- 122 -


The charge neutrality condition is


      3
C e = I  Zi d. +
     i=2  


(6.47)


   Z. C.
i=1 3   1


Substituting C1 from (6.16-i) into (6.46) and (6.47) gives


1= C   +
     e



     3
 e   i=2


3
1 d + c1(i +
i=2


Z Oi d + C,


1=3
i=


13 K'

i=2 C
     e

   K,
Z.
   c
   e


Solving for C, from (6.48) and (6.49) and equating the results,


Ce - d3       C
13     K!
        i-i
 I  z .i
i=2    C
        e


         3
1 - C  - I d,
      ei=2
      13 K' 1+  I
         -i-i
    i=2 C
         e


Using (6.8) to obtain an expression for d2, (6.50) becomes


(6.50)


equation


     q(l - Ce) - Ce d3=           qCe
     (q -i)+ +
               K*
               02

     13    K 
     Sz      1-1
     i=2   C
            e
q=
         1 3 K
    1 +  [    
        i=2 C
             .e


(6.51)


(6.52)


(6.48) (6.49)


where