A capacitor,C₁ = 5.0 $\mu$F,is charged up to 8 V.It is then connected to a second uncharged capacitor C₂ = 2.5 $\mu$F.The charge on C₁ after the system has come to equilibrium is A)26.7 $\mu$C B)13.3 $\mu$C C)20.0 $\mu$C D)40.0 $\mu$C E)6.67 $\mu$C

Question

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Explanation:The total charge in the system is conserved. Therefore, the charge on C1 after the system has come to equilibrium is:Q = C1V1 = (5.0 µF)(8 V) = 40 µCThe total capacitance of the system is:C = C1 + C2 = (5.0 µF) + (2.5 µF) = 7.5 µFTherefore, the voltage across the system is:V = Q/C = (40 µC)/(7.5 µF) = 5.33 VTherefore, the charge on C1 after the system has come to equilibrium is:Q1 = C1V = (5.0 µF)(5.33 V) = 26.7 µC

A Capacitor,C1 = 5 μ\muμ F,is Charged Up to 8 V μ\muμ

A Capacitor,C₁ = 5 $\mu$ F,is Charged Up to 8 V $\mu$