A parallel-plate capacitor with plate area 2.70 cm by 2.70 cm carrying ±0.708 nC has a plate separation of 1.50 mm. The capacitance is approximately 4.3 pF. What is the capacitance?

• A. 2.0 pF
• B. 4.3 pF
• C. 8.7 pF
• D. 13 pF

Answer: (B)

Capacitance of a parallel-plate capacitor is set by C = ε0 A / d, where A is the plate area and d is the separation (assuming a vacuum or air with εr ≈ 1). Convert to SI: the plate is 2.70 cm by 2.70 cm, so A = (2.70 cm)^2 = 7.29 cm^2 = 7.29×10^-4 m^2. The separation is 1.50 mm = 1.50×10^-3 m. Using ε0 ≈ 8.854×10^-12 F/m, C = (8.854×10^-12)(7.29×10^-4) / (1.50×10^-3) ≈ 4.30×10^-12 F = 4.3 pF. So the capacitance is about 4.3 pF, which is the option that lists 4.3 pF. The given charge value isn’t needed to find C; it would relate to Q = CV if voltage were known.