Where Does the Energy Go in an Ideal Capacitor Circuit?

In summary, the conversation discusses the distribution of charge and energy in a parallel circuit with ideal capacitors. It is noted that the total energy at the final instant is less than the initial, and the potential energy associated with the charge position may play a role. The idea of continuously removing and replacing the capacitors in parallel is also brought up. The conversation then delves into the concept of ideal systems and how they may not accurately represent real-world scenarios. The concept of energy loss through resistance and electromagnetic radiation is also mentioned. The conversation ends with the suggestion to research the "capacitor paradox" for further understanding.
  • #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.
 
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  • #2
Felipe Lincoln said:
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
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
Felipe Lincoln said:

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
gneill said:
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
rude man said:
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
gneill said:
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
rude man said:
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
I think your analogy is a good one but I would emphasize the lost energy going to another form of energy which actually is what happens in the discharging r-c network: changing electric potential energy to heat.
 
  • #10
rude man said:
I think your analogy is a good one but I would emphasize the lost energy 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
gneill said:
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
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
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
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
Oh I didn't got into LC yet, ok them, when I get there I'll see. Thanks @Gene Naden !
 
  • #17
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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Related to Where Does the Energy Go in an Ideal Capacitor Circuit?

1. What does it mean to "halve energy on capacitors"?

Halving energy on capacitors refers to reducing the amount of stored energy in a capacitor by half. This can be achieved by either decreasing the capacitance or the voltage of the capacitor.

2. Why would someone want to halve the energy on capacitors?

There are several reasons why someone may want to halve the energy on capacitors. One reason could be to decrease the risk of electrical shock or damage to sensitive electronic components. Another reason could be to reduce the overall energy consumption of a circuit.

3. How is the energy on capacitors calculated?

The energy stored in a capacitor can be calculated using the formula E = 1/2CV^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts. This means that the energy on capacitors is directly proportional to both the capacitance and voltage.

4. Can the energy on capacitors be halved without changing the capacitance or voltage?

No, the energy on capacitors cannot be halved without changing either the capacitance or voltage. In order to reduce the stored energy, one of these factors must be adjusted.

5. Are there any safety precautions to take when halving energy on capacitors?

Yes, it is important to discharge capacitors before attempting to halve their energy. This can be done by shorting the terminals with a resistor or by carefully discharging the capacitor using a specialized tool. It is also important to follow proper safety procedures and use appropriate equipment when working with capacitors.

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