Through a resistance of 100 , a 20 capacitor is connected to a...

The maximum charge on the capacitor is given by $q_{0} = CV$.

$q_{0} = 20 \times 10^{-6} \times 6.0 = 120 \mu\text{C}$

The time constant of the $RC$ circuit is $\tau = RC$.

$\tau = 100 \times 20 \times 10^{-6} = 2.0 \times 10^{-3} \text{ s} = 2.0 \text{ ms}$

The charge on the capacitor at any time $t$ during charging is given by $q(t) = q_{0}(1 - e^{-t/\tau})$.

Substituting $t = 2.0 \text{ ms}$ and $\tau = 2.0 \text{ ms}$:

$q = 120(1 - e^{-2.0/2.0}) = 120(1 - e^{-1})$

Using $e \approx 2.718$, we have $e^{-1} \approx 0.368$.

$q = 120(1 - 0.368) = 120 \times 0.632 = 75.84 \mu\text{C}$

Rounding to the nearest integer, the charge is approximately $76 \mu\text{C}$.