Potential Energy in a capacitor

  • #1
270
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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?
 

Answers and Replies

  • #2
2,268
7
well i dont 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.
 
  • #3
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ah I think this will help me a lot I'll try tomorrow i'm way to tired right now thanks a lot.
 
  • #4
4,254
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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?

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:
  • #5
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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?
 
  • #6
270
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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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