Calculating Protons after 2 Seconds - R and C Values

[chevycamaro1987]

1. A capacitor where C=1X10^-5 F and R= 1X10^6, there's 1000 electrons on the plate at t=0. How would I find the number of protons after 2 seconds?



2. half life= .7Tm, mean life=RC



3. I put the R and C values into get the mean life and got an answer of 7 seconds, and than to find the halflife, I multiplied the mean life by .7 and got 10 seconds. I just don't know where to go from there though. I know the answer is 820, but I don't know how to get that answer. Is there a formula where I can put in the number of seconds using the mean life and half life to get the answer?

 

[andrevdh]

Use the equation

$$Q = CV$$

 

[chevycamaro1987]

Use the equation

$$Q = CV$$

I tried using that formula but what do I insert for V? I put in 2 seconds and it didnt work, i tried working backwards, that didnt work either...what am i doing wrong?

 

[tim_lou]

V is voltage. Relate voltage to current and then Charge to current, you'll get a differential equation.

 

[andrevdh]

The capacitor start out with a charge of

$$1.60 \times 10^{-16}\ C$$

this means that the initial voltage over the cap will be

$$V_o = \frac{1.60 \times 10^{-16}}{C}$$

or rather... the RC discharge equation of the cap

$$V = V_o e^{-\frac{t}{\tau}}$$

can be rewritten for the decay of the charge on the cap using the relation

$$Q = CV$$