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= (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= (Q)/(C)= (52 \cdot 10^(-6) C)/(4.0 \cdot 10^(-6) F)=13 V`