Potential Energy in a capacitor

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Discussion Overview

The discussion revolves around the potential energy stored in a parallel-plate vacuum capacitor, particularly focusing on how this energy changes when the separation between the plates is altered after the capacitor has been disconnected from the potential source. Participants explore the implications of this scenario and the relevant equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to calculate the new energy stored in the capacitor after changing the plate separation, given that it was disconnected from the potential source.
  • Another participant suggests conceptualizing energy as stored in the electric field, noting that energy density is proportional to the square of the field strength.
  • A participant reiterates the initial question about energy change and proposes that the potential energy does not change when disconnected from the voltage source, while also stating that it does change with plate separation.
  • There is a discussion about the capacitance formula, with a participant questioning if it is C = kA/d, and suggesting that charge Q remains constant when the capacitor is disconnected.
  • One participant reflects on their initial misunderstanding of the question and acknowledges that the equation for capacitance should be C = A/d, while also noting the role of the constant k in relation to dielectrics.

Areas of Agreement / Disagreement

Participants express differing views on whether the potential energy changes when the capacitor is disconnected from the voltage source. Some assert that it does not change, while others argue that it does change with the alteration of plate separation. The discussion remains unresolved regarding the correct interpretation of energy changes in this context.

Contextual Notes

Participants do not reach a consensus on the implications of disconnecting the capacitor from the voltage source or the resulting changes in potential energy. There are also uncertainties regarding the area of the capacitor and how it factors into the calculations.

physstudent1
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A parallel-plate vacuum capacitor has 8.28 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.00mm. What is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?

I really don't know what to do I know that U=Q^2 / 2C = .5*C*V^2 = .5*Q*V I read through the entire chapter but I don't see anything about how the potential energy changes when the capacitor is disconnected from the potential source can anyone help?
 
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well i don't know any equations but think of the energy as being stored in the electric field of the capacitor. the energy of each infinitesimal volume of field is proportional to the square of the field strength.
 
ah I think this will help me a lot I'll try tomorrow I'm way to tired right now thanks a lot.
 
physstudent1 said:
A parallel-plate vacuum capacitor has 8.28 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.00mm. What is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed?

I really don't know what to do I know that U=Q^2 / 2C = .5*C*V^2 = .5*Q*V I read through the entire chapter but I don't see anything about how the potential energy changes when the capacitor is disconnected from the potential source can anyone help?

The potential energy doesn't change when the capacitor is disconnected from the voltage source.

The potential energy changes when the separation between the capacitor plates is changed.
Express the capacitance as a function of d, the plate separation.

Is it C = kA/d?

We don't know immediately what's going to happen to the voltage; so use the equation without voltage in it. Use the equation with Q. We do know that the charge, Q on each plate is going to stay constant.

We don't know the area of the capacitor, so lump it into the unknown constant k; C=k/d. We hope the constant will cancel out in the solution (it will).
 
Last edited:
I thought that the potential energy wouldn't change when the capacitor was disconnected from the voltage source but when I put 8.28J for my answer it was wrong?
 
ok I understand it now thanks a lot I was interpreting the question wrong. The equation should be C=A/d but i think it might have worked your way anyway (K is for dielectrics )
 

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