A parallel plate capacitor has plate area 0.01 m^2 and plate separation 0.02 m. What is its capacitance in farads? ε0 = 8.85×10^-12 F/m.

A parallel plate capacitor has plate area 0.01 m^2 and plate separation 0.02 m. What is its capacitance in farads? ε0 = 8.85×10^-12 F/m.

• A. 4.0×10^-12 F
• B. 8.85×10^-12 F
• C. 2.21×10^-11 F
• D. 4.425×10^-12 F ≈ 4.4 pF

Answer: (D)

Capacitance of a parallel-plate capacitor in vacuum follows C = ε0 A / d, where ε0 is the permittivity of free space, A is the plate area, and d is the separation. With ε0 = 8.85×10^-12 F/m, A = 0.01 m^2, and d = 0.02 m, the calculation is C = (8.85×10^-12 × 0.01) / 0.02 = (8.85×10^-14) / 0.02 = 4.425×10^-12 F. This is about 4.4×10^-12 F, i.e., 4.4 pF. The given value matches 4.425×10^-12 F, so that is the correct result.

Capacitance of a parallel-plate capacitor in vacuum follows C = ε0 A / d, where ε0 is the permittivity of free space, A is the plate area, and d is the separation. With ε0 = 8.85×10^-12 F/m, A = 0.01 m^2, and d = 0.02 m, the calculation is C = (8.85×10^-12 × 0.01) / 0.02 = (8.85×10^-14) / 0.02 = 4.425×10^-12 F. This is about 4.4×10^-12 F, i.e., 4.4 pF. The given value matches 4.425×10^-12 F, so that is the correct result.