Halved energy on capacitors

  • #1
Felipe Lincoln
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Homework Statement


I have initially a capacitor with capacitance ##C_1## with charge ##Q_1>0## and energy ##U_1>0##. I put another capacitor ##C_2## with no charge ##C_2=0## in parallel with the first. We assume this is an ideal circuit. The charge will distribute to both capacitors and the total charge will be equal the initial ##\sum Q=Q_1##.

a) Why does the total energy at the final instant is less than the initial?
b) does it have some relation with the potential energy associated with the charge position?
c) If I remove these capacitor from the circuit and associate each of them separately with another capacitors ##C_3## and ##C_4## in parallel, and keep doing this forever, will I reach energy close to zero with the same initial charge ##C_1## ?

Homework Equations


Capacitor's charge: $$Q=CV$$
Capacitor's energy: $$\dfrac{1}{2}CV^2$$

The Attempt at a Solution


We always hear that the energy is taken away by the heating and electromagnetic radiation, I'm not satisfied with that once we did all the math considering them to be ideal.
 

Answers and Replies

  • #2
gneill
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We always hear that the energy is taken away by the heating and electromagnetic radiation, I'm not satisfied with that once we did all the math considering them to be ideal.
But you didn't do the math involving the electromagnetic radiation. Declaring that there isn't any does not make it true, it only defines the limits of the physical model you are using to analyze the given system.

An analogous situation exists for gravitational potential energy. Say that you start with a cylinder of water of mass m and height h, and we assume the usual simplification of a constant uniform gravitational field. We can say that the gravitational PE is given by ##PE_o = m g h/2##. Now we move half the water to another cylinder of the same radius, so there are now two equal columns of water of height h/2. The water can be moved by any method. The total gravitational PE is now ##PE_1 = m g h/4##. Half the energy is "gone" from the system you're analyzing. It must be that the assumption that the system is closed and isolated is not strictly true; There is actually an "external" interaction with the gravitational field of the Earth.
 
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  • #3
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According to the source I looked at, if there were no resistance then the circuit (as in a superconductiong circuit) would oscillate forever, or until radiation carried off the energy. See Lincoln's post. Since there is resistance, that can carry away the "missing" energy. Here is a link: http://www.hep.princeton.edu/~mcdonald/examples/twocaps.pdf

In the reference, they derive the lost due to the resistance and it is equal to the missing energy. It is not necessary to consider electromagnetic radiation to explain the paradox, in my opinion.

You can find discussions by Googling "capacitor paradox."
 
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  • #4
Merlin3189
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The Attempt at a Solution


We always hear that the energy is taken away by the heating and electromagnetic radiation, I'm not satisfied with that once we did all the math considering them to be ideal.
Yes, I know the feeling. I was the same the first time I came across this.
But what is all this math you did? What did you find? Why doesn't it satisfy you?

Do you mean by "ideal", that there is zero resistance in the circuit? If so, did you have zero inductance as well?
If you have inductance, however small, and zero resistance, then the circuit just oscillates unless and until it loses energy somehow.

If you want to push it and have zero inductance, no resistance and no esr in the capacitor, then you're stuck with an infinite current, infinite rate of change of current and I don't know how you handle that calculation. And how do electrons get from one side of the capacitor to the other in zero time?
I expect someone can do it with relativity or QM, since the electrons are probably on both sides and everywhere else already?
But then I start to feel more comfortable with a bit of em radiating and a bit of resistive loss. So what's your take on it?
 
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  • #5
rude man
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An analogous situation exists for gravitational potential energy. Say that you start with a cylinder of water of mass m and height h, and we assume the usual simplification of a constant uniform gravitational field. We can say that the gravitational PE is given by ##PE_o = m g h/2##. Now we move half the water to another cylinder of the same radius, so there are now two equal columns of water of height h/2. The water can be moved by any method. The total gravitational PE is now ##PE_1 = m g h/4##. Half the energy is "gone" from the system you're analyzing. It must be that the assumption that the system is closed and isolated is not strictly true; There is actually an "external" interaction with the gravitational field of the Earth.
Got a tough time with this one.
Original p.e. = mgh/2
Final p.e. is mgh/4 x 2 =mgh/2.
No energy lost.
 
  • #6
gneill
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Got a tough time with this one.
Original p.e. = mgh/2
Final p.e. is mgh/4 x 2 =mgh/2.
No energy lost.
Don't forget that the mass is halved for each cylinder.
 
  • #7
rude man
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Don't forget that the mass is halved for each cylinder.
Right.
The lost p.e. goes into work done on the agent making the transfer. So that's what you had in mind? But that agent may have dissipated the lost p.e. to heat or kinetic energy; do you consider that an "external interaction with the gravitational field"?
 
  • #8
gneill
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Right.
The lost p.e. goes into work done on the agent making the transfer. So that's what you had in mind? But that agent may have dissipated the lost p.e. to heat or kinetic energy; do you consider that an "external interaction with the gravitational field"?
Well, regardless of the mechanism moving the water, the PE changes by the same amount. Change in PE is strictly a function of change of position in the gravitational field.
 
  • #9
rude man
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I think your analogy is a good one but I would emphasize the lost enery going to another form of energy which actually is what happens in the discharging r-c network: changing electric potential energy to heat.
 
  • #10
gneill
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I think your analogy is a good one but I would emphasize the lost enery going to another form of energy which actually is what happens in the discharging r-c network: changing electric potential energy to heat.
Yes, and if R → 0, there's still electromagnetic emission to carry away the energy.
 
  • #11
rude man
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Yes, and if R → 0, there's still electromagnetic emission to carry away the energy.
There would still be inductance in the circuit so we'd get oscillations. Eventually radiation would have to account for a decrease in current; that makes sense. Would be interesting to compute the current rolloff; would I'm sure depend on the resonant frequency.
 
  • #12
Merlin3189
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Isn't it odd, that in mechanics we use energy considerations as a short cut to avoid going into the detail of forces and movement? In electricity some people always want to get into complicated calculations to try to track the exact details of the route, even when we all know where we end up.
 
  • #13
Felipe Lincoln
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Oh okay, I got your point @gneill, great example tho.
I read the article you sent @Gene Naden. I'll come back to it when I studied oscillations on circuit.
Yeah @Merlin3189 I'm getting convinced that the current damps until it lost some amount of energy, I'll think of it

Okay, the capacitor seems a little less frightful to me. I will find some material to study the oscillations on circuits, it seems that this is the trick.
 
  • #14
Felipe Lincoln
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I did a research on google and tried to understand what actually is this oscillations that occurs on these circuit with two capacitors in parallel but didn't got anywhere. How is that?
 
  • #16
Felipe Lincoln
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Oh I didn't got into LC yet, ok them, when I get there I'll see. Thanks @Gene Naden !
 
  • #17
Merlin3189
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Oscillation occurs only if there is inductance, L. But there always is inductance.
Your 2 C's in series can be considered as a single C. So you are looking at an LC circuit.
In fact, since there is always some R as well, an LCR circuit. But LC will tell you what you want.
 
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