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Discharging capacitor

  1. Nov 10, 2014 #1
    $$\epsilon=IR+V_C$$

    $$\epsilon=IR+\frac{Q}{C}$$

    $$\frac{dQ}{dt}=\frac{\epsilon-Q/C}{R}$$

    $$\int\frac{dQ}{C\epsilon-Q}=\int\frac{dt}{RC}$$

    $$-ln(\frac{\epsilon C-Q}{\epsilon C})=\frac{t}{RC}$$

    in previous step im confused as to why the εC is in the denominator.

    if we have $$\int\frac{1}{x-2}dx$$

    the answer is $$ln(x-2)$$

    theres no denominator, so where is the one in the derivation popping up from?
     
  2. jcsd
  3. Nov 11, 2014 #2
    Perhaps the author used definite integrals, not indefinite integrals. Had he used the latter, there would have been a constant of integration, which seems to be missing from your equations.
     
  4. Nov 11, 2014 #3
    ##\frac{dQ}{Cε - Q} = \frac{dt}{RC}##
    ##\int^Q_{Q_0} \frac{dQ'}{Cε - Q'} = \int^t_{t_0} \frac{dt'}{RC}##
    ##ln|Cε - Q| - ln|Cε - Q_0| = \frac{t}{RC} - \frac{t_0}{RC}##
    ##ln(\frac{Cε - Q}{Cε - Q_0}) = \frac{t}{RC} - \frac{t_0}{RC}##

    Q = 0 when t = 0
    So you can set ##Q_0## and ##t_0## to 0.
     
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