Page 195 - Modern physical chemistry

8.7 Application of Gausss- Theorem to the Ionic Atmosphere 187

From result (8.53), any charge qi within a closed surface contributes q/E to the inte-
gral JE-dS. Any charge % outside the closed surface does not contribute because JdQ for
it is zero. For total charge Lqi within the closed surface, we have

fE .dS = Lqi. [8.54]
s e
This form of Coulomb's law is called Gauss's theorem.

8.7 Application 01 Gauss's Theorem to the Ionic Atmosphere

We have considered the jth ion and its neighborhood to be spherically symmetric.
Thus, the electric intensity E about the ion is considered to be directed radially. But any
element dS of a sphere centered on the ion is also directed radially. So we have
2
f E . dS = f E . dS = E f dS = E( 4nr ). [8.55]
s s
According to Gauss's theorem, the left side equals the total charge q enclosed by the
sphere of radius r divided by E; thus
!I =4nr 2 E. [8.56]
e
The change on going to a sphere infinitesimally larger is

:q =d(4nr2E). [8.57]

But if p is the charge density at distance r from the center of the jth ion, we have
dq = p4nr 2 dr. [8.58]

Eliminating dq from (8.57) and (8.58) yields

[8.59]

whence
~~(r2E)=£. [8.60]
r2 dr e
The electric intensity, the force per unit charge that would act on a test charge, is
given by
E=_df/J. [8.61 ]
dr
Inserting this into equation (8.60) leads to the form

~~[r2 dt/» __ £ [8.62]
r2 dr dr e

which can be rewritten as
[8.63]