In the given circuit diagram, when the current reaches a steady-state in the circuit, the charge on the capacitor of capacitance $$C$$ will be:

In a DC circuit, once the capacitor reaches its steady-state (is fully charged), it acts as an open circuit. This means that no current flows through the branch containing the capacitor $$C$$ and the resistor $$r_1$$. Consequently, the potential drop across $$r_1$$ is zero ($$V_{r_1} = I \times r_1 = 0 \times r_1 = 0$$).

Using Ohm's Law, the steady-state current $$I = \frac{E}{r + r_2}$$

The potential difference across $$r_2$$ is $$V_{r_2} = I \times r_2$$

$$V_{r_2} = \left( \frac{E}{r + r_2} \right) \times r_2 = \frac{E r_2}{r + r_2}$$

Since $$V_C = V_{r_2}$$ (steady state), $$V_C = \frac{E r_2}{r + r_2}$$

$$Q = C \times V$$

$$Q = C \times \left( \frac{E r_2}{r + r_2} \right)$$

$$Q = CE \frac{r_2}{r + r_2}$$