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 positiv

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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 positiv

e 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

Physics

2 answers:

Sav [3.1K] · Answer 1 · 2026-03-11T20:48:16+03:00

Answer:

0.192 C

Explanation:

The charge in a capacitor can be expressed using the formula:

Q = CtVt..................... Equation 1

Where Q represents Charge, Ct = Effective Capacitance, and Vt = Effective Voltage.

Note: When the positive plates of two capacitors are joined to the negative plates of each other, this signifies they are in series

The formula for combined capacitance in series is given by:

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

Here, we have: C1 = 4.0 mF, C2 = 6.0 mF

Substituting into equation 2:

Ct = (4×6)/(4+6)

Ct = 24/10

Ct = 2.4 mF.

Further,

Vt = V1 + V2................... Equation 4

Where V1 is the voltage of the first capacitor, and V2 is the voltage of the second.

Given: V1 = 50 V, V2 = 30 V

Vt = 50 + 30

Vt = 80 V.

Now substituting the values of Vt and Ct into equation 1 gives:

Q = 80(2.4)

Q = 192 mC

Q = 0.192 C.

As the capacitors are in series, the same charge flows through each.

Therefore, the final charge on the 6.0 mF capacitor is 0.192 C.

Maru [3.3K] · Answer 2 · 2026-03-11T20:46:52+03:00

Answer:

The 6.0 mF capacitor's final charge will be 12 mC.

Explanation:

The charge on the 4 mF capacitor is calculated as:

4 mF × 50 V = 200 mC

The 6 mF capacitor starts with:

6 mF × 30 V = 180 mC

The total charge when the negative terminals are combined is given by:

q = $q_{1} -q_{2}$

q = 200 - 180

q = 20 mC

To determine the final charge on the 6.0 mF capacitor, we need to calculate the combined voltage:

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

20 = 10 V

V = 2 V

Now for the final charge on the 6.0 mF:

q = CV

q = 6.0 mF × 2 V

q = 12 mC

Thus, the final charge on the 6.0 mF capacitor will be 12 mC.