Maximum charge capacitance C1 can withstand is Q1=C1V1 while capacitance C2 can withstand maximum charge of Q2=C2V2
When C1 and C2 are connected in series the charge on each capacitor is equal and is equal to the maximum charge the combination can withstand, Qmax=C1V1 ∵C1V1<C2V2 ... for given values
The resultant capacitance of the combination is CR=C2C1C1+C2
Qmax=6×10−3 C and CR=23×10−6 F
Maximum voltage that the combination of capacitance can withstand is Vmax=QmaxCR
Vmax=9 kV