Re: Energy density; the correct one

[Robert Cohen <Robert.Cohen@PO-BOX.ESU.EDU>]

A ball is on the side of the mountain with the surface having slope dy/dx.
The work done by gravity as the ball rolls down the mountain is m*g*dy.

Suppose the ball is nailed to the side of the mountain.  Suddenly, an
earthquake occurs.  The land "downhill" of the ball rises and the land
"uphill" of the ball lowers such that the ball remains in exactly the same
place (i.e., the y and x position did not change) yet the mountain is
suddenly transformed into a plain (i.e., the slope dy/dx is now zero).

Did work done on the ball?

Well, dy/dx changed...therefore dy must've changed...therefore m*g*dy
must've changed...therefore gravity did work?

____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301

> -----Original Message-----
> From: David Rutherford [mailto:drutherford@SOFTCOM.NET]
> Sent: Wednesday, September 25, 2002 2:06 PM
> To: PHYS-L@lists.nau.edu
> Subject: 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.