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Re: Induced dipole moments



Sorry for the last. I hit the wrong button. But:
On Mon, 1 Jan 2001, Ludwik Kowalski wrote (in part):

3) I think that a cubical cloud is mathematically easier. Try to derive
the force formula for a cloud of radius L. The left segment of your
sphere will become a meniscus lens whose volume would not be
a linear function of x. I will accept your claim "cubes are less
tractable than spheres" after I see your derivations. Perhaps you
have a good shortcut that is OK for an introductory course.
4) There was a typing error in the formula for Emax; it is obvious
that the denominator should have been L^2 and not L.
Regards, Ludwik

I agree with John's current posting. The spherical case
follows quickly from Gauss' law: There is no field from the part of the
sphere at larger radius than the point charge, and the field from the
inner part of the sphere is proportional to the enclosed volume -
a result, by the way, that was apparently known to Newton already.

You can approximate the cubical case for small displacements
perpendicular to a face. If the displacement x is much less than the
length of a side, then the field is due to the excess of charge in the
sheet perpendicular to the displacement in the interval 0->x. This
will be like the field of an infinite sheet (proportional to the surface
density of charge). Since the surface density is proportional to the
displacement, you do get an induced field proportional to the
displacement.
This argument breaks down when the displacement is of the
same order as the length of a side.
The general case of a cube cannot, as far as I can tell, be
done in closed form. Otherwise you would find it in the textbooks.
Regards,
Jack

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translation to English, it is hoped that the following translation conveys
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