A

Step 1: Understand that the equivalent capacitance of capacitors in series and parallel arrangements must be determined to find the charge.
Step 2: The equivalent capacitance is given as 0.70 µF. From the formula for capacitors in series, it can be expressed as:
$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + ... + \frac{1}{C_n}$$ where each capacitor has capacitance $C_1 = C_2 = C_3 = C_4 = C_5 = C_6 = 1 \mu F$
Step 3: Given 6 capacitors in series, the total capacitance is:
$$\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = \frac{6}{1}$$
Therefore, the equivalent capacitance is $$C_{eq} = \frac{1}{6} = 0.1667 \mu F$$ which does not match the provided value.
Step 4: Assuming two groups of three capacitors in series, gives us:
$$\frac{1}{C_1} = \frac{1}{C}\left[1 + 1 + 1\right] = \frac{3}{1}$$
Therefore, in series, each pair has: $$C_1 = \frac{1}{3} \mu F$$
Step 5: Using two of these results in a parallel arrangement:
$$C_{parallel} = C_{A} + C_{B} = 0.1667 + 0.1667 = 0.3334\mu F$$ which when calculated gives us the total of combined capacitances not matching the original problem.
Step 6: The charge on the capacitors can be calculated using the formula: $$Q = C \times V$$ where C is the equivalent capacitance, and V is the voltage.
Since $C_{eq} = 0.70 \mu F$ and $V = 600 V$, we have:
$$Q = 0.70 \times 10^{-6} F \times 600 V = 0.42 \times 10^{-3} C = 0.42 mC$$
Step 7: Since the capacitors are in series, the charge on each capacitor will be the same. Thus, the charge on each capacitor is 0.42 mC. Therefore, the answer for each capacitor is 0.42 mC as charge. Providing the resultant form in micro-coulombs gives us 420 µC that appears on each capacitor.
Final answer: Therefore, the charge that will appear on each capacitor is 420 µC.