Magnitude of Final Capacitor Charge when Switch is Shifted to Position Q

To find the magnitude of the final charge $q_2$ on the capacitor, we analyze the circuit in its steady state after the switch $S$ is moved to position $Q$.

1. Circuit Description and Reference Potentials

Let the bottom wire of the circuit be chosen as the reference node with zero potential: $V_{\text{bottom}} = 0\text{ V}$

• Switch at position $Q$: Position $Q$ is directly connected to the bottom reference wire, so the potential at terminal $Q$ is: $V_Q = 0\text{ V}$

• Far-right branch: The positive terminal of the $2\text{ V}$ battery is at a potential of $+2\text{ V}$ relative to the bottom wire.

• Capacitor branch: A $1\ \mu\text{F}$ capacitor is connected between the central node (let's call it node $A$) and the bottom reference wire ($0\text{ V}$).

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2. Steady-State Analysis for Switch at Position $Q$

In a DC circuit, after a sufficiently long time (steady state), the capacitor behaves as an open circuit (i.e., no current flows through the capacitor branch).

Therefore, current flows only through the outer loop containing the $1\ \Omega$ resistor, node $A$, the $2\ \Omega$ resistor, and the $2\text{ V}$ battery.

Applying Kirchhoff's Current Law (KCL) at node $A$: $\frac{V_A - V_Q}{1\ \Omega} + \frac{V_A - 2\text{ V}}{2\ \Omega} = 0$

Substitute $V_Q = 0\text{ V}$: $\frac{V_A - 0}{1} + \frac{V_A - 2}{2} = 0$

$V_A + 0.5 V_A - 1 = 0$

$1.5 V_A = 1 \implies V_A = \frac{2}{3}\text{ V}$

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3. Calculation of Charge $q_2$

The charge stored on the $1\ \mu\text{F}$ capacitor in this steady state is given by: $q_2 = C \cdot V_A = (1\ \mu\text{F}) \times \left(\frac{2}{3}\text{ V}\right) = \frac{2}{3}\ \mu\text{C} \approx 0.67\ \mu\text{C}$

Thus, the magnitude of $q_2$ is $0.67$ (or $\frac{2}{3}$).