Steady-state behavior of a capacitor

In a DC circuit, when steady state is reached, the capacitor behaves like an open circuit.

Hence:

• No current flows through the capacitor branch.
• The charge on the capacitor is determined by the potential difference across its terminals.

Interpret the circuit at steady state

Since the capacitor branch is open, only the resistive part of the circuit carries current.

From the given options, the capacitor voltage must be obtained by a potential divider involving the resistors in the branch across which the capacitor is connected.

The form of the correct capacitor voltage is therefore expected to be: $V_C = E\frac{\text{resistance across capacitor}}{\text{total series resistance}}$

Use the correct divider

The capacitor is across the resistor $r_2$, while the current in the conducting loop flows through $r$ and $r_2$ in series.

Therefore, the steady current in the loop is: $I = \frac{E}{r+r_2}$

So the potential difference across $r_2$ is: $V_{r_2} = Ir_2 = \frac{E}{r+r_2} \cdot r_2 = E\frac{r_2}{r+r_2}$

Since the capacitor is connected across $r_2$, $V_C = E\frac{r_2}{r+r_2}$

Find charge on capacitor

Using $Q = CV$, $Q = C V_C = C\left(E\frac{r_2}{r+r_2}\right) = CE\frac{r_2}{r+r_2}$

Match with options

This corresponds to: $\boxed{CE\frac{r_2}{r+r_2}}$

So the correct option is D.

Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.