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RE: Capacitor problem



Sun, 30 Mar 1997 09:55:51 Donald E. Simanek asks:

... Why does the result (half the energy goes elsewhere) *not* depend on
these particular details? If the only loss were radiative, you'd get 1/2
for the energy loss. If the only loss were resistive, you'd get 1/2. If
it is a combination of both, you'd get 1/2. I chose identical capacitors,
which is one reason you get 1/2, but that was only for convenience ...

Don, your parallel capacitors problem is great. And I agree with most of
your philosophical observations. But what you are asking above is trivial.
Suppose I left home with 20 dollars and came back with only 10. I can say
10 is missing, no matter what happened to it. We know how to calculate
initial and final energies and we find what is missing by arithmetic. You
must have something important in mind but I can not see it.
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Did I change the problem? Not at all. My "wires" have no resistance; they
were conceptually eliminated because plates contact each other directly.
That was an acceptable assumtion in your original problem. But how can
the resistance of a very thin metallic strips ("monoatomic" layers where
electric charges reside on each plate due to mutual attraction) be zero?
Accepting this is like trying to apply F=m*a to Brian's "immovable objects"

Joking about gravitons and neutrinos we identified two possible mechanisms
for loosing energy. How much is contributed by each mechanism does depend
on details, such as R of the layers, which you would prefer to eliminate.
We already know that one half of the initial energy is lost (arithmetic)
and we want to identify the mechanism which is likely to be dominant in
the real world. This is impossible without essential (diabolic?) details.
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: Gedanken-ing is not enough; physics is an experimental science! :
: Inspired by thinking about phys-L messages on capacitors :
: Ludwik Kowalski :
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kowalskiL@alpha.montclair.edu http://www.csam.montclair.edu/~kowalski
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