Physics Repost (Answer did not make sense)
posted by Dan .
In the circuit below, at time t = 0, the switch is closed, causing the capacitors to charge. The voltage across the battery is 12 V
The circuit:
flickr.
com/photos/
68751220@N03
/6848519773/
A)Calculate the time constant for the RC circuit
B)Calculate the time required for the voltage across the capacitor to reach 6 V
C)What will be the net charge stored in the capacitor after 10 nanoseconds?
What I did:
A) Time constant= [(C1+C2)*(R1R2)]/(R1+R2)
= [(3uF+6uF)*(4kohm*2kohm)]/(4kohm+2kohm)
= 12 ms= 0.012 sec?
TC= time constant
B) q= q0[1e^(t/TC)]
t= 8.316 ms= 0.008316 sec?
C) I don't know what to do and can't find an equation that I think would work

A) Your answer is correct  the time constant of the circuit is indeed 12ms.
B) The time required to reach the final voltage by a capacitor is roughly 4*TC. In this case,the capacitors would charge up to full battery voltage(6V)in 48 ms.
C) t=10 nanoseconds is such a short time compared to the time constant of the circuit that the capacitors shall hardly be charged. So the net charge stored would be zero coulombs at the end of 10 ns.
You can also calculate the exact amount of q using the formula q=q0[1e^(t/TC)]; just plug in q0 = Ceq*V = 9muF*6 and 10ns for t in the equation to get that small value of q. 
Thanks for your response, but I have a few questions.
For part B, why is full battery voltage 6V, and not 12V?
For part C, my question is once again about using 6V in the calculation for finding q0, as well as units. Using 6V, qo= 54 uC? For the time constant, I plugged in 12ms, so I changed 10 nanoseconds to milliseconds and plugged the ms value in. Overall, I got 4.5E5 uC 
My mistake  I wrongly considered the battery voltage to be 6V instead of 12V.
B) The capacitor voltage will reach 6V (half of its final voltage) in 0.693*TC time. You can also use the equation:
V = V0{1e^(t/RC)} where V=6 and V0=12. You'd get t= 8.316 ms
C)Find q0 by using 12V instead of 6V => q0= 108uC and use the same equation.
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