Question

find charge and voltage along each capacitor and total capacitance

Answer 1

ANSWER

Before disconnection , as the both capacitor are in parallel so potential across them will be same i.e 660V.

Now using Q=CV, the charge on 4μF is Q1=(4×10−6)×660=2.64×10−3C and the charge on 6μF is Q2=(6×10−6)×660=3.96×10−3C

When disconnected and again connected each other with terminals of unlike sign together, the total charge is  Qt=Q2+(−Q1)=(3.96−2.64)10−3=1.32×10−3C and total capacitance Ct=C1+C2=4+6=10μF

After re-connection, they are in parallel so the potential across will be same. 

Thus, common potential is Vc=CtQt=10×10−61.32×10−3=132V

Now charges become , Q1′=(4×10−6)×132=5.