The equivalent capacitance of the system is calculated as,
1
Ceq
=
1
3
+
1
4+2
1
Ceq
=
1
3
+
1
6
1
Ceq
=2µF The charge taken from source is, qeq=CeqΔV qeq=(2×10−6)(1800) qeq=3600µC The potential drop across capacitor C1 is C1=
qC1
CC1
C1=
3600×10−6
3×10−6
C1=1200V The potential drop across combination of 4 μF and 2 μF capacitors is, ΔV′=1800−1200 ΔV′=600V The charge on 4 μF capacitor is, q2=C2ΔV′ q2=(4×10−6)(600) q2=2.4×10−3C The charge on 2 μF capacitor is, q3=C3ΔV′ q3=(2×10−6)(600) q3=1.2×10−3C