Capacitors disconnected from the battery

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Homework Help Overview

The discussion revolves around the behavior of two capacitors, C1 and C2, after being charged and then disconnected from a battery. The capacitors are connected in a specific configuration, and the participants are exploring the implications for charge and potential difference across each capacitor.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the equation Q=CV to determine charge and potential difference. There are questions about the effects of connecting the positive plate of one capacitor to the negative plate of another and how this affects voltage distribution after disconnection from the battery.

Discussion Status

Some participants have attempted calculations for charge on the capacitors but express confusion regarding the results and the potential difference after the capacitors are connected. There is an acknowledgment that the potential difference changes once the capacitors are connected, indicating an ongoing exploration of the concepts involved.

Contextual Notes

Participants note that the potential difference across each capacitor is initially 9V before they are connected, but this changes after the connection, leading to confusion about the resulting values.

laduch
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Capacitors C1 = 10 µF and C2 = 23 µF are each charged to 9 V, then disconnected from the battery without changing the charge on the capacitor plates.
The two capacitors are then connected in parallel, with the positive plate of C1 connected to the negative plate of C2 and vice versa.
Afterward, what are the charge on and the potential difference across each capacitor?

The question ask for Q1, Q2(charge (µC)), and V1, V2(potential difference (V)).

I used equation Q=CV which I think is the right equation to use... but there are some conceptual problems like what happens when "C1 connected to the negative plate of C2". When battery is disconnected from the circuit, how is the distribution of the voltage related to capacitors?
 
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Yes, Q=CV is the right equation. You can use it to figure out the charges on each capacitor when they are at 9V.

what happens when "C1 connected to the negative plate of C2"

Charge will flow, until the two capacitors are at the same voltage.

When battery is disconnected from the circuit, how is the distribution of the voltage related to capacitors?
At that point, each capacitor has a 9V potential difference.
 
I applied the equation...
Q1 = C1*V1 = 23 µF * 9 V = 207 µC
Q2 = C2*V2 = 10 µF * 9 V = 90 µC
The results are incorrect.
potential difference on each Capacitor is not 9V.
Kind confusing... X(
 
laduch said:
I applied the equation...
Q1 = C1*V1 = 23 µF * 9 V = 207 µC
Q2 = C2*V2 = 10 µF * 9 V = 90 µC

Okay, good.
So that means C1 has a charge of ______ on the positive plate, and ______ on the negative plate?
Similary, C2 has ______ on the + plate and _____ on the - plate?

Next, imagine what happens if the + plate of C1 is connected to the - plate of C2 ... i.e. what will be the net charge on those 2 plates?

The results are incorrect.
potential difference on each Capacitor is not 9V.
Kind confusing... X(

It is 9V before the capacitors are connected together. It is something different after they are connected.
 

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