ELECTRICAL METHODS 487 



and 



aM2» = (1 + 00 iM« for & = 1, 2, 3 



iiV«= 01 iMjfc for ^ = 1, 2. 3 • • • • 

 siVzok-j) = (1 + 02) iN2ik-i) for ^ = 2, 3, 4 • • • • 

 1^2* = 02 iN2(k-i) ior k = 2,3,4 ■ ■ • • 



1M2 = 02 l6"o 

 25-0= (1 + 02) 16*0 



Also, c — c 



lO — 00 



These equations may be expressed in terms of one parameter So as follows : 



lN2H= 01lM2*=0l02 lAr2(*-l)= 01*02 lil/2(*-l) = (0102)SA/'2(*.2) 



or 



liV2»= (0102)",Ar2(fc-„) 



In particular, for k — ti — 1 or n = k — 1 



iN,u = ( 01 02) "^-STV^ = 01 ( 01 02) "^-NiW2 = ( 01 02) * i5"o = ( 01 02) * o5-o 

 Also, 



1M2* = ^ = -^^^^ -^^ = 0^*" 0^' -S-" 



0M2k= (1 + 00 liW2\= (1 + 00 (01*-' 02*) o5o 

 2A^2.= (1+00 lA^2fc= (1+00 (01 00" 06-0 



iSo= (1 + 0O<hS'o 



It remains now to find o^S'o itself. Consider a very small sphere surrounding the 

 real source of current So or o5"o. The total current is the sum of two currents : 

 namely, h which flows into medium through a hemisphere and /i which flows into 

 medium I also through a hemisphere. If the radius n of the sphere is negligible 

 relative to d, all terms except the first may be neglected in the expressions for Vo and 

 Vi. That is, _ „ 



jy. OOP OOP 



''^"" (r* + 5*)^~ n 

 and ^ f, 



J. , 10 p poo 



where n = radius of sphere = (r* + s*) ^ 

 The current flowing into medium is 



r 1 O K p „ „ - PO p 



/o -:: — • ZttTi- — Ztt 



po on po 



while the current flowing into medium I is 



. _ 1 dVl 2_^ oSo 



h— — ^ — • 27rri = 2ir 



pi ori pi 



The total current is thus 



so that 



I = Io + Ii = 27r(—+—)oSo 

 V Po pi / 



"^"'^ 



\ Po pi / 



p5o = y-pi(l + 0O 

 Hit 



All the potentials can now be written in terms of I, 0i, and 02. We shall only 



