The basic relationship between capacitance (C), charge (Q) and voltage (V) for a parallel-plate capacitor is
[tex]C= \frac{Q}{V} [/tex]
in our problem, [tex]C=0.40 \mu F=0.40 \cdot 10^{-6} F[/tex] and V=9.0 V, so the charge on each plate is
[tex]Q=CV=(0.40 \cdot 10^{-6}F)(9.0 V)=3.6 \cdot 10^{-6} C[/tex]
