As far as potential difference is concerned if they are in
series we have to add them. If they are in parallel the potential
difference remains the same. So we need two parallel sets to be
connected in series.
Now as we connect in parallel the effective capacitance would be
got by adding the capacitance.
If capacitors of equal value are connected in series, then its
value will be reduced to half of the individual.
So let us construct two in parallel. So effective will be 4 uF
but with 10 V
Another set of two in parallel has 4 uF with 10 V.
Now these two sets are connected in series. Hence the effective
would become 2 uF but the voltage will be 10+10 ie 20 V.
Hence the problem is solved.
So we need two sets in series