Find charge on capacitor in the given circuit at steady state.

The correct answer is:
$\frac{40}{7}\mu C$

1. Analysis of the Circuit Diagram:
Looking at the provided circuit diagram, we can observe the following components and connections:
- A DC voltage source (battery) with a potential difference of 10 V on the left.
- A bridge network consisting of four resistors: a 1 Ω resistor connected to node A, a 2 Ω resistor connected to node A, a 3 Ω resistor connected to node B, and a 4 Ω resistor connected to node B.
- A capacitor with a capacitance of C = 6 μF connected between nodes A and B.

2. Steady-State Behavior of the Capacitor:
In a direct current (DC) circuit, once steady state is reached, the capacitor is fully charged. At this stage, it blocks any further flow of direct current through its branch. Therefore, the branch containing the 6 μF capacitor behaves as an open circuit, and we can assume no current flows from node A to node B (or vice versa).

3. Calculating Node Potentials:
Since no current flows through the capacitor branch, the circuit simplified at steady state consists of two independent parallel branches connected across the 10 V source:
- Branch 1 (through node A): Consists of the 1 Ω and 2 Ω resistors in series.
- Branch 2 (through node B): Consists of the 3 Ω and 4 Ω resistors in series.
Let us define the potential of the negative (bottom) terminal of the battery as 0 V, making the potential of the positive (top) terminal 10 V.

Using the voltage divider rule, we can find the electrical potentials at nodes A and B:
- For Node A:

$V_A=10\times \frac{2}{1+2}=\frac{20}{3}\text{V}$

- For Node B:

$V_B=10\times \frac{4}{3+4}=\frac{40}{7}\text{V}$

4. Potential Difference Across the Capacitor:
The potential difference across the 6 μF capacitor is given by the difference in potential between nodes A and B:

$V_{AB}=V_A-V_B$

Substitute the values:

$V_{AB}=\frac{20}{3}-\frac{40}{7}=\frac{140-120}{21}=\frac{20}{21}\text{V}$

5. Charge on the Capacitor:
Using the formula for the charge on a capacitor (Q = C × V):

$Q=C\times V_{AB}$

Substitute C = 6 μF and the potential difference:

$Q=6\mu \text{F}\times \frac{20}{21}\text{V}=\frac{120}{21}\mu \text{C}=\frac{40}{7}\mu C$