Where is the lost energy in this example?

[abdossamad2003]

We connect the charged capacitor to the no-charged capacitor (consider the wires to be ideal R=0), the final energy is less than the initial energy of the system. Where is the lost energy? (see example blew)

 

[phinds]

Google is your friend
electrons have to jump across the gap between the ends of the connecting wires. When the electrons jump, they radiate electromagnetic waves (hence lose some of the electrical potential energy

 

[DaveE]

This is a great example of why you can't always used a lumped element circuit to represent real world situations. Or at least that you need a more complex model. The math is right (I think, didn't check it carefully).

The problem here is that the charges have to move from on cap to the other. Moving charges put energy into a magnetic field. So the circuit with just two caps isn't physically realizable. You need to add a series inductor to be (sort of) realistic. Then your solution be an oscillation that has the energy shifting between the caps and the inductors. Ideally it never stops. In reality there will be energy lost in things like resistance or radiation, so the oscillation will die out and end up as it did in the calculation you showed us.

PS: So, in summary: Conservation of charge works. Conservation of energy doesn't because you aren't looking at the entire system. Energy is only conserved if you count all of it.

 

[Ibix]

When the battery is connected it pushes electrons from one plate to the other. When you disconnect it and connect the other capacitor some of those electrons flow onto the new capacitor and some electrons flow from the other plate of the new capacitor to the positive plate of the first capacitor.

So you have current flowing round a loop. Ideally it just does what I said in the previous paragraph although, as DaveE points out, in reality there will be some sloshing back and forth. Either way, the change in the EM field as it flows is radiation which will carry away the energy.

Note that heuristically it makes sense that the energy in the two capacitors is lower - you've spread the same charge over a larger area so the repelling force between two adjacent charges falls.

 

[gleem]

Here is an analysis of the charging of one capacitor with another.
http://kirkmcd.princeton.edu/examples/twocaps.pdf

 

[berkeman]

Here is an analysis of the charging of one capacitor with another.
http://kirkmcd.princeton.edu/examples/twocaps.pdf

Nice find!

 

[Meir Achuz]

The McDonald derivation shows that the energy is lost in an RLC circuit.
The calculation is much simpler as an RC circuit with L = 0.
Since the end result is always the same, the derivation cannot depend on details of the circuit.
This is seen in the RLC derivation. This means picking L=0 will give the same answer as any other value.
R = 0 will not work because the current will never stop isolating.

 

[SredniVashtar]

I suggest reading "The two-capacitor problem with radiation" by Boykin, Hite, and Singh
Am. J. Phys. 70, 415 (2002)
DOI: 10.1119/1.1435344

 

[phinds]

I suggest reading "The two-capacitor problem with radiation" by Boykin, Hite, and Singh
Am. J. Phys. 70, 415 (2002)
DOI: 10.1119/1.1435344

Since it says it cost $40, I'll pass.

 

[DaveE]

Since it says it cost $40, I'll pass.

It can be found for free, but a bit of immorality is required. There is a copyright, I'm sure.

 

[gleem]

I couldn't let this go without saying the following. What I think needs calling attention to is when the capacitors are connected there is a charge redistribution. The charges effectively move farther apart thereby decreasing the potential energy of the distribution. In this case, how does the physics used to determine the energy of the final state "know" the actual way the energy must be lost unless it is intrinsically related to the charge configuration? Remember:
$$ Energy\: of\: a\: charge \: distribution= \frac{1}{2}\int_{A} \sigma Vda $$
For a more realistic situation, we can consider a small resistance connecting the two capacitors. For this case, the energy used can be determined by first solving the differential equation given by
$$ R\frac{dq}{dt}- \frac{Q_{0}-q}{C_{1}} +\frac{q}{C_{2}}=0$$
Where q is the charge that is transferred from C₁ to C₂ and Q₀ is the original charge on C₁.

and then using the following relation to determine the energy used.
$$ Energy(heat) = \int_{0}^{\infty }i(t)v(t)dt $$
Solving these equations the result shows that the energy lost in the resistor is
$$ Energy(heat) = \frac{1}{2} ( \frac{C_{1}C_{2}}{C_{1}+C_{2} } )V_{0}^{2}$$
The energy lost as heat is independent of the resistance and is the same as the energy lost in the OP as demonstrated by the following:

Using the method of the OP to find the energy used we have the energy of the initial and final states are
$$ U_{i} = \frac{1}{2} C_{1}V_{0}^{2} $$
$$ U _{f}=\frac{1}{2}\left ( \frac{C_{1}^{2}}{C_{1}+C_{2}} \right )V_{0}^{2}$$
So the energy lost when they are connected is
$$U_{i}-U_{f}= \frac{1}{2} C_{1}V_{0}^{2}-\frac{1}{2}\left ( \frac{C_{1}^{2}}{C_{1}+C_{2}} \right )V_{0}^{2} =\frac{1}{2}\left ( \frac{C_{1}C_{2}}{C_{1}+C_{2}} \right )V_{0}^{2} $$
When two capacitors are connected more space is available for the charges to expand reducing the space between them (reducing the charge density) and reducing the potential energy of their distribution. In the real situation, the energy of the ohmic heating and radiation reflects the work done from the charge redistribution.