Re: Induced dipole moments

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

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, 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