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# A 4.00μF capacitor and a 6.00μF capacitor are connected in parallel across a 660V supply line. The charged capacitors are disconnected from the line and from each other, and then reconnected to each other with terminals of unlike sign together. Find the final charge on each and the voltage across each capacitors.

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#### 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=3.96×10−3CWhen 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=10μFAfter re-connection, they are in parallel so the potential across will be same. Thus, common potential is Vc=QtCt=1.32×10−310×10−6=132VNow charges become , Q′1=(4×10−6)×132=5.28×10−4C and Q′2=(6×10−6)×132=7.92×10−4C

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