Simple Capacitor Charging Question
Well... Let's calculate it using Laplace transform.

$$
T(s)=\frac{\frac{\frac{1}{sC}\cdot R_{2}}{\frac{1}{sC}+ R_{2}}}{\frac{\frac{1}{sC}\cdot R_{2}}{\frac{1}{sC}+ R_{2}}+R_{1}}=\frac{R_{2}}{R_{1}+R_{2}+sC\cdot R_{1}R_{2}}=\frac{R_{2}}{R_{1}+R_{2}}\cdot \frac{1}{1+sC\cdot \frac{R_{1}R_{2}}{R_{1}+R_{2}}}
$$
You can see, that we have a resistor divider and RC filter with time constant equal:

$$
\tau = C\cdot \frac{R_{1}R_{2}}{R_{1}+R_{2}}=0.5s
$$

So in 0.5s time, you will have a 0.63% of input voltage multiplied by resistor divider - 2V * 0.5 * 63% = 0.63V