Re: Capacitor charge, were is it?

To: tesla-at-pupman-dot-com
Subject: Re: Capacitor charge, were is it?
From: "Edward V. Phillips" <ed-at-alumni.caltech.edu>
Date: Sat, 26 Oct 1996 10:04:22 -0700 (PDT)

"There is a classic problem in which two identical capacitors are connected
with a switch. Before the switch is thrown one capacitor has a certain
voltage, the other none. If we give numbers, let's say C=1uF and V=1000V on
the first capacitor. The energy is (cv^2)/2 = 0.5 joules. Now after the
switch is closed both capacitors will have v/2 = 500V across them. This can
actually be done. But now the energy in the system is half the original
value.
(1uf*500^2)/2 = 0.125 joules/cap times two caps = 0.250 joules. Were is the
other half?"
	The rest of the energy was dissipated in resistive losses while
the current flowed.  If there were no resistance in the circuit there
would be an undamped train of oscillations due to the inductance of
the interconnecting leads.  No laws of nature violated.
"The classic things I've been taught have me confused with the actual
workings of nature. If two charged plates (air dielectric) have a certain
charge Q and then a dielectric is inserted between them with a K>1, the
voltage should decrease since the charge hasn't changed but the capacitance
has increased."
	An even simpler experiment is to consider a capacitor (say
two spheres) which is charged to some voltage.  If you separate
the plates (spheres, etc) the voltage difference between them rises,
while if you bring them closer together the voltage difference is
reduced.  Conservation of charge holds.  People have built high-
voltage electrostatic generators in which a rotating variable
capacitor was charged in the high-capacitance state, and then
discharged as the capacitance was reduced and the voltage raised.
Ed Phillips