Capacitor A is charged to a potential of 100 V and, capacitor B is charged to a potential of 75 V. What are the charges on A and B after key is closed Fig.

qA=5×100=500μCqB=10×75=750μCNet charge =750μC−500μC=250μC(∵Negative plate of A is connected to the positive plate of B)Common potential, V=2505+10=25015voltNow, qA′=C1V=5×25015=2503μCand  qB′=C2V=10×25015=5003μC