Capacitors and RC Decay

Right away when I think of charge I think of Q = CV (since its asking for coulombs). I am thinking of using the question Q = CV(1-e-t/RC). The thing is that what is the capacitance if I'm using this... and what is t if I have the time constant already?

Q = CV

First we must find C, because we have a given V(10V)

Your time constant(RC or tau) is given at 100 seconds.

Your resistance is given at 20M ohms.

tau = RC 100 s = 20Mohm * C C = 5uF

Q = CV = 5uF * 10V = 50 microcoulombs

t, is the time in seconds. That second equation you mentioned is solving for instantaneous charge on the plates.

Q = CV(1-e(-t/tau) = 5uF * 10V(1 - e(-100s/20M*5uF)

The 100 seconds is your time constant, so tau over tau equals one, which completely wipes away the rest of the equation and you are left with Q = CV.

You need to find the capacitance of the capacitor (that's C in the equation Q = CV), and then you can use Q = CV to find the charge on its plates as you said.

The definition of the time constant is RC, or the resistance times the capacitance. Since (time constant) = RC, you can find the capacitance by dividing the time constant by the resistance:

(100 s)/(20 Mohms) = 2 times 109 farads

Now use Q = CV = (2 x 10^9 F)(10 V) = 2 x 1010 coulombs

I have to say I'm not absolutely sure about that answer, especially since a) I'm just learning about RC circuits myself in school (we did finish the unit, though, so I do KNOW the stuff) and b) that's a HECK of a lot of charge on a capacitor. Hope I helped, though!

C=100/20*106
=5*10(-6) (F)

Q=C*V=10*5*10(-6)