A 3.0 μF capacitor is 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?

A) 11 μC

B) 15 μC

C) 19 μC

D) 26 μC

E) 79 μC

Please help,

Thank you!

Option E is correct, 79 μC

Explanation:

Given that,

A capacitor of capacitance C1

C1 = 3 μF

The capacitor is charge to 40V

V1=40V

Another capacitor of capacitance

C2 = 5 μF

Charge to 18V,

V2 =18V

When the positive terminal of one is connected to the negative terminal of the other, then, this is a series connection.

The capacitor are connected in series.

Series connection have the same charge.

Equivalent capacitance for series connection is give as

Ceq = C1 + C2

Ceq = 3μF + 5μF

Ceq = 8μF

Charge in a capacitor is given as

q = CV

Then, charge on Capacitor 1.

q1 = C1V1

q1 = 3μF × 40V

q1 = 120μC

Charge on capacitor 2

q2 = C2V2

q2 = 5μF × 18V

q2 = 90μC

The total charge in the circuit is

Qeq =q1 + q2

Qeq =90 + 120

Qeq = 210 μC

Then, the combine voltage in the circuit is

Qeq = CeqV

Then, V= Qeq/Ceq

V= 210 μC / 8 μC

V=26.25Volts

Then the charge on the 3μF is

q =CV

q =3 μC × 26.25 V

q = 78.75 μC

q ≈ 79 μC

Final answer is E.

The question is not typed properly! Complete question along with answer and step by step explanation is provided below.

Question:

A 3.0 μF capacitor is 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?  

A) 11 μC  

B) 15 μC  

C) 19 μC  

D) 26 μC  

E) 79 μC  

Given Information:

Capacitor 1 = C₁ = 3.0 μF

Capacitor 2 = C₂ = 5.0 μF

Initial voltage across capacitor 1 = V₁ = 40 V

Initial voltage across capacitor 2 = V₂ = 18 V

Required Information:

Charge on Capacitor 1 = q₁ = ?

The correct option is A

Charge on Capacitor 1 ≈ 11 μC

Explanation:

Since the positive plate of one capacitor is connected to the negative plate of other capacitor then it means they are connected in series.

The change in charge is given by  

Q = q₁ - q₂

Where q = CV

Q = C₁V₁ - C₂V₂

Q = 3*40 - 5*18

Q = 120 - 90  

Q = 30 μC

Since capacitors are connected in series

ΔV₁ = ΔV₂

Since ΔV = q/C

q₁/C₁ = q₂/C₂

q₁/3 = q₂/5

q₁ = 3/5q₂

The total final charge is

Q = q₁' + q₂'

30 = 3/5q₂' + q₂ '

30 = 8/5q₂ '

q₂' = 30*5/8

q₂' = 18.75 μC

Therefore, the final charge on capacitor 1 is

q₁' = 3/5q₂'

q₁' = 3/5(18.75)

q₁' = 11.25 μC

That is close to option (A) 11 μC

q₁' ≈ 11 μC

power out to power in less than 100%