Two capacitors of capacitance 2 μF and 6 μF are charged with batteries of 100 V and 300 V seperately. Now the capacitors are connected such that the like charged plates are joined together. The total charge on the system is:

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16×10−4C

18×10−4C

20×10−4C

24×10−4C

answer is C.

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Detailed Solution

Total charge Q=Q1+Q2=C1V1+C2V2=2×10−6×100+6×10−6×300=200×10−6+1800×10−6=2000×10−6C=20×10−4C.

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