Capacitor circuit time constant problem

[helppls]

Homework Statement

How long does it take in time constants ( = RC) for the charge on the capacitor in the circuit
shown below to reach one quarter its initial value?

Homework Equations

C=Q/V
time constant=RC
Vc = Vo * (e)^(-t/RC)

The Attempt at a Solution

From C=Q/V
I get
Q=VC
so for the final charge, Qf, to be 1/4 of it's value
Qf= 0.25 * Vf * C
rearrange to get
Vf = 4 * Qf / C
then I assume I can ignore Qf and C and plug in 4 to the discharging equation:

4= 1 * e^(-t/RC)
then take the natural log of both sides

ln(4) = -t/RC
t= ln(4) * (-RC) = 1.39s
the problem: my friend got a different answer: ln(1/4)s , which makes sense because it yields -1.39s (and is then multiplied by -1) , so I am wondering if I did something incorrectly?

I also am not sure if either of us did it correctly because it is asking for the answer in time constants? That being said it should happen between 1 and 2 time constants because at 1 it would be at 36%, and 2 it would be at 13%.

thanks in advance for your help.

 

[cnh1995]

t= ln(4) * (-RC) = 1.39s

That is incorrect. Your friend got it right.

Qf= 0.25 * Vf * C

No. It should be
Qf= 0.25*V_initial*C.

 

[helppls]

ahhhh, thank you!