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.
Answer
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.
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