ELECTRICAL METHODS 485 



solution of the problem. The boundary conditions to be satisfied at the boundary 

 of medium and medium I are : 



Vo=Vx 

 and 



— = — at 2 — 0, r arbitrary 



pa OS p\ oz 



Likewise at the boundary of medium I and medium II, 



Fi = F2 

 and 



— — — = ;r — 2Xz — a, r arbitrary 



pi OS pi oz 



Each of these four conditions results in a number of equations involving the un- 

 known parameters Sa, Mt, and Nm. The first condition at ^ = 0, namely Vq = Vi, 

 requires that 



00 00 00 



CkSo , V 0-M"2fc _ l5'o , V iMzfc , V lA^'s* 



r ^[r+{2kdy]'^ r Z^ [k + (2kdyV^' Zj [r + (2M)=]''^ 



k=l k=l k=l 



For this equality to hold for all values of r the corresponding terms of the summation, 

 i.e., those having the same denominators, must be equal. This gives 



00 -^ 16 

 0M21C = iM2k + 1N2W for ^ = 1, 2, 3 • • • 



„. . ,.^. 1 9Fo_ 1 dVx . ^, ^ 



Ihe second condition, — — — , requires that 



po oz pi OS 



00 



I ] oSoZ 1 V olVhk(2kd — z) f _ I ] iSoZ 



po j [r^ + ."=]% ^Ir+iZkd-zryn p,) [r + z"-r 



( k = i J 2=0 ( 



00 CO ) 



Y xM2u(2kd-z) _ Y _J:hjA2kd±z)__ I 

 Zd [r'+ (2kd-sy]V^ Zi [r'+ (.2kd + sy]^i ( 



k=] k=l ) z—0 



or 



CO i ^^ "O ) 



1_ Y 0M2. (2kd) l_ \ Y J42, (2kd) Y _^Nj^i2kd)_l 



po Z, [r" + (2kdr]y^ Pi 1 Z ['-=+(2^)==]% 2^ [r + {2kdr-rH 



k=l { k=l k=l J 



The corresponding terms in both members of the equation have the same denominators ; 

 hence 



— oM2fc = - iMik 1N21C 



po pi Pi 



The boundary conditions at z = d yield another set of equations between the con- 

 stants. The condition Fi = F2 at 5 = cf requires that 



OD 00 



iSo _^ Y iMlh + V lA'':* 



[r' + d']'^ Z^ [r' + (2k — ly d"-y^ Z^ [r' + (2k + \y d']'^ 



00 



2O0 , X'' 2A'2* 



^^1 



[r' + d'V' Zj [r + (2k + ly d"]'^ 



k=l 

 Equating corresponding terms of the summation yields 



iSo + 1M2 = 25'o for k = 1. 

 and 



iA/'2fc + iA^2(*-i) = 2A%(fc-i) for ^ = 2,3,4 • • 



