The relationship between capacitance C, charge Q and voltage V for a capacitor is
[tex]V= \frac{Q}{C} [/tex]
In our problem, the charge accumulated on the capacitor is [tex]Q=36 \mu C= 36 \cdot 10^{-6} C[/tex], while its capacitance is [tex]C=4 \mu F=4 \cdot 10^{-6} F[/tex]. So, the voltage applied is
[tex]V= \frac{36 \cdot 10^{-6} C}{4 \cdot 10^{-6} F} =9 V[/tex]
