is there any violation of law of conservation of energy? any emission of e m waves?does the energy appear as magnetic field?
2006-11-04 18:49:53 · 3 answers · asked by RRB 1 in Science & Mathematics ➔ Physics
Assuming the plates are also superconductors and the dielectric between them is vacuum, the only place for the energy to go is by electromagnetic radiation. The connecting superconducting wire will have inductance so you will have a resonant "tank" circuit with a very high "Q" factor. What you get is damped oscillations at the resonant frequency. Depending on dimensions, it could take a while for all the energy to radiate away, but radiate it will.
2006-11-04 19:34:03 · answer #1 · answered by hevans1944 5 · 0⤊ 0⤋
Same thing happens as when you short out a capacitor with any conductive material -- quick surge of current through the conductor from the negative plate to the positive plate until the two charges are equalized. There will also be a very brief pulse of electromagnetism as happens with any conductor having current flow through it.
2006-11-04 19:18:25 · answer #2 · answered by Chug-a-Lug 7 · 0⤊ 0⤋
Capacitor will rapidly discharge.
2006-11-04 18:55:34 · answer #3 · answered by Spacekid 2 · 0⤊ 0⤋