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Re: Toward the equilibrium



brian whatcott wrote:

At 17:28 1/3/01 -0500, Ludwik wrote:
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.

1) Do charges travel to final destinations through the entire
volume or do they travel mostly near the surface?

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.
Ludwik Kowalski

You will be amused, I expect, by the stark contrast between
the conceptual approach that Bob sketched out and the
intuitive one I provide here: with no disrespect, my estimate
could hardly be less accurate in comparison!

Engineers often need to estimate magnitudes - in the case of
capacitors, even physicists in metrology crave computable
component geometries.

There is a useful geometry for dual concentric conductive
spheres:
it is Capacitance (picofarads) = 111.27 R1R2/(R2-R1)
where R1 is inner radius, R2 outer radius in meters.

We suppose Ludwik to be standing in a room whose ceiling is
10 feet high (3m) and whose walls are 14 ft apart (4.27m)
and we take R1 as 0.5m and R2 as 1.9m

The capacitance is then about 75 pF
and we check that the equilibrium potential will be about
Q/C = 1E-6 C/75E-12 F = 13 kV

We can easily visualize the step rise at the point of
(arcing) contact. It will be several times this equilibrium
potential over the smaller initial capacitance. It is not
difficult to imagine the electric wave ballooning from the contact
point around the surface. As the wave expands, the potential
falls, so that when the wave front finally reaches the
antipodean point, there is possibly less than the equilibrium
potential at the starting point.
The wave continues to return from the antipodes, doubling the
potential until after one circumferential transit time, there
is a reasonable settling at the equilibrium value over the
surface - the wave propagates at a quite high fraction of c
I'd expect, perhaps 0.85, so that the time in question is
pi D/(3E8 0.85) = 12 nanoseconds, about.....

(This is well within the viewing range of a fast oscilloscope)

Perhaps many waves will start propagating at t=0 along the
meridians (assuming the pole is defined as a point at which
the charge is initially concentrated). The time to reach the
electrostatic equilibrium (constant potential everywhere in the
sphere) is presumably equal to the time at which the wave
amplitude becomes zero. This may be several orders of
magnitude longer than the time of one round trip, especially
when the resistivity is very low. Therefore the time of 12 ns
is the lower limit of what we want to estimate. But it is a
much more realistic limit than 10^-19 s.

Waves are not charge carriers. The redistribution of the
total charge is driven by the constantly changing electric
field within the volume of the sphere. (OK, PERHAPS
THIS CAN BE CALLED A WAVE.) That field E is due
to the delivered charge, initially at one point. Thus the
field is radial at t=0 (the center is at the point of impact,
not at the center of the sphere) and charges start drifting
along the initial field lines. At the end of the process
E becomes zero everywhere within the sphere.
Ludwik Kowalski