Discharging capacitor

  • Thread starter iScience
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  • #1
466
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$$\epsilon=IR+V_C$$

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

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

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

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

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?
 

Answers and Replies

  • #2
479
20
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.
 
  • #3
479
20
##\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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