So this is a homework question and I am having trouble figuring out what they are asking.
'The potential difference (voltage) across the capacitor at time t > 0 is given by $V_C(t) = q(t)/C$. The quantity RC has the dimensions of time and is often called the time constant for the circuit. How many time constants does it take for a capacitor to charge to 90% of the applied voltage, V0? Justify your answer'
So change in V, or $V_C$, or $\delta{V}$ is 90%. In other words we have $0.9V=q(t)/C$?
I have found in a previous question that $q(t)=V_0C(1-e^{-\frac{t}{CR}})$ so
$$0.9V=\frac{V_0C(1-e^{-\frac{t}{CR}})}{C}$$
However I am not sure where to go from here, if I am even on the right track at all.
If you look at the capacitor voltage curve, you notice that somewhere between $2$ and $3$ time constants, we have $90\%$ charge. We should be able to figure this out generally when not given the resistor and capacitor value.
We have the unknown:
$$\mbox{Time Constant} = \tau = RC$$
Using:
$$\large 0.9V=\frac{V_0C(1-e^{-\frac{t}{CR}})}{C} = V_0(1-e^{-\frac{t}{CR}}) = V_0(1-e^{-\frac{t}{\tau}})$$
We want to solve for $t$, so we have:
$$t = - \tau \ln \left(-\frac{0.9 V-V_0}{V_0}\right)$$
However, the voltage across the capacitor is $.9$ of the the voltage source $V_0$, so we can rewrite this as:
$$t = - \tau \ln \left(-\frac{0.9 V-V_0}{V_0}\right) = - \tau \ln \left(-\frac{0.9 V_0-V_0}{V_0} \right) = - \tau \ln (0.1) = -(-2.30259)\tau = 2.30259 \tau$$
In other words, it will take $2.30259$ time constants to charge to $90\%$.