Question

Given that a large Van de Graaff generator's terminal stores 8.00 mC of charge at a voltage of 12.0 MV, what is the capacitance?

a) 6.67 × 10^(-7) F
b) 1.50 × 10^(-6) F
c) 8.33 × 10^(-7) F
d) 1.20 × 10^(-6) F

Answer 1

The capacitance of a large Van de Graaff generator's terminal, which stores 8.00 mC of charge at a voltage of 12.0 MV, is calculated using the formula C = Q/V. The correct capacitance is found to be 6.67 × 10^-7 F, which corresponds to option (a).

Step-by-step explanation:

To find the capacitance of a large Van de Graaff generator's terminal, we can use the relationship defined by the formula C = Q/V, where C is the capacitance, Q is the stored charge, and V is the voltage. In this particular case, the generator's terminal stores 8.00 mC (or 8.00 × 10-3 C) of charge at a voltage of 12.0 MV (or 12.0 × 106 V).

Using the formula, the capacitance (C) can be calculated as:

C = Q/V

C = (8.00 × 10-3 C) / (12.0 × 106 V)

C = (8.00 / 12.0) × 10-3-6 F

C = 0.667 × 10-9 F

C = 6.67 × 10-7 F

Therefore, the capacitance of the Van de Graaff generator's terminal is 6.67 × 10-7 F, which is option (a).