ELECTRICAL PROSPECTING 13 
The change in the field due to the layer of high conductivity is: 
1 Se it (d)en#O7—b" 9 To(nr) dd (26) 
0 
where 
2k?2ko2(d?2— k?2)(1—tanh h(\?— k?)}/2) 
ON) SS SS g (YF 
$( ) [22(A2— Ro?) 24 ho?(A2— k?) 1/2] [ &2(X2— Ro?) 1/24 Ro2(X2— k?)*/?tanh h(Q2— k2) 12] ( 
7) 
In order to calculate the effect of the layer on the normal field, one could cal- 
culate the components of this change in field and compare them with the 
normal field components. A simpler way to compare them is to compare the 
power flow associated with the change in field to the normal power flow. The 
changes in the field components are: 
(E,)o a CED) = ii (A) (A? eae ho?) 1/2¢-20°— ko?) 27 (Xr) dd 
0 
iky?. ¢° ) 
(H5)o ae (H3)n = alll (dA) E20) 274 (Ar) ad . 
uw Jo 
The power flow of this field through a horizontal plane in the air is: 
P= a { [(E)o— (E)a][(He)o— (Ha)n]*rar 
; (30) 
Tv 

Ro? ko 
i (Ro? — n2)1/2 | (a) | 23g). 
ue) 0 
The asterisk indicates that the complex conjugate of (Hy)o—(H¢)zn is to be 
taken. 
For all but very high frequencies, k? is very much greater than ky?. Under . 
this condition, the power flow is approximately 
4rkoo[(1 — R)? + I?] 1 
= —___ —________ log —— (31) 
(2)®yo| k|(1—-R+1) ~ R-I 
where R and 7 are real, and 

i+ 
1/2 
tanh h| k| A 
=R+il. (32) 
The normal power flow from a homogeneous earth of conductivity @ is 
4k? | k| 4 ko (Ro? — d?2)1/2\8dn 
oS COU 
pu i | R2(A2 — Ro2)1/2 + o2(A2 — 2) 1/2 | 2 
which becomes for k2>k,? 
P, = 8rko'/3yw. (34) 
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