Calculating Charge on Capacitor: 6.0 uF, 12V, and 100ohm Resistor

[SAT2400]

Hard problemT_T Charge on capacitor??

Homework Statement

A 6.0 uF capacitor is charged up to 12V and subsequently connected through a witch to a 100ohm resistor. At t=0, the switch is closed. What's the charge on the capacitor 6.0ms(=.0006sec which is RC) after the switch is closed??

Homework Equations

Q= Qmax e^(-t/RC)

Please tell me some more equations like this(Q= Qmax e^(-t/RC), I =Imax e^(-t/RC)...etc..
do i use these when what is discharging??!or charging?!)

The Attempt at a Solution

-t/RC is -1 ?? is it right?!

I don't know what to do T_T/...

Help needed ASAP Thank you very much

 

[ApexOfDE]

Charging: q = CV(1 - exp(-t/RC))
Discharging: q = q0 * exp(-t/RC)

In this case, i think you should use discharging eq and there is miscalculation in t/RC.

 

[SAT2400]

hmmm.. the answer is 3.3 x 10^-9C

I have no idea how to get this answer...

I do know that I should use Discharging Equation...

But I don't get the answer...

Please help me again! Thank you soo much!

 

[nasu]

6 ms = 0.006 s (you have an extra zero).