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Circuits with capacitators

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The figure (attached) shows a simplified circuit for a photographic flash unit. This circuit consists of a 9.0 V battery, an R = 45.0 k resistor, a 140 µF capacitor, a flashbulb, and two switches. Initially, the capacitor is uncharged and the two switches are open. To charge the unit, switch S1 is closed; to fire the flash, switch S2 (which is connected to the camera's shutter) is closed. How long does it take to charge the capacitor to 5.5 V?

(See attachment for diagram)

I do not know how to do circuits with capacitors in them. I tried treating the circuit like it was in parallel, but it didn't work. I also tried using calculus and got 5.9 x 10-3 seconds, but it was wrong.

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Suppose, at any given moment of time t, the charge on the capacitor (C = 0.14x10^{-3}F) is
Q(t).
Then the voltage on the capacitor is
V_c = Q(t) / C

Given this voltage, if the switch S_1 is closed and the switch S_2 is open (the charging stage), the current is
I = (V_b-V_c) / R,

where V_b = 9 V is the voltage of the battery.
Since
I = dQ/dt,
we now have a differential equation for Q as a function of time:

...

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