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 = ?
Answer:
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