Question

The two plates of a capacitor hold +2500 μC and -2500 μC of charge, respectively, when the potential difference is 960V. What is the capacitance?

Answer 1

The capacitance of the capacitor is given by

$$C=\frac{q}{V}=\rm\frac{2500*10^{-6}\: C}{960\: V}=2.6*10^{-6}\: F=2.6\: \mu F$$