Then 



where dA is the increment of area which in our case is rdrd<^. Thus 



/ J,rdrdv + / / Jsrclrdc/" = I, + l2 

 The integration on If simply yields the factor 2 T. Then 



T A ^ ^ „A\i \ /I 3d , I5d 35d , % j 

 I,- 4-7r(JadVoy^ r {- - ^^ +-g^ - ^g^ + ■■) dr 



= 47r(TaVo{l--^+|---|- + ^- ■■), (18) 



Likewise 



, . 4ngaVo / . I 3r^ , ISr" 35r^ , ., 

 ^2- ^2 / ru ^d^"^ Sd'* I6d« + jar 



- 47fcJaVo ( 2 - -Q- +-|g -|28 + ■■) • (19) 



1 = I,+l2= 47roaVo[(l --^ + ^-^+ ) 

 ^^2 8 ^16 128 ^ 'J- 



We notice that the two infinite series are exactly the same, except for sign, after the 1 of the first 

 series. Thus all terms cancel each other leaving only the term 1. Hence 



I = 47To-aVo amperes (20) 



is the total current flowing between the two electrodes. 



Resistance 



The net resistance between the electrodes is the total applied potential divided by the total 

 current. 



„ 2Vo 2Vo I , ,,,> 



R - "T~ - A^^ M - o^^ ohms (21) 



Discussion 



It is to be noted that the problem is not essentially altered if the two electrodes are half 

 submerged on the surface of an infinite body of water. The only changes will be in the total current 

 and resistance. The total current will be one-half of the value in equation (20): 



I = 2TaaVo amperes (22) 



and the total resistance will then be twice the value given in equation (21): 



R= ohms. (23) 



If as. 



These conclusions are drawn fronn the fact that any plane through the Z-axis divides the field into two 

 independent regions, for the electric field at all points in the plane is parallel to the plane. Thus, if 

 one of the regions is air instead of water, it will not affect the behavior of the field and current in the 

 other region. 



20 



