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Re: Induced dipole moments

I agree with John Mallinckrodt about the source of the error.
Referring to:

-->The net force of the +Q is thus equal to the force from
the leftmost layer. The distance d between the center of
that layer and +Q is 0.5*L, for any x (as long as +Q is inside
the cube). According to Coulomb's Law that net force is

F = -k*Q*q/(0.5*L)^2 = -4*k*Q*q/L^2 (L = constant !)

John wrote:

Here is the source of your error: You are treating the
leftmost layer as a point charge.

Yes, this was the killing oversimplification. A summation
of longitudinal components over the volume of the leftmost
layer is unavoidable, no matter what is the shape of the
rigid cloud. Thus the problem can not be solved without
calculus. It seems to me that the simplest possible geometry
would be neither the cube nor the sphere; it would be a
cylinder whose axis coincides with the direction of E.

Another step in the direction of simplification, if one is
desperate to produce an analytical (rather than numerical)
solution for the restoring force, is to make the diameter
of the cylinder negligibly small in comparison with its
length L. Then the 3-dim integration is reduced to the
1-dim integration. Most calculus-based textbooks deal
with this in the Gauss Law chapter. In the third edition
of Tipler, for example, the force is shown to be

F=-k*Q*q/[d*(d-l)]

where l is the length of the left-most segment (not to be
confused with the length of the cloud L), d is the distance
of the +Q (proton) from the left side of that segment and
q is the net charge of the segment.

Let x be the distance of the proton from the effective
negative charge q (in the center of the left segment, as
before). It is clear that the dependence of F on x is now
slightly more complicated than in my incorrect formula
(q, d and l are now functions of x) but it can easily be
written down in a closed form.

Is this worth doing in a calculus-based course? It depends
on the situation. But one does not have to be quantitative
in using a "rigid cloud" model. A mental subdivision of
the cloud into three regions is already sufficient to show
why the effective negative charge increases with x. This
may be sufficient to explain how the restoring force can
be increasing when the proton is pushed away from the
center by an external field E. Thanks to all who helped.
Ludwik Kowalski
P.S.
John is our expert on Interactive Physics. Perhaps he can
produce a good one-dimentional IP simulation showing
that F increases when x becomes larger. I suspect that
a discrete line of ten equidistant electrons could be a
sufficient approximation of a continuos distribution.
But this is not obvious.

John Mallinckrodt wrote:

On Mon, 1 Jan 2001, Ludwik Kowalski wrote:

Suppose the positive charge Q is point-like while the negative
charge Q is a cubical cloud.

...

b) Suppose the proton was pulled from its equilibrium
position by a distance x. How does the attractive force
(between +Q and -Q) depend on the distance x? My answer
is F=-8*k*rho*Q*x (in SI units), where rho is the charge
density in the cubical cloud (rho = magnitude of Q divided
by the volume, L^3) and k=9e9. I know it is not prudent to
post something without waiting another day or two. But I
can not resist; after all it is the first day of a new year.
Please correct me, if necessary.

Ludwik,

This can't be correct. The answer will surely depend on the
direction of the displacement and the necessary integrations will
result in ugly looking analytical expressions if they can be
performed at all.

Your cubical charge distribution is artificial and, in fact,
*creates* these needless mathematical difficulties. It isn't hard
to show that the potential energy of a point charge Q a distance r
< R from the center of a (more realistic) uniform spherical charge
distribution of total charge -Q and radius R is

U = - (k*Q^2/2*R)[3-(r/R)^2]

and that the restoring force is

F = -dU/dr = -(k*Q^2/R^3)*r

As Leigh noted yesterday, this result is closely related to the
interesting (if unrealistic result) that a particle dropped into a
straight frictionless tunnel through a uniformly dense planet
executes SHM with a period identical to that of low altitude
orbiting satellite.

c) Suppose a uniform electric field E is applied along the
x direction. Find the new equilibrium position, X.

Balance of forces gives

X = (R^3/k*Q)*E

a result that could be obtained purely from dimensional analysis.

Here is my way of reasoning:

-->Draw a square representing a cubical cloud (of size L)
with the proton at the center. Draw an identical square
below. This time the charge +Q is at a distance x (for
example, on the right from the center). The cube can now
be subdivided into three vertical layers.

--> The right layer whose thickness is 0.5*L-x, the
left layer whose thickness is 2*x and the central layer.
The force on the +Q from the right layer and the force on
on it from central layer cancel each other.

-->The net force of the +Q is thus equal to the force from
the leftmost layer. The distance d between the center of
that layer and +Q is 0.5*L, for any x (as long as +Q is inside
the cube). According to Coulomb's Law that net force is

F = -k*Q*q/(0.5*L)^2 = -4*k*Q*q/L^2 (L = constant !)

Here is the source of your error: You are treating the leftmost
layer as a point charge.

where q is the net charge in the left layer. The volume of
the left layer is L*L*2*x and its net q is q=2*L^2*x*rho.

--> Therefore, F=-(8*k*Q*rho)*x, is directly proportional
to x. So much for the dependence of the restoring force on
the distance x.

--> Now what happens when the electric field E is applied
along the x axis? The force on proton is now Q*E while the
force on the cube is -Q*E. The center of +Q will be separated
from the center of the cube up to the distance X at which the
magnitude of the restoring force is Q*E. Right?

--> Therefore 8*k*Q*(Q/L^3)*X = Q*E and the equilibrium
distance, X, is given by:

X=[(L^3)/(8*k*Q)] * E = Const*E

I'm pretty confident that a correct treatment for the cube will
yield a value for X that is a nonlinear and possibly even
discontinuous function of E. The discontinuities might result
from catastrophic changes in the configuration of maximal
stability as E increases.

Nevertheless, it is interesting (but not *so* surprising from a
dimensional analysis point of view) to note that this is identical
to the correct answer for a spherical distribution with a diameter
equal to L.

Note that X must not exceed 0.5*L because the assumptions
(about the three layers) are no longer applicable. How large
field is needed to destroy the stability of this structure? It
must be larger than Emax=4*k*Q/L^2, as can be seen from
the above relation (replace X by 0.5*L).

But, leaving alone the previous error, you haven't provided a
demonstration that your assumed structure would be stable in any
event. I could easily imagine, for instance, that the most stable
configuration, at least for some range of strengths for E, would
involve a displacement along a body diagonal of the cube.

I'd stick with spheres!

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm