A capacitor of 50$$\mu$$F is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is _________ $$\mu$$C.

We need to find the steady-state charge on the upper plate of the capacitor connected in the given circuit network.

1. Analyze the Circuit in Steady State

In a DC circuit, once the circuit reaches a steady state, a capacitor becomes fully charged and acts as an open circuit (infinite resistance). Therefore, no current flows through the branch containing the capacitor $$C = 50\ \mu\text{F}$$.

The entire current ($$I$$) supplied by the $$6\text{ V}$$ battery flows through the three series-connected resistors:

• $$R_1 = 2\text{ k}\Omega$$ (top resistor)
• $$R_2 = 2\text{ k}\Omega$$ (middle resistor)
• $$R_3 = 2\text{ k}\Omega$$ (bottom resistor, connected in parallel with the capacitor)

2. Calculate the Total Circuit Current ($$I$$)

The total equivalent resistance ($$R_{\text{eq}}$$) of the series circuit path is:

$$R_{\text{eq}} = R_1 + R_2 + R_3$$

$$R_{\text{eq}} = 2\text{ k}\Omega + 2\text{ k}\Omega + 2\text{ k}\Omega = 6\text{ k}\Omega = 6 \times 10^3\ \Omega$$

Using Ohm's law, the steady-state current running through the main loop is:

$$I = \frac{V}{R_{\text{eq}}} = \frac{6\text{ V}}{6 \times 10^3\ \Omega} = 10^{-3}\text{ A} = 1\text{ mA}$$

3. Determine the Potential Difference Across the Capacitor ($$V_C$$)

The capacitor is connected in parallel across the bottom $$2\text{ k}\Omega$$ resistor ($$R_3$$). Thus, the potential difference ($$V_C$$) across the plates of the capacitor equals the voltage drop across $$R_3$$:

$$V_C = I \times R_3$$

$$V_C = (1 \times 10^{-3}\text{ A}) \times (2 \times 10^3\ \Omega) = 2\text{ V}$$

Looking at the polarity, the upper plate of the capacitor connects to a point higher up in the potential ladder relative to the ground terminal ($$0\text{ V}$$), meaning the upper plate accumulates a positive charge.

4. Calculate the Charge on the Upper Plate ($$Q$$)

The magnitude of the charge accumulated on the capacitor is calculated using the formula:

$$Q = C \times V_C$$

Substitute the given value of capacitance ($$C = 50\ \mu\text{F}$$) and the derived voltage drop:

$$Q = 50\ \mu\text{F} \times 2\text{ V} = 100\ \mu\text{C}$$

Final Whole Number Answer: $$100$$