Question

What is the voltage of a battery that will charge a 4.0 μf capacitor to ± 36 μc ?

Answer 1

The relationship between capacitance C, charge Q and voltage V for a capacitor is

`V= (Q)/(C)`
In our problem, the charge accumulated on the capacitor is
`Q=36 \mu C= 36 \cdot 10^(-6) C`, while its capacitance is
`C=4 \mu F=4 \cdot 10^(-6) F`. So, the voltage applied is

`V= (36 \cdot 10^(-6) C)/(4 \cdot 10^(-6) F) =9 V`