Re: non-dissipative circuitry

["John S. Denker" <jsd@MONMOUTH.COM>]

Chuck Britton wrote:

> ok, I understand your mechanical non-dissipative examples BUT:
>
> Back to the twin capacitor test.
>
> Conservation of charge seems to ME to be powerful enough to define
> the final state of the system.
>
> Final state requires that the two voltage differences are equal AND
> the total amount of separated charge remains unchanged.
>
> Where is my thinking going astray????

I believe the crucial nasty trap here is to
assume that when a "charge" Q appears on the
second capacitor the same "charge" Q had to
come out of the first capacitor.

Conservation of charge is a wonderful thing,
but it doesn't apply here.

Alas, the term "charge" is ambiguous.  When we
speak of "charging" a battery or capacitor,
it does not involve the sort of charge that
is conserved.  Microscopically speaking, the
capacitor has positive elementary charges on
one plate and negative elementary charges on
the other plate.  The overall object is
electrically neutral.  It doesn't have any
net elementary charge at all;  the property
that it does have should perhaps be called
something else, perhaps _bicharge_ or something
like that.  We put a bicharge on a battery or
a capacitor.  When a battery runs down, we
might say it is disbicharged.  There is no
law of conservation of bicharge.


Here's how to apply this idea:  The initial setup
is:

          _________  |  ________
         |           |          |
         |           |          |
         |           |          |
         |           $          |
       ===== C1      $ L       === C2
         |           $          |
         |           |          |
         |___________|__________|


Throw the switch as shown below, hooking the
inductor across the first capacitor.  This is
non-dissipative.  The current in the inductor
starts building up.  This results in a magnetic
energy E1 stored in the inductor.

          __________    ________
         |          \           |
         |           |          |
         |           |          |
         |           $          |
       ===== C1      $ L       === C2
         |           $          |
         |           |          |
         |___________|__________|


At time T quickly throw the switch, simultaneously
disconnecting the inductor from the first
capacitor and connecting it to the second:

          _________    _________
         |            /         |
         |           |          |
         |           |          |
         |           $          |
       ===== C1      $ L       === C2
         |           $          |
         |           |          |
         |___________|__________|


Due to the inductance, current continues to
flow in the inductor.  After a quarter cycle
of the LC resonance, the current will go to
zero, whereupon there is no
energy stored in the inductor.  100% of
the energy E1 has been transferred to the
second capacitor.  Disconnect the inductor
from both capacitors at this point, as shown
in the first diagram.  The bicharge gained by
the second capacitor will in general not be
the same as the bicharge lost by the first
capacitor.  The current flowing in the inductor
at time T minus epsilon equals the current
flowing in the inductor at time T plus epsilon,
but this current flows for an amount of time
that depends on the size of C2.

Bicharge is not conserved!