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) ⇒20×10−6=C1+C2V⇒Commom Potential=V=20×10−6(2+8)×10−6=2 volt Thus, q2=VC2=16μC