Answers
Capacitors C1 and C2 are in parallel. Hence their net capacitance C12 = (0.40 + 0.40) 
F = 0.80 F
Similarly, Net capacitance of C3 and C4 in parallel, C34 = (0.20 + 0.60) F = 0.80 F
Thus we have two equal capacitances C12 and C34 connected in series across 12 V supply. Hence potential difference acros each capacitance C12 and C34 is 6 V. As C1 and C2 are in parallei and C3 and C4 in parallel, therefore
P.D. across C1 = P.D.
Across C2 = P.D.across C3 = P.D. across C4 = 6 V
Charge on C1 = Charge on C2 = (0.4 F) x 6 V = 2.4 C
Charge on C3 = (0.20 F) x 6 V = 1.2 C
Charge on C4 = (0.60 F) x 6 V = 3.6 C
Energy stored in C1 = Energy stored in C2 = (CV2) / 2 = (0.40 F) x 36 V 2 / 2 = 7.2 J
Energy stored in C3 = (CV2) / 2 = (0.20 F) x 36 V 2 / 2 = 3.6 J
Energy stored in C4= (CV2) / 2 = (0.60 F) x 36 V 2 / 2 = 10.8 J
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