A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 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 6.0-mF capacitor? Group of answer choices

0.192 C

Explanation:

The charge in a capacitor is given as,

Q = CtVt Equation 1

Where Q = Charge, Ct =  Effective Capacitance of the capacitors, Vt =  Effective Voltage.

Note: If the capacitors are connected to each other with the positive plate of each connected to the negative plate of the other means that the capacitors are connected in series

The combined capacitor in series is given as,

1/Ct = 1/C1 + 1/C2

Where C1 = Capacitance of the first capacitor, C2 = Capacitance of the second capacitor.

Ct = C1C2/(C1+C2) Equation 2

Given: C1 = 4.0 mF, C2 = 6.0 mF

Substitute into equation 2

Ct = (4×6)/(4+6)

Ct = 24/10

Ct = 2.4 mF.

Also,

Vt = V1 + V2 Equation 4

Where V1 = Voltage in the first capacitor, V2 = Voltage in the second capacitor.

Given: V1 = 50 V, V2 = 30 V

Vt = 50+30

Vt = 80 V.

Substitute the value of Vt and Ct into equation 1

Q = 80(2.4)

Q = 192 mC

Q = 0.192 C.

Since both capacitors are in series, The same quantity of charge flows through them.

Hence the final charge on the 6.0 mF capacitor = 0.192 C

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