Answer
20 micro Coulomb will be charge on C2.
Work Step by Step
See the attached figure.
As per given data, all capacitors have 10 micro farad capacitance and voltage of battery is 10V.
1: Series capacitors have same charge and share voltages. Parallel capacitors have same voltage and share charge. C2 capacitor is in series with C3 and Both are parallel with C4.
Net capacitance will be:
C2 and C3 net Capacitance is = 5 micro farad
C2,C3 and C4 is= 15 micro farad.
The equivalent capacitance of C2,C3,C4 is in series with C5.
Therefore total capacitance of the branch is 15 micro farad in parallel with 10 micro farad, = 6 micro farad.
Therefore total charge on equivalent capacitance of 6 micro farad (due to C2,C3,C4,and C5) supplied by battery of 10V is
Q=CV
=10*6=60 micro Coulomb.
Since, C2 ,C3 and C4 have equivalent capacitance of 15 micro farad and are in series with C5 of 10 micro farad. The charge of 60 micro Coulomb will be on C5 60 and Ceq ( C3,C2,C4) 60 micro Coulomb as both are in series.
Therefore , the charge on C2 and C3 will be half of C4 as C2 ,C3 have Ceq of 5 micro farad while as C4 in parallel with both have 10 micro farad.
Charge on C4 + Ceq(C2,C3)=60 micro Coulomb
C2 =20 micro Coulomb
