Question

A .30uf capacitor is connected to 9.0v battery what is the charge on the playe of the compacitor

Answer 1

The relationship between the capacitance, the charge on the capacitor and the potential difference applied across its plates is

`C= (Q)/(V)`
where C is the capacitance, Q is the charge and V is the potential difference.

In our problem,
`C=0.30 \mu F` and
`V=9.0 V`, so if we re-arrange the previous formula we can find the charge stored in the capacitor:

`Q=CV=(0.30 \mu F)(9.0 V)=2.7 \mu C`