminized optical flats 15 cm in diameter are separated by a gap of 0.04 mm, forming a capacitor. what is the capacitance in picofarads

Answer 1

Answer: 3978 pF

Explanation:

Capacitance can be defined as ability to store charge. It is the ratio of charge  over electric potential.

$C =\epsilon_o \frac{A}{d}$

where, A is the area of the plates and d is the distance between plates forming the capacitor.

diameter of the plates, di = 15 cm =0.15 m

radius of the plate, r = di/2 = 0.075 m

Area of the plates, A = πr² = 0.018 m²

Distance between the plates, d = 0.04 mm = 0.00004 m

Permittivity, ∈₀ = 8.84 × 10⁻¹² F/m

Capacitance, $C = 8.84 \times 10^{-12} F/m \times \frac{0.018 m^2}{0.00004 m} = 3978 \times 10^{-12} F = 3978 pF$