Where Does the Missing Energy Go in a Capacitor Circuit?

[fuzzylogic]

I got this from Brainteaser Physics.

[PLAIN]http://img820.imageshack.us/img820/4614/lostenergy.jpg

 

[phyzguy]

One way to look at this is to ask yourself what will happen if there is no resistance and the circuit is completely lossless. What do you think will happen?

 

[Mapes]

I disagree with the assertion, "When the switches are closed, the charge Q becomes equally divided between the capacitors." I assert that the charge will transfer back and forth like water sloshing from side to side of a U-shaped pipe.

 

[RonL]

Your charge current does not double.

 

[fluidistic]

I disagree with the assertion, "When the switches are closed, the charge Q becomes equally divided between the capacitors." I assert that the charge will transfer back and forth like water sloshing from side to side of a U-shaped pipe.

I'm a bit confused. When the electrons move, they do it with an acceleration, right? Don't they radiate EM waves and therefore the circuit "loses" energy in form of photons?

 

[D H]

When the electrons move, they do it with an acceleration, right? Don't they radiate EM waves and therefore the circuit "loses" energy in form of photons?

Bingo. So even in the case of a lossless circuit the circuit will eventually settle to a state where the conserved charge is evenly divided between the two capacitors.

 

[fluidistic]

Bingo. So even in the case of a lossless circuit the circuit will eventually settle to a state where the conserved charge is evenly divided between the two capacitors.

Ah thank you... I get it now.

 

[D H]

Just to add to my last post: The energy isn't "lost". It just isn't contained in the circuit anymore.

 

[fuzzylogic]

good answer on the radiative loss. but I'm curious what if there's some resistance in the wire, how do you calculate the loss? I guess I'm puzzled by the fact that without considering the mechanism of energy loss, the calculation shows exactly half of the original energy remains.

 

[D H]

Charge is also a conserved quantity. The charge isn't / cannot be stored in the wires. Assuming non-leaky capacitors, there is nowhere for the charge to go other than to the ends of the capacitors. Conservation of energy dictates that the "lost" energy is not really lost. It just isn't in the circuit anymore. The circuit is not an isolated system. The conservation laws often present a very useful shortcut to solving all kinds of problems. Exactly what happens that causes the conserved quantity to remain constant is a bit irrelevant.

Note: Real capacitors do leak charge across the gap, so the ultimate fate of even the single capacitor (open) circuit is to be storing zero energy.

 

[fuzzylogic]

ok. I guess I'm slightly bothered by this thought:
if you have another circuit that is the same except for the fact that there's a resistor connecting the caps, it's kinda surprising that the energy loss from both circuits are still the same. furthermore, in the resistive circuit, the total radiative and thermal energy dissipation is also exactly half of the original energy.

 

[D H]

The circuit presumably is a closed system, so charge is conserved. It is not an isolated system, so there is no reason to expect that energy is conserved. How exactly energy is dissipated is a bit irrelevant. That it is dissipated (somehow) is all one needs to know.

 

[fuzzylogic]

I understand what you've saying, just that I have yet to reconcile with the fact that even when another path of energy loss is introduced, the total energy loss is always half that of the initial. in the first circuit, radiative loss accounts for all of them. in the second, it's split between radiation and heat such that the total sum is half the total energy. until you do the calculation, you might thought more energy is lost when the resistor is added.

 

[phyzguy]

Exactly half of the energy is lost because you've doubled the total capacitance and the total charge is conserved. It's like a block sliding downhill. If it's at rest at the top of the hill and at the bottom of the hill, the energy dissipated will always equal the potential energy lost, regardless of the path that it takes sliding down the hill. There's nothing magic about 1/2 - if you had three capacitors, or different capacitor values, then the energy dissipated would be different.

 

[D H]

Conservation of charge tells you the final configuration of the system. All that adding lossy elements does is to change the way energy is lost and to change the length of time it takes to get to that final configuration. It will not change how much energy is lost.

 

[fluidistic]

I understand what you've saying, just that I have yet to reconcile with the fact that even when another path of energy loss is introduced, the total energy loss is always half that of the initial. in the first circuit, radiative loss accounts for all of them. in the second, it's split between radiation and heat such that the total sum is half the total energy. until you do the calculation, you might thought more energy is lost when the resistor is added.

No. The resistor only makes the final static equilibrium happen faster. The "heat" you talk about is precisely EM waves or photons.

 

[fluidistic]

Charge is also a conserved quantity. The charge isn't / cannot be stored in the wires. Assuming non-leaky capacitors, there is nowhere for the charge to go other than to the ends of the capacitors.

I was about to ask the question: what if there are 1 free electron in each plates at start? Where would they go? Obviously they can't split into 2.
I guess they might stay in the wire, but I'm unsure. Maybe it depends of the 3 dimensional configuration of the setup... what do you think?