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Re: Toward the equilibrium



As always, it is good to know exactly what the question is. Let me state
three variations of Leigh's original question:

1) A charge of +1 uC is held at a small distance delta above the surface
of a conducting sphere. At t=0, the charge is brought the rest of the way
to the surface. (More or less the version I had in mind)
2) The 1 uC charge is carried with some significant speed toward the
conductor, arriving at t=0. (Brian's view?)
3) An insulator has a charge of 1 uC placed at one spot on the surface.
At t=0, the insulator suddenly becomes a conductor. (Closest in spirit to
Leigh's original query?)

Further, I am assuming the +1 uC is treated as a "point charge". Note that
(1) could be viewed as the the low speed limit of (2).

I was thinking of (1) when I wrote my response, and for that situation I
stand by my answer, but let me rephrase - "No significant travel of
electrons will be needed ** after t=0 ** to establish equilibrium."

If Tim concurs on the magnitude of the potential on the 1 meter sphere in
the steady state - about 13 kV for the room size I had in mind -
he will presumably agree that there is a surface displacement of
electrons
to comply with this field.

Let's start with the original sphere. According to Jackson (2nd Ed, p 56),
for a point charge "q" a distance "y" from the center of a conducting
sphere of radius "a", the induced charge (by the method of images) is

q' = -(a/y)q = -(a/(a+delta))q = -q

y' = a^2/y = a^2/(a+delta) = a

So there is a charge ~ -q a distance ~ 2 delta away. When the charge
completes its trip, it just recombines with handy -q and no redistribution
of the remaining charge on the conductor is needed. The +1 uC required to
create the 13 kV is already in place, spread evenly over the surface.

(Actually, this is derived for a grounded sphere, but I submit that, except
for the -q "image", the remaining +q needed to make the sphere initially
neutral is spread evenly over the surface.)

*************************
Hey! I just thought of a simpler line of reasoning. In electrostatics,
there can be no E field in a conductor. In the limit that the charge is
brought slowly (<<c) E ~ 0 the entire time, including t=0. If E=0 at t=0,
F=0 and the charges in the conductor don't move.

In the "electrostatic limit" the charges will all redistribute before t=0.
Sphere or coax, it doesn't matter. Granted, all this an idealization only
a physicist could love. In real life the charge has to move to the
conductor, and the last part of the trip will presumably involve some
arcing. The faster the motion or the more shielded the conductor, the
longer the time to reach equilibrim. But those detail go considerably
beyond what I think Leigh was asking
*************************


Tim Folkerts
FHSU