Two capacitors, C1 = 27.0 µF and C2 = 35.0 µF, are connected in series, and a 21.0 V battery is connected across them.

Question

(a) Find the equivalent capacitance, and the energy contained in this equivalent capacitor.
equivalent capacitance=

µF
total energy stored=

(b) Find the energy stored in each individual capacitor.
energy stored in C1 J
energy stored in C2

Answer 1

The charge is the same on both
Q = C V
so C1V1 = C2 V2
V2 = (C1/C2)V1 and V1 + V2 = 21
let C1 = 27*10^-6
let C2 = 35*10^-6
then C1/C2 = .77
then V2 = .77 V1
1.77 V1 = 21
V1 = 11.9 volts
V2 = 9.1 volts

I think you can take it from there