In the following circuit $$C_1 = 12$$ $$\mu F$$, $$C_2 = C_3 = 4$$ $$\mu F$$ and $$C_4 = C_5 = 2$$ $$\mu F$$. The charge stored in $$C_3$$ is ______ $$\mu C$$.

Let the battery of emf $$8 \text{ V}$$ be connected between the upper junction $$X$$ (positive) and the lower junction $$Y$$ (zero potential). The network between $$X$$ and $$Y$$ consists of two series branches kept in parallel and a bridging capacitor $$C_3$$ joining the mid-points of the two branches.

Branch-1 (left): $$C_1 = 12 \,\mu\text{F}$$ in series with $$C_2 = 4 \,\mu\text{F}$$. Branch-2 (right): $$C_4 = 2 \,\mu\text{F}$$ in series with $$C_5 = 2 \,\mu\text{F}$$. The mid-points of the two branches are the plates joined by $$C_3 = 4 \,\mu\text{F}$$.

Step 1 : Equivalent capacitance of each series branch
For two capacitors in series, $$C_{\text{eq}} = \frac{C_a\,C_b}{C_a + C_b}$$.

Left branch: $$C_{L} = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3 \,\mu\text{F}$$.

Right branch: $$C_{R} = \frac{2 \times 2}{2 + 2} = \frac{4}{4} = 1 \,\mu\text{F}$$.

Step 2 : Charge flowing through each branch
Because the two branches are in parallel across the same 8 V, $$Q_{L} = C_{L}\,V = 3 \times 8 = 24 \,\mu\text{C}$$ $$Q_{R} = C_{R}\,V = 1 \times 8 = 8 \,\mu\text{C}$$.

Step 3 : Potentials of the mid-points
In a series combination the charge on each capacitor is the same, so

Left branch: Voltage across $$C_2$$ (lower capacitor) $$V_{C_2} = \frac{Q_L}{C_2} = \frac{24}{4} = 6 \text{ V}$$. Therefore the potential of its upper plate (mid-point $$A$$) with respect to $$Y$$ is $$6 \text{ V}$$.

Right branch: Voltage across $$C_5$$ (lower capacitor) $$V_{C_5} = \frac{Q_R}{C_5} = \frac{8}{2} = 4 \text{ V}$$. Therefore the potential of its upper plate (mid-point $$B$$) with respect to $$Y$$ is $$4 \text{ V}$$.

Step 4 : Potential difference across $$C_3$$
$$\Delta V_{3} = V_A - V_B = 6 \text{ V} - 4 \text{ V} = 2 \text{ V}$$.

Step 5 : Charge on $$C_3$$
Using $$Q = C\,\Delta V$$, $$Q_3 = 4 \,\mu\text{F} \times 2 \text{ V} = 8 \,\mu\text{C}$$.

Hence, the charge stored in $$C_3$$ is $$\mathbf{8 \,\mu C}$$.