How Does Charge Redistribute When Capacitors Are Connected in Series?

[xxkylexx]

1. Capacitors C1 = 4.0 µF and C2 = 2.0 µF are charged as a series combination across a 95 V battery. The two capacitors are disconnected from the battery and from each other. They are then connected positive plate to positive plate and negative plate to negative plate. Calculate the resulting charge on each capacitor. ___ µC (C1)
___ µC (C2)
2. Q = CV , 1/Ceq = 1/C1 + 1/C2
3. First, I calculated the charge on each capacitor while they were in series connected to the battery-- Which is 380 and 190, respectively. I then calculated the equivalent capacitance to be 4/3, and think that Q1 + Q2 (the charge after they are hooked together) should equal 380, but I can't ever seem to get a correct answer. I know how to solve this if they are in parallel, but this series configuration is screwing me up.Any help is appreciated.Thanks so much,
Kyle

 

[xxkylexx]

Anyone able to assist with this one?

 

[m2003]

Hello Kyle,

I can understand your confusion with the series configuration of capacitors. Let me try to explain it in a simpler way.

When capacitors are connected in series, the equivalent capacitance is given by the formula Ceq = 1/C1 + 1/C2. In this case, the equivalent capacitance would be 1/Ceq = 1/4 + 1/2 = 3/4. This means that the equivalent capacitance is 4/3 µF.

Now, let's calculate the charge on each capacitor when they are connected in series to the battery. Using the formula Q = CV, we get Q1 = 4 µF * 95 V = 380 µC and Q2 = 2 µF * 95 V = 190 µC.

After disconnecting the capacitors from the battery and connecting them positive plate to positive plate and negative plate to negative plate, the equivalent capacitance remains the same, but the charge on each capacitor changes. This is because the charge is redistributed between the two capacitors.

To calculate the new charge on each capacitor, we can use the formula Q = CV. Since the equivalent capacitance is 4/3 µF, the charge on each capacitor would be Q = (4/3 µF) * 95 V = 380/3 µC = 126.67 µC.

Therefore, the resulting charge on each capacitor would be 126.67 µC for C1 and C2.

I hope this explanation helps you understand the concept of connecting capacitors in series. Let me know if you have any further questions.

Best,