A  3 μF capacitor is charged to a potential of 300 V and a 2 μF capacitor is charged to 200 V. The capacitors are then connected in parallel with plates of opposite polarity joined together. What amount of charge will flow through the connecting wire when the plates are so connected?

Charge on the first capacitor  Q1=C1V1=3×10−6×300=900×10−6 C Charge on the second one  Q2=C2V2=2×10−6×200=400×10−6 C When plates with opposite polarity are joint, charge will be  900−400=500μCThe capacity will be  C=C1+C2=5 μFCommon potential  =500μC5=100 V Final charge on first capacitor  =300μCCharge flow  =900μC−300μC=600μC