Question

The two plates of a capacitor hold 2500 mc and -2500 c of charge, respectively, when the potential difference is 850 v. What is the capacitance?

Answer 1

The capacitance of the capacitor can be calculated using the formula C = Q/V. With a charge of 2500 mc (microcoulombs) and a potential difference of 850 V, the capacitance is found to be approximately 2.94 µF.

Step-by-step explanation:

The question asks about calculating the capacitance of a capacitor when two known quantities are provided: the charge on the plates and the potential difference between them. To find the capacitance, we can use the formula C = Q/V, where C is the capacitance, Q is the charge, and V is the potential difference.

In this case, we have a charge of 2500 mc (which is actually 2500 µC, since 'mc' is likely a typo and should be 'microcoulombs') or 2.5 × 10⁻³ C, and the potential difference is 850 V. Substituting these values into the formula gives us:

C = (2.5 × 10⁻³ C) / (850 V) = 2.94 × 10⁻⁶ F or 2.94 µF.

Hence, the capacitance of the capacitor is approximately 2.94 µF.