Conference rusure::math

    Electric fields can't act on a chargeless particle.
    I.E. a chargeless particle would not experiance any force.
                                                   
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I think the root of the confusion here is that Morrison is
talking about *measuring* the field, and you are talking about
*calculating* it.  If you wish to measure the field, it is 
important that the test charge be kept small because it can
induce charges and polarizations that change the original field.
Your statement about calculating the field is correct.

re .1, .2:

  Aha.  I think I see.  Let me see if I can restate in my own words.

  It is the total field (i.e. that due to ALL charges actually
present) that is subject to Gauss law.  Conceptually, we can measure the field
at a particular point by calculating the E field at that point due to all other
charges.  However, if a test charge were actually at that point, the
total E field would be different due to the effect of the test charge.
The presence of the test charge would cause other free charge to move around
until Gauss law was satisfied anew.  But this new charge and field distribution
would be different from that when the test charge was absent.

  For example, (if I understand this correctly) physics books always state that
the charge distribution on a conducting sphere is uniform and that the
force on a point charge at a distance r from the sphere is therefore
Q1*Q2/(4*pi*epsilon*r^2).

  It seems that this calculated force isn't quite right because the presence of
the point charge would modify the charge distribution on the conducting
sphere and therefore that the field due solely to the charge distribution on
the sphere (i.e. the field that causes the force on the point charge) is
DIFFERENT from that of a uniform distribution.

  Is this a correct interpretation ?

  If so, this IS a subtlety that had escaped me.

  This looks like an example of "you can't observe something without disturbing
it".

	Technically the field is the derivative of the force experienced
with respect to the particle's charge, E = dF/dq, thus the *limit* of
F on q at a given point as q->0. Any q ^= 0 of course experiences F ^= 0
and disturbs the field as well; as q -> 0 this disturbance, and F, -> 0
as well, but the ratio of F/q approaches some nonzero limit we call the
"field strength".

	Same problem with gravity, any other field...

re .4:

>	Technically the field is the derivative of the force experienced
> with respect to the particle's charge, E = dF/dq, thus the *limit* of
> F on q at a given point as q->0. Any q ^= 0 of course experiences F ^= 0
> and disturbs the field as well; as q -> 0 this disturbance, and F, -> 0
> as well, but the ratio of F/q approaches some nonzero limit we call the
> "field strength".

>	Same problem with gravity, any other field...

With gravity, it's somewhat different, I think (at least for solids).
Because solids are generally treated as rigid bodies, the distribution of
mass doesn't shift in response to an external gravitational field, or
at least the effect is generally neglected.  So, this behavior
(i.e. mass or charge moving around to a new equilibrium configuration)
would probably not be considered in a first order analysis of gravity
effects.

On the other hand, for fluids, the situation seems more analogous to that of
electric charge.  For example, ocean water on the earth's surface can shift in
response to the gravitational forces of the moon and other planets.
It would seem then that to calculate tidal effects such (fluid) mass shifting
would have to be accounted for.

I'd bet that the computation of the resulting charge or fluid distributions
isn't easy.