Solving Discharging Capacitor: Why εC in Denom.

[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 I am 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?

 

[PFuser1232]

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.

 

[PFuser1232]

$\frac{dQ}{Cε - Q} = \frac{dt}{RC}$
$\int^Q_{Q_0} \frac{dQ'}{Cε - Q'} = \int^t_{t_0} \frac{dt'}{RC}$
$ln|Cε - Q| - ln|Cε - Q_0| = \frac{t}{RC} - \frac{t_0}{RC}$
$ln(\frac{Cε - Q}{Cε - Q_0}) = \frac{t}{RC} - \frac{t_0}{RC}$

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