Question

A 650 × 10–4 F capacitor stores 24 × 10–3 of charge.

Answer 1

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