Question

When a cylindrical capacitor is given a charge of 0.500 nC, a potential difference of 20.0 V is measured between the cylinders. What is the capacitance of this system?

Answer 1

The capacitance of the system

$$C=\frac{q}{V}=\frac{0.500*10^{-9}}{20.0}=25*10^{-12}\:\rm F=25\: pF$$