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Q.

A  2μF capacitor  C1 is first charged to a potential difference of 10 V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C2  of  8μF. The charge in  C2 on the equilibrium condition is ______μC . (Round off to the Nearest integer)

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answer is 16.

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Detailed Solution

q=q1 +q2 (Law conservation of charge) 20×106=C1+C2VCommom Potential=V=20×106(2+8)×106=2 volt Thus, q2=VC2=16μC

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