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

  1. Dec 10, 2006 #1
    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. jcsd
  3. Dec 10, 2006 #2


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    Use the equation

    [tex]Q = CV[/tex]
  4. Dec 10, 2006 #3
    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?
  5. Dec 10, 2006 #4
    V is voltage. Relate voltage to current and then Charge to current, you'll get a differential equation.
  6. Dec 11, 2006 #5


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    The capacitor start out with a charge of

    [tex]1.60 \times 10^{-16}\ C[/tex]

    this means that the initial voltage over the cap will be

    [tex]V_o = \frac{1.60 \times 10^{-16}}{C}[/tex]

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

    [tex]V = V_o e^{-\frac{t}{\tau}}[/tex]

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

    [tex]Q = CV[/tex]
    Last edited: Dec 11, 2006
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