ELECTRICAL PROSPECTING 5 
o,0V,/dz = o20V2/dz. (3.4) 
(3) No current flow out of the surface of the earth 
0V,/az = 0 at z= 0. (3.5) 
(4) The conditions near the electrode 
I 
= OS OW (3.6) 
Qmay(r? + 22) 12 
(5) The conditions at °% 
Vi—0as ro 
(3.7) 
V2— 0 as 712 OO. 
This problem has usually been solved by the method of images, which results 
in an infinite series. An integral solution which is often more useful! may be 
obtained from the following solution of V?7V=0 
V = [[AQder + fOe*Voonyan (4) 
where f; and fe may be any functions of \ subject to the condition that the 
integral converges. Thus if 

IT [--) 
vale if [(A(d) — 1)e* + Ae ]Jo(ar)ar (5) 
24a 1 0 
and 
Vee il B(d)e™J o(dr) dd (6) 
where A (A) and B(A) are arbitrary functions of ) all the conditions (3) will be 
satisfied except the conditions at the boundary z=a. Of course, A(A) and 
B(A) must be chosen in such a manner that the integrals converge. These 
functions are determined by substituting V; and V2 given by Eqs. (5) and 
(6) in the boundary conditions (3.3) and (3.4). The solution for A (A) is 

1 
A(\) = are (7) 
where 
b i 02 — Oj 
02 ap O1 
On the surface of the earth, z =0, the potential is 
algae f a oe i | To(dr)dr (8) 
2moiJo0 L1 + ke 
1 This method is especially suitable when a number of horizontal layers are present. 
149 
