Re: Potential of a point charge

[John Denker <jsd@AV8N.COM>]

I wrote:
> | Equivalently you could set dr = +ds provided
> | you write the integral as
> |
> |              / to r
> |             |
> |             |
> |            / from infinity
> | . . .

Quoting "RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>:
> in order to set dr = + ds, you should integrate from r to infinity,
> correct?
>
> In which case I think you have a typo in the above and should have the
> limits reversed in the integral.

I suppose it depends on what we think the integrand is (F dot dx
or -F dot dx) and what we think the integral represents (potential
gain or potential drop).

In any case, I think Rick's basic point is a good one:  There are
endless opportunities for confusion about the signs.  Contributing
factors include:
 -- you can integrate F dot dx or -F dot dx
 -- you can swap limits of integration
 -- you can calculate potential gain or potential drop
 -- etc......

It's unwise to be dogmatic about the "right" choice at each of
these steps, because only the product of all the choices matters.

In general, a good rule for staying out of trouble (which I alas
violated in my previous note) is to write down a complete equation
(not a fragment) and be absolutely clear about the meaning of all
the elements of the equation.

==============

Note: in general, the voltage difference is 1/|r_A| - 1/|r_B|
for any points A and B whatsoever (except r=0).  In particular,
one point need not be radially outward from the other.  So in
general we must not assume ds=dr or -ds=dr ... rather ds follows
an arbitrary curvy path between A and B.