2 capacities of 2.0 mf and are in paralllel and the group is connected in series w/ a 3.0mf capacitor across a

Question

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600 volts line. find the change and the voltage of each capacitor

Answer 1

Q3, V3 pertain to 3 mf cap, Q2(1), Q2(2), V2(1), V2(2) pertain to the 2 mf caps.
Total capacitance Ct = 1/(1/3 + 1/(2+2)) = 1.71438 mf
Q = VC
Total charge Qt = 600 * Ct = 1028.57 mC
Since I, and therefore dQ/dt, is equal in all series elements of a series circuit,
Q3 = Qt = 1028.57 mC
Q2(1) + Q2(2) = Qt
Q2(1) = Q2(2) = Qt/2 = 514.29 mC
V = Q/C
V3 = Qt / 3 = 342.757 v
V2(1) = V2(2) = Qt / (2+2) = 257.143 v
Check: V3 + V2(1) = 600 v