An electricity and magnetism problem by Abhay Tiwari

There are actually TWO answers to this problem, as it is not indicated how the potential differences (given) are measured. To find the initial charge on each capacitor, we multiply it's capacitance by the potential difference. Therefore, $C_1$ holds 5 Coulombs while $C_2$ holds 30 Coulombs.

Case I: IF the left side of each capacitor in the open circuit is positive, then when the switch is closed there will be a net charge of 25 Coulombs ( $-$ 5 + 30) on the plates facing the switch, and -25 Coulombs on the opposite plates. The potential differences across each capacitor now must be equal, but since $C_2$ has three time the capacitance of $C_1$ , it must hold three times the charge. Therefore, it holds $\frac{3}{4}$ of $25$ = $75\div4$

Case 2: IF the left side of the first capacitor is positive, but the left side of the second capacitor is negative, then the plates that face the switch will carry a net charge of -35 Coulombs ( $-$ 5 + $-$ 30) while the opposite plates carry +35 Coulombs. Again, $C_2$ carries three times the charges so it holds $\frac{3}{4}$ of $35$ = $105\div4$