Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: Toward the equilibrium



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)
brian whatcott <inet@intellisys.net> Altus OK
Eureka!