Question

Consider a parallel-plate capacitor made up of two conducting plates with dimensions 38 mm x 29 mm. If the separation between the plates is 0.75 mm, what is the capacitance, in pF, between them?

Answer 1

The capacitance of a parallel-plate capacitor with given dimensions and separation can be calculated using the formula C = ε₀ * (A / d). Plugging in the values, the capacitance is approximately 12,970 pF.

Step-by-step explanation:

The capacitance of a parallel-plate capacitor can be calculated using the formula:

C = ε₀ * (A / d)

where C is the capacitance, ε₀ is the permittivity of free space (ε₀ = 8.85 x 10^(-12) F/m), A is the area of each plate, and d is the separation between the plates.

Plugging in the given values: A = (0.038 m)(0.029 m) = 0.001102 m² and d = 0.00075 m, we can calculate the capacitance:

C = (8.85 x 10^(-12) F/m) * (0.001102 m² / 0.00075 m) = 1.297 x 10^(-8) F = 12.97 nF

To convert the capacitance from farads (F) to picofarads (pF), we can use the conversion factor: 1 F = 10^(12) pF. Therefore, the capacitance is approximately 12,970 pF.