The correct option is **B**. Let's break down the problem step-by-step to determine the charge stored on the capacitor $C$ after a very long time. ### Step-by-Step Derivation #### 1. Analyze the Circuit at Steady State When the switch $S$ has been closed for a very long time, the circuit reaches a steady state. In a DC circuit at steady state, a capacitor acts as an open circuit. This means that no current flows through the branch containing the capacitor $C$ and resistor $R_2$. #### 2. Determine the Current Flow Since no current flows through the branch containing $R_2$ and $C$, the entire current from the battery will flow through the branch containing only resistor $R_1$. The internal resistance $r$ of the battery is in series with this active branch. The effective resistance of the circuit for current flow will be the sum of the internal resistance $r$ and the resistance $R_1$, as the other branch effectively has infinite resistance: $$R_{\text{eq}} = R_1 + r$$ #### 3. Calculate the Total Current from the Battery Using Ohm's Law, the total current $I$ flowing from the battery is: $$I = \frac{\mathcal{E}}{R_{\text{eq}}} = \frac{\mathcal{E}}{R_1 + r}$$ #### 4. Determine the Voltage Across the Capacitor The capacitor $C$ is in series with $R_2$, and this combination is in parallel with $R_1$. Since no current flows through $R_2$ at steady state (due to the capacitor being fully charged and acting as an open circuit), there is no voltage drop across $R_2$: $$V_{R_2} = I_C \cdot R_2 = 0 \cdot R_2 = 0$$ Therefore, the voltage across the capacitor $C$ ($V_C$) is equal to the potential difference across the parallel combination of $R_1$ and the $(R_2+C)$ branch. This potential difference is simply the voltage drop across $R_1$: $$V_C = V_{R_1} = I \cdot R_1$$ Substituting the value of $I$: $$V_C = \left( \frac{\mathcal{E}}{R_1 + r} \right) R_1 = \frac{\mathcal{E} R_1}{R_1 + r}$$ #### 5. Calculate the Charge on the Capacitor The charge $Q$ stored on a capacitor is given by the formula $Q = C V_C$. Substituting the expression for $V_C$: $$Q = C \left( \frac{\mathcal{E} R_1}{R_1 + r} \right) = \frac{C \mathcal{E} R_1}{R_1 + r}$$ Thus, the charge stored on the capacitor $C$ is **$\frac{C \mathcal{E} R_1}{R_1 + r}$**. --- ### Analysis of Incorrect Options: * **Option A ($Q = C \mathcal{E}$):** This would be true if the capacitor was directly connected across the battery with no internal resistance and no other resistors, which is not the case here. * **Option C ($Q = \frac{C \mathcal{E} R_1}{R_1 + R_2 + r}$):** This expression incorrectly suggests that $R_2$ carries current and contributes to the voltage division across the external circuit, which is false at steady state for the capacitor branch. * **Option D ($Q = C \mathcal{E} \left( \frac{R_1}{R_1 + R_2} \right)$):** This would be applicable if $R_1$ and $R_2$ were part of a voltage divider across the battery, and $C$ was parallel to $R_1$, with no internal resistance, which is not the configuration here.