Correct Option: (C) Let's analyze the circuit at steady state. Step 1: Understand the behavior of the capacitor at steady state in a DC circuit. At steady state in a direct current (DC) circuit, a capacitor acts as an open circuit. This means that once the capacitor is fully charged, no current flows through the branch containing the capacitor. Consequently, the current from the cell will flow exclusively through the resistive paths. Step 2: Identify the path for current flow at steady state. According to the circuit description: * The cell with EMF $\mathcal{E}$ is connected in series with resistor $R_1$. * This series combination is then connected across a parallel combination of resistor $R_2$ and capacitor $C$. Since the capacitor $C$ acts as an open circuit at steady state, the current from the cell will flow through $R_1$ and then entirely through $R_2$. The capacitor $C$ will store charge until the potential difference across its plates equals the potential difference across $R_2$. Step 3: Calculate the total equivalent resistance of the circuit for current flow. At steady state, the current flows through $R_1$ and $R_2$, which are effectively in series with respect to the cell (as no current goes through the capacitor branch). The total equivalent resistance $R_{eq}$ is: $$R_{eq} = R_1 + R_2$$ Substituting the given values: $$R_{eq} = 2 \, \Omega + 3 \, \Omega = 5 \, \Omega$$ Step 4: Calculate the total current flowing in the circuit. Using Ohm's law, the total current $I$ supplied by the cell is: $$I = \frac{\mathcal{E}}{R_{eq}}$$ Substituting the values: $$I = \frac{10 \, \text{V}}{5 \, \Omega} = 2 \, \text{A}$$ Step 5: Determine the voltage across the capacitor. At steady state, the capacitor is fully charged, and the voltage across it, $V_C$, is equal to the voltage across the resistor $R_2$, because they are connected in parallel. The voltage across $R_2$, $V_{R_2}$, is given by Ohm's law: $$V_{R_2} = I \cdot R_2$$ Substituting the calculated current and the value of $R_2$: $$V_{R_2} = (2 \, \text{A}) \cdot (3 \, \Omega) = 6 \, \text{V}$$ Therefore, the voltage across the capacitor is $V_C = 6 \, \text{V}$. Step 6: Calculate the charge stored on the capacitor. The charge $Q$ stored on a capacitor is given by the formula: $$Q = C \cdot V_C$$ Substituting the given capacitance $C = 2 \, \mu \text{F} = 2 \times 10^{-6} \, \text{F}$ and the calculated voltage $V_C = 6 \, \text{V}$: $$Q = (2 \times 10^{-6} \, \text{F}) \cdot (6 \, \text{V})$$ $$Q = 12 \times 10^{-6} \, \text{C} = 12 \, \mu \text{C}$$ Thus, the magnitude of the charge stored on the capacitor is $12 \, \mu \text{C}$. Final check of other options: (A) $6 \, \mu \text{C}$: This would imply $V_C = Q/C = (6 \times 10^{-6}) / (2 \times 10^{-6}) = 3 \, \text{V}$, which is not the voltage across $R_2$. (B) $10 \, \mu \text{C}$: This would imply $V_C = Q/C = (10 \times 10^{-6}) / (2 \times 10^{-6}) = 5 \, \text{V}$, which is incorrect. (D) $20 \, \mu \text{C}$: This would imply $V_C = Q/C = (20 \times 10^{-6}) / (2 \times 10^{-6}) = 10 \, \text{V}$. This voltage would only be across the capacitor if it were directly connected across the cell's EMF (assuming no internal resistance and $R_1=0$), which is not the case here. The resistor $R_1$ causes a voltage drop, so the voltage across $R_2$ (and thus $C$) must be less than the cell's EMF.