Answer:
[tex]Q=81.72\times10^{-6}C[/tex]
[tex]I=2.1\times10^{-5}A[/tex]
Explanation:
The maximum charge on the capacitor will be, at the end of the process, given by the formula (and for our values):
[tex]Q=CV=(6.81\times10^{-6}F)(12V)=81.72\times10^{-6}C[/tex]
The maximum current on the resistor will be, at the beginning of the process, given by the formula (and for our values):
[tex]I=\frac{V}{R}=\frac{12V}{5.8\times10^{5}\Omega}=2.1\times10^{-5}A[/tex]