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Re: Potential of a point charge



John M.

You forgot the QED at the end.

:-)

Joel R

| -----Original Message-----
| From: Forum for Physics Educators [mailto:PHYS-L@lists.nau.edu] On Behalf
| Of John Mallinckrodt
| Sent: Thursday, February 26, 2004 1:20 PM
| To: PHYS-L@lists.nau.edu
| Subject: Re: Potential of a point charge
|
| I think this confusion is widespread, but I find that it pretty much
| evaporates if one keeps in mind two things:
|
| 1) The integral involves a dot product of the VECTOR E_vec with the
| VECTOR ds_vec.
|
| 2) When integrating "radially" the vector ds simply becomes the
| vector (dr r_hat) where r_hat is a unit vector in the positive r
| direction (i.e., radially outward) and dr is a SCALAR with an
| algebraic sign that is determined AUTOMATICALLY by the limits of the
| integral. If the lower limit is larger than the upper limit, than dr
| IS negative. One need not "make adjustments" for this fact by
| putting in a minus sign somewhere; its all "in there" already.
|
|
| Thus, to find the absolute potential at a distance R from a point
| charge, we simply apply the definition
|
| delta V = V_B - V_A = - int( E_vec dot ds_vec from A to B)
|
| to a radial path from "infinity" (where V = V_A = 0) to R where (V = V_B)
|
| V = - int( [k q r_hat/r^2] dot [dr r_hat] from infinity to R)
|
| = - k q int( [dr/r^2] [r_hat dot r_hat] from infinity to R)
|
| = - k q int( dr/r^2 from infinity to R)
|
| = - k q (-1/r evaluated from infinity to R)
|
| = k q/r
|
| --
| John Mallinckrodt mailto:ajm@csupomona.edu
| Cal Poly Pomona http://www.csupomona.edu/~ajm