Charge and Potential Difference on 2 Capacitors

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Homework Help Overview

The problem involves two capacitors, C_1 and C_2, with given capacitances and initial voltages. After being disconnected from their batteries, the capacitors are connected in such a way that their positive plates are joined and their negative plates are joined. The discussion centers around determining the potential difference across each capacitor and the charge on each after this connection.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants express confusion regarding the problem setup, particularly about the absence of batteries and the implications of connecting the capacitors. Questions arise about the definitions of charge and potential difference, as well as the relationships between capacitance, voltage, and charge. Some participants suggest calculating the charge on each capacitor based on their initial conditions and exploring the conservation of charge.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some guidance has been offered regarding the calculation of charge using the capacitance and voltage, and the concept of conservation of charge is being examined. However, there is no explicit consensus on the approach to take.

Contextual Notes

Participants are grappling with the implications of connecting the capacitors after disconnection from their power sources, and there is uncertainty regarding the relevant equations and definitions. The problem does not provide explicit values for the final potential difference or charge, leaving room for interpretation and exploration.

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Homework Statement


A C_1 = 2.40 \mu(micro) F capacitor is charged to 861 V and a C_2 = 6.40 \mu(micro) F capacitor is charged to 662 V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other.

What will be the potential difference across each? [Hint: charge is conserved.]

What will be the charge on each?

Homework Equations


1/CT = 1/C1+1/C2+...+1/CN
E = CV2 / 2

C = 2E / V2

V = sqrt(2E / C)


The Attempt at a Solution



I am confused as to what the question is saying. There will be no battery and the 2 capacitors are connected? Please help!
 
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What are E and V? And what do you mean on E=CV2/2? 2/2=1. So is this C*V? Or CV2/2 Or CV2/2?

The charge is conserved. What is the total charge on the connected plates?

What is the resultant capacitance?

The plates connected are on the same potential. How is the capacitance, potential difference (voltage) and charge related?


ehild
 
Im not sure if I have the right equations here. I am totally clueless on this problem
 
But you can think and look after the definition of capacitance, don't you? You can get the charge on the capacitor from the voltage and capacitance. First calculate the charge on both capacitors.

ehild
 
The capacitors are going to be connected in parallell, which means that the voltage over 1 will equal the voltage over the other one.

Also, charge will be conserved. So the total charge you put on each capacitor will equal to the total charge that is over BOTH capacitors when they are connected. You will find a simple equation in your book that looks like this:

Q=CV

So calculate the charge on each capacitor using the capacitance and the voltage applied, and add them together. That will be your Q_{final}. Then calculate the new capacitance of the 2 capacitors in parallel and your voltage will then be given by the same equation above :)
 

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