Re: Potential of a point charge

[John Denker <jsd@AV8N.COM>]

Quoting Rick Tarara <rtarara@SAINTMARYS.EDU>:

> 1) delta PE = (negative) of Work done by the field
>
> 2) so that   F = -qE

Balls roll *down* the hill.

> 3) so that   delta V = - integral (E dot ds)
>
> 4)  BUT now dr = -ds (this is what took me a while to see)

That's because you are integrating *from* infinity back toward
some particular r.  Equivalently you could set dr = +ds provided
you write the integral as

             / to r
            |
            |
           / from infinity

> 5) so that delta V = - integral (Edr)
>
> 6)  which is (for point charge)   deltaV = - integral (kQ/r^2) dr
>
> 7)  which integral generates another negative so  delta V = kQ/r

... the correctness of which is easily verified by differentiation.
Balls roll down the hill.


> That dr = -ds is always my stumbling block because most texts don't really
> spell it out.  HR&K just jump from E dot ds to Edr.

Probably their limits of integration are swapped relative to yours.