Step-by-step Solution

Part 1: Switch S is at position P for a long time.

1. Analyze the circuit configuration: When the switch S is connected to position P, the circuit consists of a 2V battery, a 1kΩ resistor, a 2kΩ resistor, and a 1μF capacitor. The 2V battery and the 1kΩ resistor are in series. This combination is connected across the parallel combination of the 2kΩ resistor and the 1μF capacitor.

2. Analyze the steady-state condition: When the switch has been at position P for a long time, the circuit reaches a steady state. In a DC steady-state, a capacitor acts as an open circuit because it is fully charged and no more current can flow through it.

3. Calculate the steady-state current and voltage: Since the capacitor branch acts as an open circuit, the current from the 2V source flows through the 1kΩ and 2kΩ resistors, which are in series with each other. The total resistance in the current path is: $R_{total} = R_1 + R_2 = 1 \, k\Omega + 2 \, k\Omega = 3 \, k\Omega = 3000 \, \Omega$ The steady-state current, I, flowing through the resistors is given by Ohm's law: $I = \frac{V}{R_{total}} = \frac{2 \, V}{3 \, k\Omega} = \frac{2}{3} \, mA = \frac{2}{3} \times 10^{-3} \, A$

4. Determine the voltage across the capacitor: The capacitor is connected in parallel with the 2kΩ resistor. Therefore, the voltage across the capacitor ($V_C$) is equal to the voltage across the 2kΩ resistor ($V_{R2}$). $V_{C1} = V_{R2} = I \times R_2$ $V_{C1} = \left( \frac{2}{3} \times 10^{-3} \, A \right) \times (2 \, k\Omega) = \left( \frac{2}{3} \times 10^{-3} \, A \right) \times (2 \times 10^3 \, \Omega) = \frac{4}{3} \, V$ Alternatively, using the voltage divider rule, the voltage across the 2kΩ resistor is: $V_{C1} = V_{source} \times \frac{R_2}{R_1 + R_2} = 2 \, V \times \frac{2 \, k\Omega}{1 \, k\Omega + 2 \, k\Omega} = 2 \, V \times \frac{2}{3} = \frac{4}{3} \, V$

5. Calculate the charge q1 on the capacitor: The charge q1 on the capacitor is given by the formula $q = C \times V$. $q_1 = C \times V_{C1} = (1 \, \mu F) \times \left(\frac{4}{3} \, V\right) = \frac{4}{3} \, \mu C$

6. Final Answer: The magnitude of q1$ is $\frac{4}{3} \, \mu C$. As a decimal, this is approximately 1.333... $\mu$C. The question asks for the magnitude of $q1, which is 1.33 when rounded to two decimal places.

Note on the second part of the problem (for completeness): When the switch is moved to position Q, the 1V source is connected across the parallel combination of the 2kΩ resistor and the 1μF capacitor. After a long time, the capacitor is fully charged to the voltage of the source, so $V_{C2} = 1V$. The charge would be $q_2 = C \times V_{C2} = 1 \, \mu F \times 1 \, V = 1 \, \mu C$. The question, however, only asks for the magnitude of q1.