674 



ELECTRICAL METHODS 



[Chap. 10 



and m sin a, we note that m cos a acts at P as a vertically polarized sphere. 

 Hence, the potential at P, according to eq. (10-196), is 



Fi = 



2 



h cos 



(X2 + /l2)3/2 



The horizontal doublet m sin a is the equivalent of a horizontally polarized 

 sphere. With x instead of h as its distance from the suri"ace, we have for 

 the potential of this doublet at P 



y _ ER^ X sin a 



The total potential at P is the sum of the potentials due to the components 

 of the dipole: 



V = 



E^ 



2 



Fora = 45°, 7 = 



ER^ (h + x) 



{h cos a -\- X sin a) 



(X2 + /l2)3/2 ■ 



; and for a = 90°, 7 = 



(10-20a) 



ER' 



2 



2\/2 (a;' + h^fi^' "" ' ■ 2 (x2 + /i2)3/2 



The point of maximum potential for a — 45° is given by x = 

 h{dz\/l7 — 3)/4; 0.281/1 is a maximum and — 1.781/i a minimum point. 



At both points the current density 

 is zero (see Fig. 10-24a). By re- 

 peated differentiation, the points of 

 maximum and minimum current 

 density are found at x = 0.55/i 

 (max.), X = —0.425/1 (min.), and 

 X = —2.1/1 (max.) (see Fig. 10- 

 246). For a = 90°, the pomt of 

 maximum potential is at x = 

 ±h/\/2; and the current density 

 at X = is a minimum and at 

 X = ± 1.22/1 is a maximum. 



Simple relations may be derived 

 for the self-potential of an ore body 

 considered as a polarized bar with 

 a negative current source on its 

 upper end and a positive source on 

 its lower end. Assume the ore body to be located in the xz plane, so 

 that the vertical distance of its upper end from the surface is hi , that of 

 its lower end is /12 , its projection on the x axis is a, and the distances of its 

 ends from a surface point are n and r2 . The coordinates of this point are 

 X and y and the 0-point of the system is assumed to be directly above the 



600 



Fig. 10-246. Current density curves 

 for polarized spheres of various angles 

 of polarization. 



