The capacitance of a capacitor is given by the following equation: [tex]C= \frac{Q}{V} [/tex] (1) where Q is the charge stored on the capacitor and V the voltage across its plates.
In this problem, we have a capacitor of [tex]C=4.0 \mu F=4.0 \cdot 10^{-6}F [/tex] which is charged with a charge equal to [tex]Q=52 \mu C= 52 \cdot 10^{-6} C[/tex] If we rearrange equation (1) and we use these data, we can find the voltage of the battery at which the capacitor was connected: [tex]V= \frac{Q}{C}= \frac{52 \cdot 10^{-6} C}{4.0 \cdot 10^{-6} F}=13 V [/tex]
