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Re: electrostatic shielding



There is a serious problem here. If by "very large" you mean a
finite sheet,
then all (hypothetically infinitessimally thick) flat sheets of the
same shape have the same geometry, whether one considers them to be
large or not. If you mean an infinite sheet, then it is difficult to
see what you mean by "isolated".
May we speak of, say, a charged disk and consider your questions?

Leigh


Yes. (In fact, I originally had both "infinite" and "isolated" in my
description, but decided the two were not compatible. I don't think I
necessarily had infinitesimal thickness in mind.)
______________________________________
Fred Lemmerhirt
Waubonsee Community College
Sugar Grove, Illinois
flemmerhirt@mail.wcc.cc.il.us <mailto:flemmerhirt@mail.wcc.cc.il.us>
http://chat.wcc.cc.il.us/~flemmerh/physics.html
<http://chat.wcc.cc.il.us/~flemmerh/physics.html>

-----Original Message-----
From: Leigh Palmer [SMTP:palmer@SFU.CA]
Sent: Friday, February 02, 2001 11:29 AM
To: PHYS-L@lists.nau.edu
Subject: Re: electrostatic shielding

>I guess John Mallinckrodt's reminder of the uniqueness principle
finally
>provides a convincing requirement that the charge on the outside of
the
>sphere must remain uniformly distributed, provided that
rearrangement of the
>inner surface charge alone can cancel the off-center point charge's
field
>within the conducting shell. (As Bob Sciamanda pointed out, no one
here
>seems to have shown that this is possible, but it surely seems
reasonable.)
>
>Now can I extend this result to another situation?:
>
>Suppose a very large isolated flat conducting sheet carries a net
negative
>charge. Will the charge be uniformly distributed on both surfaces,
>producing a uniform field pointing straight toward the sheet on
both sides?
>And if a positive point charge is brought near one side of the
sheet, will
>the charge only on that side of the sheet be rearranged and the
field on
>that side "distorted", while the charge and field on the opposite
side
>remain uniform?