Answers
from the cirucit c1 and c2 are in parallel combination
so c12 = c1+c2 = 11.1*10^-6+9.62*10^-6 = 20.72*10^-6 F
c12 and 0.3*10^-6 F are in series then
c tot = 0.3*10^-6*20.72*10^-6/(0.3+20.72)*10^-6
c tot = 0.296*10^-6 F
total charge q tot = c tot*v = 0.296*10^-6*1.57 = 0.465*10^-6 C
in a series combination charge is same across the capacitors
q12 = q3 = 0.465*10^-6 C
voltage across c12 is v12 = 0.465*10^-6/20.72*10^-6
v12 = 0.0224 V
voltage across c3 is v3 = 0.465/0.3 = 1.55 V
charges on c1 and c2
q1 = 11.1*10^-6*0.0224 = 0.249*10^-6 C
q2 = 9.62*10^-6*0.0224 = 0.215*10^-6 C
b) energy
E1 = q1^2/2C1 = (0.249*10^-6)^2/2*11.1*10^-6 = 2.79*10^-9 J
E2 = q2^2/2c2 = (0.215*10^-6)^2/2*9.62*10^-6 = 2.4*10^-9 J
E3 = (0.465*10^-6)^2/2*0.3*10^-6 = 0.36*10^-6 J
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