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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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Duplicate of – Olin Lathrop Jan 16 '14 at 19:44

4 Answers 4

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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


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.

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No it's not, since this magnetic field is not maintained. This answer is just plain wrong. – Olin Lathrop Jan 16 '14 at 19:43
@OlinLathrop, the answer is just plain right. See, for example:‎. The missing energy is radiated away, i.e., transferred to the electromagnetic field. – Alfred Centauri Jan 16 '14 at 22:40
A little energy will be radiated away, but that will be tiny in the scheme of things. If you had all truly ideal component, then the inductance of the wire would form a tank circuit with the capacitors and the system would ring. A tiny fraction of the energy would get radiated each cycle, so the ringing would eventually die down, but that's not what you said, and this is certainly not where the energy would go in a real system since there would be real resistance that would dissipate it. – Olin Lathrop Jan 16 '14 at 23:22
@OlinLathrop, are you deliberately being obtuse? Did I not write "if we assume ideal capacitors and ideal short circuits" so isn't that the context of my answer and, further, the "missing energy paradox"? There's nothing mysterious about ohmic dissipation of energy and, thus, no 'paradox'. Did you not read the following from the link: "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." – Alfred Centauri Jan 16 '14 at 23:30
Agreed with @OlinLathrop. Circuit theory allows us to use "ideal" elements only as far as it doesn't fall into a paradox (two different current sources in series, two different potential sources in parallel). Then an "ideal short circuit" must be be recognized as at best ideal inductor: if for no other reason charge carriers have mass. Sure there are radiative losses, but the OP doesn't seem concerned that eventually the energy will dissipate in a real system. Instead the energy in the magnetic field from the moving charge becomes static charge in the capacitors. – Lenzuola Aug 27 at 14:58

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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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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total charge stored in capacitor is given by, Q=CV then total energy suplied by batery to the capacitor is given by, U=QV Then u =cv.v energy stored in capacitor= 1/2cv.v then energy loss in capacitor=cv.v-1/2cv.v = 1/2cv.v

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protected by Qmechanic Jan 2 '14 at 8:16

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