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(B) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.Therefore,the entire

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$\frac{C E R_1}{R_1+r}$

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(B) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.Therefore,the entire current $I$ flows through the resistor $R_1$ and the internal resistance $r$ of the battery.The total resistance of the circuit is $R_{total} = R_1 + r$.The current in the circuit is $I = \frac{E}{R_1 + r}$.The potential difference across the resistor $R_1$ is $V = I R_1 = \frac{E R_1}{R_1 + r}$.Since the capacitor is connected in parallel with the resistor $R_1$,the potential difference across the capacitor is equal to the potential difference across $R_1$.Thus,the potential difference across the capacitor is $V_C = \frac{E R_1}{R_1 + r}$.The charge on the capacitor is given by $Q = C V_C$.Substituting the value of $V_C$,we get $Q = C \left( \frac{E R_1}{R_1 + r} \right) = \frac{C E R_1}{R_1 + r}$.

In the circuit given,the charge on the capacitor is

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