The charge on C1 and C2 must be equal by conservation of charge because the node between them is isolated. The voltage of C1 and C2 must sum to 6V. Use q=CV and solve for the voltages.
Reworked by RM:
The same current flows in C1 & C2.
As charge is, by definition, proportional to current (Q = I x t) then
the charge on C1 and C2 must be equal.
But, also by definition Charge = capacitance x Voltage (Q = C x V).
Or, rearranging, V = Q/C.
So, for equal charges in each, capacitor voltage will be inversely proportional to capacitance.
The voltage of C1 and C2 must sum to 6V. Use q=CV and solve for the voltages.
The relationship between charge Q, voltage V and capacitance C is given by the expression
Q = C x V.
Rearranging, V = Q/C.
Charge Q is defined as the summation of current with time ie Q = i x t
As an identical current must flow C1 and C2, they both experience the same current x time profile so their charges are equal.
But from above, Vcap = Q/C.
ie for equal charges, cap voltages will be inversely proportional to capacitor sizes.
So, in this example, the voltages on C1 & C2 will be inversely proportional to capacitor sizes so VC1 = 2 x VC2.
By inspection, VC2 = 2, VC1 = 4.