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 $$\left(C_{eq} = \frac{C_1 C_2}{C_1 + C_2}\right)$$:

$$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$$

Using Kirchhoff's loop rule for series charging:

$$E - \frac{q}{C_1} - \frac{q}{C_2} - iR = 0 \implies E - \frac{q}{C_{eq}} - R\frac{dq}{dt} = 0$$

Integrating with initial condition $$q(0) = 0$$:

$$q(t) = C_{eq}E\left[1 - \exp\left(-\frac{t}{RC_{eq}}\right)\right]$$

Since the capacitors are connected in series, the same charge flows through both:

$$q_1(t) = q(t) = C_{eq}E\left[1 - \exp\left(-\frac{t}{RC_{eq}}\right)\right]$$