Capacitor charge??

Question

A capacitor is charged to 1 coulomb; the capacitance is 6.00×10-5 farads. Then a switch is closed which puts the capacitor in a closed circuit with a resistor; the resistance is 8.50×101 ohms. Calculate the charge on the capacitor after 9.00 milliseconds. (1 ms = 0.001 s)

Please help I have no idea where to start :(

Answer 1

In this case, the capacitor will behave as a battery. The resistor will dissipate electrical energy according to the function:

Q(t) = Q₀e^(-t / τ)

The time constant is:

τ = RC = 85.0 Ω x (6.00 x 10^-5 F) = 5.1 ms

At t = 9.00 ms, the new charge on the capacitor is:

Q = 1 C x e^(-0.009 s / 0.0051 s)

= 0.171 C