In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
[tex]Q(t) = Q_0 (1-e^{-t/\tau})[/tex]
where t is the time, [tex]Q_0 = CV[/tex] is the maximum charge on the capacitor at voltage V, and [tex]\tau = RC[/tex] is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.
1) Using [tex]R=39.1 \Omega[/tex] and [tex]C= 65.7 mF=65.7\cdot 10^{-3}F[/tex], the time constant of the circuit is:
[tex]\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s[/tex]
2) To find the charge on the capacitor at time [tex]t=1.95 \tau[/tex], we must find before the maximum charge on the capacitor, which is
[tex]Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C[/tex]
And then, the charge at time [tex]t=1.95 \tau[/tex] is equal to
[tex]Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C[/tex]
3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be [tex]Q_0 = 0.59 C[/tex]. We can see this also from the previous formule, by using [tex]t=\infty[/tex]:
[tex]Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C[/tex]
