Final answer:
The plate area required to provide a capacitance of 1.50 pF in a parallel-plate capacitor with a plate separation of 2.80 mm and air as the dielectric material is 4.26 x 10^-10 m².
Step-by-step explanation:
To find the plate area required to provide a capacitance of 1.50 pF in a parallel-plate capacitor with a plate separation of 2.80 mm and air as the dielectric material, we can use the formula for capacitance:
C = (ε₀ * A) / d
where C is the capacitance, ε₀ is the permittivity of free space (8.85 x 10^-12 F/m), A is the plate area, and d is the plate separation.
Rearranging the formula to solve for A:
A = (C * d) / ε₀
Substituting the given values:
A = (1.50 x 10^-12 F * 2.80 x 10^-3 m) / (8.85 x 10^-12 F/m)
A = 4.26 x 10^-10 m²
Therefore, the plate area required is 4.26 x 10^-10 m².