Thank you for reporting, we will resolve it shortly
Q.
In the following circuit, the switch $S$ is closed at $t=0$. The charge on the capacitor $C_1$ as a function of time will be given by $\left( C_{eq} = \frac{C_1 C_2}{C_1 + C_2} \right)$
Since the two capacitors are joined in series, charge that passes through each will be same.
And equivalent capacitor is given by $C_{e q}=\left[\frac{C_{1} C_{2}}{C_{1}+C_{2}}\right]$.
We know that growth of charge with time is given by $Q=Q_{0}\left[1-\exp \left(-\frac{t}{C_{ eq } R}\right)\right]$.
Charge on the capacitor $C_{1}$ is given by $Q_{0}=C_{e q} E .$ Therefore,
$Q=E C_{ eq }\left[1-\exp \left(-\frac{t}{C_{ eq } R}\right)\right]$
