Capacitance of a parallel-plate capacitor is C = ε0 A / d. If A = 0.01 m^2 and d = 0.005 m, what is C approximately? (ε0 = 8.85×10^-12 F/m)

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Multiple Choice

Capacitance of a parallel-plate capacitor is C = ε0 A / d. If A = 0.01 m^2 and d = 0.005 m, what is C approximately? (ε0 = 8.85×10^-12 F/m)

Explanation:
Capacitance for a parallel-plate capacitor grows with the plate area and shrinks as the separation increases, which is captured by C = ε0 A / d. Here A/d = 0.01 / 0.005 = 2, so C = ε0 × 2. With ε0 ≈ 8.85 × 10^-12 F/m, C ≈ 2 × 8.85 × 10^-12 = 1.77 × 10^-11 F. The value 1.77 × 10^-11 F matches the given option.

Capacitance for a parallel-plate capacitor grows with the plate area and shrinks as the separation increases, which is captured by C = ε0 A / d. Here A/d = 0.01 / 0.005 = 2, so C = ε0 × 2. With ε0 ≈ 8.85 × 10^-12 F/m, C ≈ 2 × 8.85 × 10^-12 = 1.77 × 10^-11 F. The value 1.77 × 10^-11 F matches the given option.

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