Question

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

Answer 1

0.192 C

Step-by-step 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

Answer 2

The final charge on the 6.0 mF capacitor would be 12 mC

Step-by-step explanation:

The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC

The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC

Since the negative ends are joined together the total charge on both capacity would be;

q =
`q_(1) -q_(2)`

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q = 12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC