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Re: Toward the equilibrium



Ohanian (AJP, Nov 1983, pg 1020) treats the approach to equilibrium as a
three-step process:
1) the "expulsion" of unbalanced free charge from the interior volume of
the conductor,
2) the "expulsion" of fields and currents from the interior,
3) the damping out of oscillating surface currents and wave fields.

1) follows a relaxation equation with time constants like10^-14 sec (for
Cu when finite mean free paths are taken into account )
2) follows a diffusion equation, with damping times of the order of 10^-4
sec for Cu.
3) is geometry dependent.

Regarding 3) Ohanian remarks (referring to a specific "beer-can"
geometry ) :

" . . . the relaxation time will be shorter than the light-travel time
h/c. This is not in conflict with causality because no signal needs to
travel from one end of the conductor to the other - the signal is already
present at the initial instant, in the electric fields of the initial
charge configuration."

Bob

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor

----- Original Message -----
From: "Ludwik Kowalski" <KowalskiL@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Saturday, January 06, 2001 08:45 AM
Subject: Re: Toward the equilibrium


In his first reply Bob Sciamanda wrote:

Using Maxwell's equations, the continuity equation , and the
definition of
conductivity, texts routinely derive an exponential decrease of rho in
a
region of constant conductivity. The relaxation time is shown to be
epsilon/conductivity. If memory serves, this is about 10^-19 sec for
copper. This is the simple, classical model, and even ignores mean
free
path considerations.

Bob was responding to this:

Suppose that at t=0 a charge of 1 microC is deposited at one
point of a copper sphere whose diameter is one meter. I used
to say that "the equilibrium is going to be established very
quickly, perhaps in a couple of nanoseconds or so. Why?
Because it can not be faster than 3.3 ns; the time light
needs to cover the distance of 1 m in air.
. . .
2) How to calculate the time needed to establish equilibrium?
I suspect it is much longer than ~3 ns but I do not know how
to estimate it realistically.

I got nearly the same (unrealistically small) order of magnitude on
the assumption that the charge equilibration takes place through the
volume (rather than along a very thin skin). It goes like this:

1) The capacitance of an isolated sphere is about 50 pF (assuming
the nearest wall is very far away)

2) The resistance is roughly the same as for a one-meter copper
cube, or about 2*10^-8 ohms.

3) Therefore, the RC constant is roughly 10^-18 seconds.

Note that the answer does not depend on the size of the sphere
because C is proportional to the radius while R is inversely
proportional to the radius. This is very unrealistic; how can
the equilibration time be the same for one meter and for 1000
km? Something is basically wrong. Do we have to use special
relativity to calculate small time constants?

Will I be able to read e-mail from where are am going today?
Ludwik Kowalski