Determine the charge on each capacitor 0.20\ ms after closing...

The two capacitors are connected in parallel. Their equivalent capacitance is:

$C_{eq} = 2.0\ \mu\text{F} + 2.0\ \mu\text{F} = 4.0\ \mu\text{F}$

The time constant of the circuit is:

$\tau = R C_{eq} = 25\ \Omega \times 4.0 \times 10^{-6}\ \text{F} = 100 \times 10^{-6}\ \text{s} = 0.10\ \text{ms}$

The maximum total charge on the equivalent capacitor is:

Q_\max = C_{eq} V = 4.0\ \mu\text{F} \times 6.0\ \text{V} = 24.0\ \mu\text{C}

The total charge at time $t = 0.20\ \text{ms}$ is given by:

Q(t) = Q_\max (1 - e^{-t/\tau})

$Q(0.20\ \text{ms}) = 24.0 (1 - e^{-0.20/0.10}) = 24.0 (1 - e^{-2})$

Using $e^{-2} \approx 0.1353$:

$Q = 24.0 (1 - 0.1353) = 24.0 \times 0.8647 \approx 20.75\ \mu\text{C}$

Since the two identical capacitors are in parallel, they share the total charge equally. The charge on each capacitor is:

$q = \dfrac{Q}{2} = \dfrac{20.75}{2} \approx 10.37\ \mu\text{C}$