Question

When a battery is connected to the plates of a 3.00-µF capacitor, it stores a charge of 27.0 µC. What is the voltage of the battery?

Answer 1

Q=27$\mu C$ = $27 \times 10^{-6}$ C

C= -3.0 $\mu F$ = -3.0 $\times 10^{-6}$ F

V= voltage of the battery, C= system capacitance , Q= charge stored by the capacitor

$Q= CV$

$V=\frac{Q}{C}= \frac{27\times 10^{-6} C}{3.0\times 10^{-6}F}$

V=9.0V