Magnitude of Initial Capacitor Charge when Switch is at Position P

To find the magnitude of the initial charge $q_1$ on the capacitor when the switch $S$ is at position $P$ in DC steady state, we follow these steps:

1. Circuit Analysis in Steady State

In DC steady state, a capacitor acts as an open circuit, meaning no current flows through the branch containing the $1\ \mu\text{F}$ capacitor.

Let the potential of the bottom reference wire be $0\text{ V}$.

• With switch $S$ at position $P$, the node to the left of the $1\ \Omega$ resistor is connected to the positive terminal of the $1\text{ V}$ battery. Thus, its potential is $V_P = 1\text{ V}$.
• The node to the right of the $2\ \Omega$ resistor is connected to the positive terminal of the $2\text{ V}$ battery. Thus, its potential is $V_R = 2\text{ V}$.

2. Current Calculation

Since no current flows into the capacitor branch, the current $I$ flows in a single loop through the $2\ \Omega$ and $1\ \Omega$ resistors from the higher potential ($2\text{ V}$) to the lower potential ($1\text{ V}$):

$I = \frac{V_R - V_P}{R_1 + R_2} = \frac{2\text{ V} - 1\text{ V}}{1\ \Omega + 2\ \Omega} = \frac{1}{3}\text{ A}$

3. Potential at the Capacitor's Top Plate

Let $V_A$ be the electric potential at the node between the $1\ \Omega$ resistor, $2\ \Omega$ resistor, and the top plate of the capacitor:

$V_A = V_P + I \cdot R_1 = 1 + \left(\frac{1}{3}\right)(1) = \frac{4}{3}\text{ V}$

Since the bottom plate of the capacitor is connected to $0\text{ V}$, the potential difference across the capacitor is:

$V_C = V_A - 0 = \frac{4}{3}\text{ V}$

4. Magnitude of Charge $q_1$

The charge $q_1$ stored on the capacitor is:

$q_1 = C \cdot V_C = (1\ \mu\text{F}) \times \left(\frac{4}{3}\text{ V}\right) = \frac{4}{3}\ \mu\text{C} \approx 1.33\ \mu\text{C}$

Final Answer: The magnitude of $q_1$ is $1.33$ (or $\frac{4}{3}$).