Question

a capacitor that has air between its plates is connected across a potential differences of 12.0 v and stores 48.0 mu c. what is the capacitance of capacitor?

Answer 1

The capacitance

$$C=\frac{q}{V}=\rm\frac{48.0*10^{-6}\:C}{12.0\: V}=4.00*10^{-6}\: F=4.00\: \mu F$$