Question

What is the voltage of a battery that will charge a 4.0 μf capacitor to ± 52 μc ?

Answer 1

The capacitance of a capacitor is given by the following equation:
$C= \frac{Q}{V}$ (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
$C=4.0 \mu F=4.0 \cdot 10^{-6}F$
which is charged with a charge equal to
$Q=52 \mu C= 52 \cdot 10^{-6} C$
If we rearrange equation (1) and we use these data, we can find the voltage of the battery at which the capacitor was connected:
$V= \frac{Q}{C}= \frac{52 \cdot 10^{-6} C}{4.0 \cdot 10^{-6} F}=13 V$