Determine the charge on each capacitor 0.20 ms after closing switch S , as shown in the figure.

Step-by-step solution

The two capacitors are connected in parallel. Their equivalent capacitance is: C_ eq = 2.0 F + 2.0 F = 4.0 F The time constant of the circuit is: = R C_ eq = 25 4.0 10⁻⁶ F = 100 10⁻⁶ s = 0.10 ms The maximum total charge on the equivalent capacitor is: Q_ max = C_ eq V = 4.0 F 6.0 V = 24.0 C The total charge at time t = 0.20 ms is given by: Q(t) = Q_ max (1 - e^ -t/ ) Q(0.20 ms ) = 24.0 (1 - e^ -0.20/0.10 ) = 24.0 (1 - e⁻²) Using e⁻² 0.1353 : Q = 24.0 (1 - 0.1353) = 24.0 0.8647 20.75 C Since the two identical capaci