The total charge on the system of capacitance $$C_1 = 1\mu F$$, $$C_2 = 2\mu F$$, $$C_3 = 4\mu F$$ and $$C_4 = 3\mu F$$ connected in parallel is (Assume a battery of $$20 \text{ V}$$ is connected to the combination)

Four capacitors $$C_1 = 1\,\mu F$$, $$C_2 = 2\,\mu F$$, $$C_3 = 4\,\mu F$$, and $$C_4 = 3\,\mu F$$ are connected in parallel across a $$20 \text{ V}$$ battery. We need to find the total charge on the system.

For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances:

$$C_{eq} = C_1 + C_2 + C_3 + C_4 = 1 + 2 + 4 + 3 = 10\,\mu F$$

Using $$Q = C_{eq} \times V$$:

$$Q = 10\,\mu F \times 20 \text{ V} = 200\,\mu C$$

The correct answer is Option A: $$200\,\mu C$$.