A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V 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 3.0-µF capacitor? *

Explanation:

Given that,

Capacitor 1

Capacitance C1 = 3.0-µF

Voltage V1 = 40V

Capacitor 2

Capacitance C2 = 5.0-µF

Voltage V2 = 18V

The two capacitor are connected such that, the positive place is connected to the negative plate of the other, I.e series connection

Charge on each capacitor before connection

Q1 = C1V1 = 3.0-µF × 40

Q1 = 120 -µC

Q2 = C2V2 = 5.0-µF × 18

Q2 =90 µC

So, when they are connected together in series, the charge with combine together to have a single end equal charge since series connection have the same charge

Therefore,

1/Qeq = 1/Q1 + 1/Q2

1/Qeq = (Q1+Q2)/Q1Q2

Qeq = Q1Q2/(Q1+Q2)

Qeq = 51.43 -µC

So the charge on the will be

Q = Qeq/2

Q = 51.43/2

Q = 25.71

Q≈ 26 -µC

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.