# Discharging capacitor

1. Nov 10, 2014

### iScience

$$\epsilon=IR+V_C$$

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

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

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

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

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. Nov 11, 2014

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. Nov 11, 2014

$\frac{dQ}{Cε - Q} = \frac{dt}{RC}$
$\int^Q_{Q_0} \frac{dQ'}{Cε - Q'} = \int^t_{t_0} \frac{dt'}{RC}$
$ln|Cε - Q| - ln|Cε - Q_0| = \frac{t}{RC} - \frac{t_0}{RC}$
$ln(\frac{Cε - Q}{Cε - Q_0}) = \frac{t}{RC} - \frac{t_0}{RC}$
So you can set $Q_0$ and $t_0$ to 0.