Question

At (= 0, switch S is closed. The charge on the capacitor is
varying with time as $Q=Q_0(1-e ^{-at}). $ Obtain the value of $Q_0 $
and $\alpha $ in the given circuit parameters.

Answer 1

$Q_0 $ is the steady state charge stored in the capacitor.
$Q_0=C $ [PD across capacitor in steady state]
= C [steady state current through $R_2](R_2) $
$ =C\bigg (\frac {V}{R_1+R_2}\bigg ).R_2 $
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, Q_0= \frac {CV R_2}{R_1+R_2} $
$\alpha \, is \, 1/ \tau_c \, or \, \frac {1}{CR_{net}} $
Here, $R_{net} $ is equivalent resistance across capacitor after
short circuiting the battery. Thus,
$R_{net}= \frac {R_1R_2}{R_1+R_2} $ (As $R_1 \, and \, R_2 $ are in parallel)
$\alpha = \frac {1}{C\bigg (\frac {R_1R_2}{R_1+R_2}\bigg )}= \frac {R_1+R_2}{CR_1R_2} $