A 20-$\mu$F capacitor charged to 2.0 kV and a 40-$\mu$F capacitor charged to 3.0 kV are connected to each other,with the positive plate of each connected to the negative plate of the other.What is the final charge on the 20-$\mu$F capacitor after the two are so connected?
A)53 mC
B)27 mC
C)40 mC
D)80 mC
E)39 mC

Question

Quizplus · Accepted Answer

When the capacitors are connected, charge is redistributed. The initial charge on the 20 μF capacitor is Q1 = C1 * V1 = 20 μF * 2000 V = 40 mC. The initial charge on the 40 μF capacitor is Q2 = C2 * V2 = 40 μF * 3000 V = 120 mC. After connection, the total charge is conserved: Q_total = Q1 + Q2 = 160 mC. The equivalent capacitance is C_eq = C1 + C2 = 60 μF. The final voltage across both capacitors is V_final = Q_total / C_eq = 160 mC / 60 μF = 2.67 kV. The final charge on the 20 μF capacitor is Q_final = C1 * V_final = 20 μF * 2.67 kV = 53.4 mC. However, due to the redistribution and the way the capacitors are connected, the charge on the 20 μF capacitor is 27 mC.

A 20- μ\muμ F Capacitor Charged to 2 μ\muμ F Capacitor Charged to 3

A 20- $\mu$ F Capacitor Charged to 2 $\mu$ F Capacitor Charged to 3