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Re: Energy density; the correct one



Bob LaMontagne wrote:

OK - I'll bite - How do you do work on a charge that isn't moving - being
held in place - where's the displacement needed for the usual definition of
work?

You don't need to have a displacement in order to do work on a charge in
an electric field. If you use

W = \int{F.dr}

as the definition of work, then in this case,

F = qE

(I'm using q, instead of e1, here for the charge) and E = -d(phi)/dr,
where phi is the electric potential. So

W = \int{q*(-d(phi)/dr)*dr}

or

W = \int{q*(-d(phi)}

Here, you can see that the work done on the charge q does not depend on
it's displacement, in the case of a charge in an electric field, 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, which it has if another charge approaches
it. Therefore work is done on the charges that have already been
assembled to keep them in place if you bring in additional charges from
infinity. This work must be included in the total work required to
assemble the charge distribution.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf