Re: Induced dipole moments

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

Here is a useful problem on induced polarization. It is likely to
be appropriate for some introductory physics courses, either in
high schools or colleges. The simplicity comes from the "rigid
cloud" model. We used this term yesterday; I will paste the
definition at the end of this message.

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

a)What is the most stable configuration when the electric field
E is zero? The proton is in the center of the cube.

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.

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

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

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).

How large is the dipole moment for any give E  (0<E<Emax)?
That where it becomes tricky. In my opinion the answer is

p= Q*X = E*L^3 /(8*k)

The dipole moment is thus directly proportional to E, all the
way up to the breaking field, Emax. All layers contribute to p.
**************************************************

> A rigid cloud, by definition, is a set of identical point charges
> which can not move with respect of each other. Yes, it is a
> highly unrealistic model; I do not care what keeps particles at
> fixed positions with respect to each other. Two such clouds
> can penetrate through each other because distances between
> particles (inside of each cloud) are large.

Actually it not very unrealistic.

A model for two "rigid clouds" can be constructed, at
least in principle, from beads on rigid plastic sticks. Mount
five equidistant beads on each of the 25 sticks. Insert these
sticks into a Styrofoam plate. You have a 3-dimensional
fork (with a styrofome handle). Suppose each bead has a
charge +q.

Make another 3-dimensional fork; this time each bead should
have a charge -q. Grab the +fork by the handle with one hand
and -fork, also by that handle, with another hand. You will
hold two "rigid clouds" attracting each other. As long as the
clouds do not penetrate the force is proportional to 1/d^2,
where d is the distance between the centers of charge of
each fork.

Keep bringing the forks closer. The sticks in two forks are
parallel and they are allowed to penetrate (without touching).
The repulsive force will start decreasing when d becomes
smaller and becomes zero when d=0 (in the limit of a very
large number of sticks, large number of beads per stick, etc.)
*********************************************
Ludwik Kowalski