🚀 go-pugleaf

In article <2992@phoenix.Princeton.EDU> tycchow@phoenix.Princeton.EDU (Timothy Yi-chung Chow) writes:
}The following problem appears in many books on elementary physics,
}elementary electromagnetism, and elementary circuit theory:
}
}Two capacitors, with capacitances C1 and C2 respectively, carry
}charges Q1 and Q2 respectively.  Find the total energy stored.
}Now connect the positive plates and connect the negative plates.
}Show that the total energy stored decreases.  Where has the
}energy gone? 
}
}The energy calculations are trivial.  What intrigues me is the
}question of the lost energy.  The answer usually given is that
}a current flows when the capacitors are connected, and this
}causes energy to be dissipated as heat.  However, suppose we
}idealize the problem and assume zero resistance.  Surely energy
}is still lost somehow.  How do we explain this?
}
}My own tentative explanations include:
}
}1. It is meaningless to "idealize" the problem by assuming zero
}   resistance, because then an infinite current would flow.
}
}2. Charge would move from the plates of the capacitor that was
}   initially at the higher potential to the other capacitor, but
}   would not stop flowing the instant that the voltages equalized.
}   Instead, it would "overshoot" and charge would keep flowing,
}   although it would be slowed down by the opposing voltage.
}   Eventually the current would stop, and then it would reverse.
}   We would get an alternating current.  Eventually the energy
}   would get dissipated via electromagnetic radiation emitted by
}   the accelerating charges.  (The analogy of a pendulum slowed
}   down by friction seems appropriate.)
}
The energy is generally lost in the form of a brilliant flash and a loud bang
that scares hell out of you (if the capacitors are fairly large).  The heat
generated is frequently sufficient to weld the two wires together.

If you mange to short the leads quickly enough, current will oscillate back
and forth between the capacitors until the damping from the residual
resistance dissipates the required energy. (see 2, above)

-- 

Norm   (strong@tc.fluke.com)