Given that e = 9.0 v, r = 98 ω and c = 23 μf, how much charge is on the capacitor at time t = 4.0 ms

Let charge across the capacitor be Q, current through the circuit be I.

Voltage difference across the resistor = rI

Voltage difference across the capacitor = Q/c

Loop rule: net voltage change through a loop must be zero, so

9 = rI + Q/c. Since I = dQ/dt,

r dQ/dt + Q/c = 9

Solving, Q = 9c (1 - e^ (t/rc)). Plug in the numbers from the problem for the numerical answer.