Compute the charge on the capacitor

[fubag]

Homework Statement

http://img259.imageshack.us/img259/1535/question2az5.th.jpg

Compute the charge on the capacitor 10.0 ms after the switch is thrown from position 2 back to position 1.I know from previous parts of this problem that the Voltage across the resistor before the switch is thrown is 8.28 V. Afterwards it is 9.72V.

I figured that the charge q before switch is thrown is 133 micro coulombs.

Homework Equations

The Attempt at a Solution

I tried using the new voltage after the switch is thrown to find q but it didn't work. Can someone please help me out?

 

[fubag]

C = Q/V;

and I found the voltages using the equations relating to RC charging and discharging.

but I am still not getting the right answer.

 

[kamerling]

I can't tell you what you did wrong, because you didn't show your work. Didn't the problem
give R, C and E?

 

[PaintballerCA]

What kind of equation governs a capacitor?

 

[mikelepore]

You're looking for an equation with time t in it. You're going to need that to use your given value of 10.0 ms.

 

[fubag]

A capacitor with C = 1.30×10−5 F is connected as shown in the figure View Figure with a resistor with R = 990 Omega and an emf source with E = 18.0 V and negligible internal resistance. Initially the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2, so that the capacitor begins to charge. After the switch has been in position 2 for 10.0 ms, the switch is moved back to position 1 so that the capacitor begins to discharge.I used the equations for V and I for discharging and charging capacitors.

V = E(1-e^(-t/RC)) charging
V = V_oe^(-t/RC) discharging

I = (E/R)e^(-t/RC) charging
I = (V_o/R)e^(-t/RC) dischargingI then used C = Q/V to find Q for the first part where before switch is thrown...hope this helped clarify some things, sorry for the lack of info