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

To find the capacitance of the described capacitor, we use the formula for a parallel-plate capacitor (C = ε0A/d) with the area calculated from the diameter and convert the resulting value to picofarads. The calculated capacitance is approximately 165.7 pF.

Step-by-step explanation:

To find the capacitance of a parallel-plate capacitor, we apply the formula:

C = ε0A/d

where:

• C is the capacitance,
• ε0 is the vacuum permittivity (ε0 = 8.854 x 10-12 F/m),
• A is the area of one of the plates,
• d is the separation between the plates.

Given that the diameter of the optical flats is 15 cm, the radius r is 7.5 cm or 0.075 m. The area A is therefore πr2. The separation d is 0.04 mm or 0.04 x 10-3 m. Plugging these values into the formula, we get:

C = (8.854 x 10-12 F/m) x π x (0.075 m)2 / (0.04 x 10-3 m)

When you calculate this, you will get the capacitance C in farads. Convert this value to picofarads by multiplying it by 1012, as 1 pF equals 10-12 F.

The actual calculation gives you a capacitance which is approximately:

C ≈ 165.7 pF

Answer 2

Answer: 3978 pF

Step-by-step explanation:

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

`C =\epsilon_o (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 * 10^(-12) F/m * (0.018 m^2)/(0.00004 m) = 3978 * 10^(-12) F = 3978 pF`