Final answer:
The charge that accumulates on the plates of a 6.0-mF air-filled capacitor charged to a potential difference of 40 V is 240 mC.
Step-by-step explanation:
The charge that accumulates on the plates of a 6.0-mF air-filled capacitor when it is charged until the potential difference across its plates is 40 V can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. Substituting the given values, we get Q = (6.0 × 10^{-3} F) × (40 V) = 0.24 C, or 240 mC. Therefore, the charge that accumulates on the plates of the capacitor is 240 mC, which corresponds to option C.