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Work; the correct one



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.

--
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