Answer: 63% of the final charge
Explanation:
When a capacitor is connected to a battery/power supply, the capacitor charges following the exponential law:
[tex]Q(t)=Q_0 (1-e^{-\frac{t}{\tau}})[/tex]
where
[tex]Q_0[/tex] is the final charge of the capacitor, which is
[tex]Q_0 =CV_0[/tex]
where C is the capacitance and [tex]V_0[/tex] the potential difference of the battery
t is the time
[tex]\tau[/tex] is the time constant of the circuit
Re-writing the equation,
[tex]Q(t)=CV_0 (1-e^{-\frac{t}{\tau}})[/tex]
After a time equal to one time constant,
[tex]t=\tau[/tex]
Therefore the charge on the capacitor will be
[tex]Q(\tau)=CV_0 (1-e^{-\frac{\tau}{\tau}})=CV_0(1-e^{-1})=0.63CV_0[/tex]
Which means 63% of the final charge.
