Re: Energy density; the correct one

[David Rutherford <drutherford@SOFTCOM.NET>]

Bob LaMontagne wrote:
> Whoa!
>
> David Rutherford wrote:
>
> > W = \int{q*(-d(phi)/dr)*dr}
> >
> > or
> >
> > W = \int{q*(-d(phi)}
>
> IMHO - The scond equations is incorrect! - because the second dr in the first
> equation is 0.

Please check your local calculus book

(d(phi)/dr)dr = d(phi)

Plug this into the first equation and you'll get the second equation.
The second equation is just as valid as the first and it has no dr's.

> Actually performing \int involves a path,

Obviously, not.

> not just a change in phi
> for reasons independent of dr.
>
> You can change the energy of q by changing potential (by moving other charges),
> but it's not work done on q itself, V has changed because of energy input
> elsewhere in the q1-field-q2 configuration.
>
> Maybe I'm seeing this totally wrong - I'd appreciate a clarification if I am.

I don't know how I can show you any more clearly. Just look at the
second equation

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

It says that work does not depend on a path (for a field that is the
gradient of the potential). It only depends on the change in potential.
Therefore, work is done on a particle in keeping it in the same position
if the potential at that position changes, even if the particle doesn't
move.

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

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