A capacitor is charged. It is then connected to an identical uncharged capacitor using superconducting wires. Each capacitor has 1/2 the charge as the original, so 1/4 the energy - so we only have 1/2 the energy we started with. What happened? my first thoughts were that the difference in energy is due to heat produced in the wire. It may be heat, or it may be that this is needed to keep equilibrium.
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2$\begingroup$ Duplicate of electronics.stackexchange.com/q/29535/4512 $\endgroup$– Olin LathropCommented Jan 16, 2014 at 19:44
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4 Answers
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Short answer: this is a textbook example of the limitations of ideal circuit theory. There seems to be a paradox until the underlying premises are examined closely.
The fact is that, if we assume ideal capacitors and ideal superconductors, i.e., ideal short circuits, there appears to be unexplained missing energy.
What's not being considered is the energy lost to radiation at the moment the two capacitors are connected together.
At the moment the capacitors are connected, in accord with ideal circuit theory, there should be an impulse (infinitely large, infinitely brief) of current that instantaneously changes the voltage on both capacitors.
But this ignores the self-inductance of the circuit and the associated electromagnetic effects. The missing energy is transferred to the electromagnetic field.
From the comments:
This answer is just plain wrong. – Olin Lathrop
and
Agreed with @OlinLathrop. - Lenzuola
If you find yourself in agreement with the comments above, consider the following excerpt from the exercise "A Capacitor Paradox" by Kirk T. McDonald, Joseph Henry Laboratories, Princeton University:
Two capacitors of equal capacitance C are connected in parallel by zero-resistance wires and a switch, as shown in the lefthand figure below.
Initially the switch is open, one capacitor is charged to voltage V0 and the other is uncharged. At time t = 0 the switch is closed. If there were no damping mechanism, the circuit would then oscillate forever, at a frequency dependent on the self inductance L and the capacitance C. However, even in a circuit with zero Ohmic resistance, damping occurs due to the radiation of the oscillating charges, and eventually a static charge distribution results.
And then, in problem 2:
Verify that the “missing” stored energy has been radiated away by the transient current after the switch was closed, supposing that the Ohmic resistance of all circuit components is negligible.
answered Sep 7, 2012 at 12:19
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$\begingroup$ Thank you Farhan for fixing the link to Kirk McDonald's exercise! $\endgroup$ Commented Jan 25, 2023 at 23:58
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1$\begingroup$ There's a particularly nice way to analyze this problem: just introduce a resistance $R$ in series with the capacitors. You will find that for any value of $R$, the resistor dissipates half of the initial electrostatic energy, which demonstrates that the problem is pathological without the resistor. $\endgroup$ Commented Feb 4 at 16:56
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$\begingroup$ @DanielSank, this is true within the context of ideal circuit theory, and I'm almost certain I've written an answer or three to this effect. But this problem is also nice in that it illustrates the limitations of ideal circuit theory, i.e., the approximation holds as long as the resistance R dominates the loss mechanisms. $\endgroup$ Commented Feb 5 at 3:35
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You're sloshing charge around!
You've set up an LC circuit (if there's a current there's an L - if for no other reason the electron's mass), so when the capacitors are equally charged, the current is at it's maximum. Currents have an energy associated with them! If you work it out with the current term included, you'll see that the current term accounts for your missing energy
I don't think that the ideal circuit theory is seriously at fault here. Even a conceptual, ideal, conductor must have an inductance associated with the charge carrier's mass! The only thing ideal circuit theory doesn't capture well here are the radiative losses.
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Indeed, there will be at least some losses in the superconducting wires: first, as far as I know, losses in superconductors only vanish for zero frequency, second, initial high current can exceed the critical current of the superconductor.
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This is a Gedankenexperiment. If such an experiment results in a paradox, the experiment is set up the wrong way. And the answer is given. There is no electrical connection that doesn't show an inductance. So we should construct the setup as simple as possible. There are two sets of metal plates of no length in parallel at a distance forming two capacitors. There are two conductors of diameter zero connecting two plates of two capacitors respectivly. No more requisites are needed as those ideal capacitors don't show inductance and those two wires show inductance but no capacitance. Now we formulate the boundary condition that current is zero, one capacitors voltage is zero, one is non zero and we can simply show that this is an LC-oscillator so we will see sinusoidal convertion of electrostatic field energy to magnetic field energy and vice versa.
We do not introduce a super conductor nor electromagnetic field nor radiation or any kind of object that creates losses. All of this is not part of said Gedankenexperiment.
A problem arises when you introduce a switch to have this boundary condition "realized", to be able to charge one capacitor.
A switch can only be closed by bringing together two connections which can only meet when there is an area. So if you bring the switches contact in proximity they form a capacitor and a current will start to flow. As the capacitance of the contacts at a initial distance can not be zero and as the distance must reach zero to close the contact, the capacity of this capacitor reached infinity and all the energy stored in this capacitor will be dissipated. as this charged capacitor stores energy and a short circuit will not be consistant with this condition.
