672 



ELECTRICAL METHODS 



[Chap. 10 



curves of a degree of approximation which is satisfactory in practice. It is 

 evident, of course, that actual ore bodies are far from having the shapes 

 assumed here. 



Assume that the electrical field distribution is equivalent to that of a 

 vertically polarized doublet and that E is the potential difference between 

 its terminals, corresponding to the electrical charges e and —e (Fig. 10-23). 

 Then the potential at a point P is Vp = e/n - e/r2 = ein — ri)/rir-i . 

 If the diameter 2R is assumed to be small compared ^vith r, then cos 



Fig. 10-23. Vertically polarized sphere. 



di ^ (r2 — ri)/2R ^ cos 6, and r2 — n = 2R cos 6, so that Vp = 

 2eR cos d/riTi . As n ^ n , 



2eR cos 6 



Vp = 



(10-19a) 



Since the moment of the electric doublet is 2eR = m, Vp = m cos 6/r . 



E 

 The potential on the sphere^" for an angle d \& - • cos, 6; therefore, 



E 



— ' cos d = m cos ^/R^ so that m ^ ER^/2. The expression for Vp 



r''p2 



then becomes^^ Vp = -^^ cos d or, since cos d = A/Vx^ + h^, 



" 2 (a;2+/i2)3/2 



(10-196) 



To find the maximum potential, differentiate (10-196) mth respect to x: 



^^ = - f ER'hxix' + hT'^\ (10-19c) 



ax 



2" See A. Petrowsky, Inst. Prakt. Geophys. Bull. No. 1, 87 (1925). 



SI Eq. (10-196) is obtained also by applying the theory of images, that is, by 

 assuming a reflected sphere above the earth's surface. See E, Poldini, Univ. de 

 Lausanne BuU. No. 61, 21 (1938). 



