Chap. 10] 



ELECTRICAL METHODS 



719 



due to the image of —7 at C'^ is kIpi/2T{2d — a), where k = 

 (P2 — Pi)/{p2 + Pi)- The total potential at Pi is therefore 



^' 27r [\a 2a) '^^\2d-h 2a 2d - a)] ' 

 Similarly, the potential at Pa is 



^' " 27L\2^ ~ a) "^ ^ V2j+~a ~ 2d - 2a) J 

 Hence we have for the potential difference: 



Fi - 72 = 



[PL 

 2Tra 



1 +4A; 



i^-i) 



/J 



(10-40a) 



(10-406) 



(10-40c) 



so that the apparent resistivity becomes 



d/ 1 



P« = Pi 



1 +4fc 



4^-4 4^-1 



(10-40d) 



(6) In case 2 (one current electrode in the second medium) the potential 

 difference Vi — V^ due to the source / at Ci and its image at C( will be 

 the same as in case 1. However, the potential in medium (pi) due to the 

 sink — / at C2 in medium (pa) must be considered as being due to a source 

 — 2pi//(p2 + pi) (see eq. (10-32<i), and the potential at Pi will be the sum 

 of the potentials due to a source 7 at Ci , to an image A;7 at Ci , and to a 

 source (/: — 1) 7 at C2 : 



= 1 r (^ ^ fe \ P2(fc - 1) 1 



' 2^ L'' \« 2d + 2a/ 2a J* 



Fi = 



The potential at P2 is accordingly 



(10-40e) 



L\ (L^ Jl\ 4- P2(fc - 1) 1 



l',c[f^\2a'^ 2d + a)^ a J 



^' = %rV'\2-a^2d + 

 From the difference the apparent resistivity is 



/c/. . 1 2 



Ps = Pi 



(10-40/) 



(c) In case '3 (two electrodes in each medium) the potential at Pi is 

 due to the source 7 at Ci , A;7 at Ci , and — 2pi7/(p2 + pi) = {k — 1)7 at 



