Re: E field near charged disk

["John S. Denker" <jsd@MONMOUTH.COM>]

At 11:29 AM 1/26/01 -0500, Eugene Mosca wrote:
> The E field on the axis of a uniformly charged disk is given by
>
> E = (2k*pi*sigma)(1-[1+(R/x)^2]^-0.5)  (Tipler 4e, p694)
>
> where R is the radius of the disk.  Taking the limit as the ratio R/x
> approaches infinity gives the field for an infinite uniformly charged disk.
> This field is given by
>
> E = 2k*pi*sigma
>
> A plot of E versus x for second equation (infinite disk) is a straight line
> with zero slope.  However, a plot of E versus x for the first equation
> (finite disk) gives a curve whose slope does not appear to approach zero for
> x << R.

Right, when I differentiate the first equation I get slope = 1/R
independent of x for small x.

This is perfectly consistent with the previous result;  the slope goes to
zero as R becomes large.

> My question is, why not?  It seems to me it should.

Well, there's some good physics in this.

1)  We say that electrostatics is a "long range force" and we are seeing
that here.  The test particle, even at small x, can feel the charge even
when it is a distance R away.

2)  Assuming of course constant charge density, the infinite disk has a lot
more total charge than the finite disk -- infinitely more.  That extra
charge contributes to the field.  The only reason it doesn't make an
infinite contribution is because it suffers a mechanical disadvantage (a
factor of sin(x/R)) because it isn't pulling in the optimal direction.

Write out an estimate of the field contributed by the charge outside R:
                                               (basic
                                            electrostatics)
  |field| goes like  integral       sin(x/r)     1/r^2     2 pi r dr
                        (from R       (geometry)             (measure for
                   to infinity)                        polar coordinates)

Without the sin(x/r) that would be logarithmically divergent.