Re: There's work, and then there's work

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

David Rutherford wrote:

> ... how do the charges continue on to the other plate of the capacitor,
> in order to equalize the potentials of the plates, if they've expended
> all of their kinetic energy in the resistor?

This question reveals a number of elementary misconceptions.  For one
thing, as I explained with a detailed numerical example in a previous
posting, the charges never have *any* kinetic energy that is
attributable to the flow of electrical current for a practical
standpoint.

For another, the charges do not "continue on to the other plate of
the capacitor."  The sea of conduction electrons merely shifts en
masse toward the initially positive plate by a distance that might
typically measure no more than a few thousandths of a millimeter.
That's all it takes to completely eliminate the polarization charge
on the capacitor.

> If what you say is true, you could install a galvanometer downstream
> from the resistor and it would show zero current during the
> discharging of the capacitor.

Again, no.  And your use of the word "downstream" hints at yet
another elementary misconception.  A galvanometer placed at any
position in the circuit will measure the same current since the
conduction electrons in its vicinity move by the same amount as they
do *everywhere* along the circuit.

> I assume that [zero current "downstream" from the resistor] is not
> what's observed.

*Here* you are right!  That is *not* what is observed.  If it were
observed, we would have to (and, because physics is first and
foremost an *experimental* science, we *would*) radically alter or
abandon what has otherwise been a fantastically successful model of
current electricity for more than a century.

For more information, please refer to any introductory physics textbook.

--
   A. John Mallinckrodt        http://www.csupomona.edu/~ajm
   Professor of Physics      mailto:ajm@csupomona.edu
   Physics Department         voice:909-869-4054
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