NetNews Usenet Archive 1993 #1/NN_1993_1.iso/spool/sci/physics/fusion/3230

Text File | 1993-01-10 | 1.8 KB | 39 lines

Newsgroups: sci.physics.fusion
Path: sparky!uunet!spool.mu.edu!umn.edu!noc.msc.net!ns!ns!logajan
From: [e-mail] (John Logajan)
Subject: Falling thru a voltage gradient
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Originator: logajan@ns
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Organization: Network Systems Corporation
Date: Sun, 10 Jan 93 05:17:49 GMT
Lines: 26

I have a question.  Given a very large conductive flat surface, say a circular
plane of copper sheeting one mile in circumference, and charged with a finite
number of electrons (X coulombs), we know that the voltage gradient will
drop off very slowly with distance for points along the axial center of the
plane.

Suppose we deposit a free electron one millimeter above the surface of the
plane, on the axial center line.  It should be repelled upward away from the
electrons distributed on the surface.  Being close to some electrons
distributed near the axial center, the "test" electron should undergo strong
repulsive forces, yet as it moves away, the voltage gradient hardly changes,
since the gradient represents the vector sum of all unbalanced charges on
the entire surface of the plane, and the vector sum changes significantly
only when the "test" electron moves significantly farther away from *ALL* the
surface charges. (i.e. the "test" electron movement from 1mm to 10 meters
is an increase in distance of a factor of 10,000 for electrons on the axial
line, but only 60 mm farther from the electrons 800 meters away on the
plane's edge, or an increase of only a factor of 1/10,000th.)

Don't we then have enormous electron acceleration (or at least force) without
seeing commensurate change in the local field strength?   Why am I confused
by this?

--
- John Logajan MS010, Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428
- [e-mail], 612-424-4888, Fax [phone]