Subject:
Work; the correct one
Date:
Fri, 27 Sep 2002 09:26:48 -0400
From:
"Emory F. Bunn" <ebunn@lfa222122.richmond.edu>
Reply-To:
ebunn@richmond.edu
To:
drutherford@softcom.net
You submitted the article below to the moderated newsgroup
sci.physics.research. I am unable to accept your article, as the
newsgroup does not accept "overly speculative" postings.
Since this post is based on a notion of work that is different from
that used everywhere else in the physics community, it is overly
speculative. I absolutely, unequivocally promise you that no work is
done on a particle that undergoes no displacement.
You are of course free to post your article to another newsgroup such
as sci.physics.electromag if you choose.
Ted Bunn
sci.physics.research co-moderator
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To: sci-physics-research@moderators.isc.org
From: David Rutherford <drutherford@softcom.net>
Newsgroups: sci.physics.research
Subject: Work; the correct one
Date: Wed, 25 Sep 2002 15:49:24 -0700
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You don't need to have a displacement in order to do work on a charge q
in an electric field E. If you use
W = \int{F.dr}
as the definition of work W, then in this case, the force F is
F = qE
The electric field can be written E = -d(phi)/dr, where phi is the
electric potential. So
W = \int{F.dr} = \int{qE.dr} = \int{q(-d(phi)/dr)dr} (1)
where "\int" means integral. But
d(phi) = (d(phi)/dr)dr (2)
so if you insert (2) into the last equation in (1), you get
W = \int{q(-d(phi)} (3)
Equation (3) is more fundamental than the last equation in (1) for the
work done on a charged particle in an electric field. Here, you can
see that the work done on the charge q does _not_ depend on it's
displacement, but only on the change in potential at its location. Work
is done on a charge to keep it in place, even if it has not moved, if
the potential at its location has changed.