Charge on Capacitor After 4.00ms: Solving the Problem

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A capacitor is charged to 1 coulomb; the capacitance is 9.00×10-5 farads.
Then a switch is closed which puts the capacitor in a closed circuit with a resistor; the resistance is 9.00 ohms.

Calculate the charge on the capacitor after 4.00 milliseconds. (1 ms = 0.001 s).

I know some equations relating capacitance, charge, and voltage, but I don't understand how to approach this problem.
 
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Look up "RC time constant" and the associated equation for a discharging capacitor.
 
Time Constant is T=RC;

but I don't know how to use this equation to solve my problem.
 
That's just the definition of the time constant, not the capacitor discharge equation that uses it. Read this: Charging a Capacitor (Charging and discharging are inverse operations.)
 
Still not making sense. I don't know the value of Q0 or V. Someone please help.
 
Q0 is given in your first post; you don't need V. Don't you have a textbook?

[tex]Q = Q_0 e^{{-t}/{RC}}[/tex]