Voltage across the capacitors
The charge built up in C1 becomes subsequently shared with C2 and this changes the "common" voltage from 10 volts to 6.8 volts.

Note that the initial difference in voltage between the two capacitors (10 volts) becomes 6.8 volts i.e. any difference in voltage (such as when it happens the next cycle) is subject to a reduction of 0.68.

So, the next time this happens, C2 is at 6.8 volts and the difference between this and 10 volts is 3.2 volts so, subjecting this new difference voltage to a multiplication by 0.68 means that it becomes 2.176 volts.

In other words C2 charges a further 2.176 volts from 6.8 volts to 8.977 volts.

Or, put another way, at the final hurdle the total charge is 0.01 + 0.003196 = 0.013196 coulombs. So divide this by total capacitance (1470 uF) to get 8.977 volts.