The relationship between the capacitance, the charge on the capacitor and the potential difference applied across its plates is
[tex]C= \frac{Q}{V} [/tex]
where C is the capacitance, Q is the charge and V is the potential difference.
In our problem, [tex]C=0.30 \mu F[/tex] and [tex]V=9.0 V[/tex], so if we re-arrange the previous formula we can find the charge stored in the capacitor:
[tex]Q=CV=(0.30 \mu F)(9.0 V)=2.7 \mu C[/tex]
