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.

Your awser would be a acetone

Hello there,it will only be somewhat eroded (25%)hope this : ))~top