
The charge density r of the electrons then is 


r =
– 
3z · e^{ }
4p · R^{3} 



In an electrical field E a force
F_{1}
acts on charges given by 





We will now drop the
underlining for vectors and the mauve color
for the electrical field strength E
for easier readability. 

The positive charge in the nucleus
and the center of the negative charges from the electron "cloud" will
thus experience forces in different direction and will become separated. We
have the idealized situation shown in the image above. 

The separation distance
d will have a finite value because the separating force of the
external field is exactly balanced by the attractive force between the centers
of charge at the distance d. 


How large is his attractive force? It
is not obvious because we have to take into account the attraction between a
point charge and homogeneously distributed charge. 


The problem is exactly analogous to
the classical mechanical problem of a body with mass m falling
through a hypothetical hole going all the way from one side of the globe to the
other. 


We know the solution to that problem
(see the link): The
attractive force between the point mass and the earth is equal to the
attractive force between two point masses if one takes only the mass of the volume inside the sphere given
by the distance between the center of the spreadout mass and the position of
the point mass. 


Knowing electrostatics, it is even
easier to see why this is so. We may divide the force on a charged particles on
any place inside a homogeneously charged sphere into the force from the
"inside" sphere and the force
from the hollow "outside" sphere.
Electrostatics teaches us, that a sphere charged on the outside has no field in the inside, and therefore no
force (the principle behind a Faraday cage). Thus we indeed only
have to consider the "charge inside
the sphere. 

For our problem, the attractive force
F_{2} thus is given by 


F_{2} = 
q(Nucleus) · q(e in d)
4p e_{0} ·
d ^{2} 




with q(Nucleus) = ze and q(e in
d) = the fraction of the charge of the electrons contained in the
sphere with radius d, which is just the relation of the volume of
the sphere with radius d to the total volume . We have 


q(e in d) = ze · 
(4/3) p · d
^{3}
(4/3) p · R ^{3} 
= 
ze · d ^{3}
R ^{3} 




and obtain for F_{2}: 


F_{2} = 
æ
ç
è 
(ze) ^{2}
4 pe_{0} ·
R ^{3}

ö
÷
ø 
· d 



We have a linear force law akin to a
spring; the expression in brackets is the "spring constant". Equating
F_{1} with F_{2} gives the
equilibrium distance d_{E}. 


d_{E}
= 
4 pe_{0} · R ^{3} · E
ze 



Now we can calculate the induced dipole moment m, it is 


m =
ze · d_{E} = 4
pe_{0} ·
R ^{3} · E 



The polarization P
finally is given by multiplying with
N, the density of the dipoles; we obtain 


P = 4 p
· N · e_{0} · R
^{3} · E 



Using the definition P =
e_{0} · c
· E we obtain the dielectric
susceptibility resulting from atomic polarization, c_{atom} 


c_{atom} = 4 p · N · R ^{3}



Let's get an idea about the numbers
involved by doing a simple exercise: 


