# Capacitor dealing with exponential forms, please email me or IM so i know you replied

1. Dec 10, 2006

### 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 in to 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?

Last edited: Dec 10, 2006
2. Dec 10, 2006

### andrevdh

Use the equation

$$Q = CV$$

3. Dec 10, 2006

### chevycamaro1987

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?

4. Dec 10, 2006

### tim_lou

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

5. Dec 11, 2006

### 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

$$Q = CV$$

Last edited: Dec 11, 2006