A650 × 10–4 f capacitor stores 24 × 10–3 of charge. what is the potential difference between the plates? a. 0.0016 v b. 0.089 v c. 0.37 v d. 2.7 v

C. 0.37V. A capacitor of 650x10⁻⁴F that stores 24x10⁻³C has a potential difference of 0.37V between its plates.

The key to solve this problem is using the capacitance equation C = Q/Vᵃᵇ, where C is the capacitance, Q the charge stored in the plates, and Vᵃᵇ the potential difference between the plates.

A 650x10⁻⁴F capacitor stores 24x10⁻³C, clear Vᵃᵇ for the equation:

C = Q/Vᵃᵇ > Vᵃᵇ = Q/C

Solving

Vᵃᵇ = 24x10⁻³C/650x10⁻⁴F = 0.37V

it is the second option.