THE SMOOTH SOUNDING GRAPH. A Manual for Field Work ... - BGR

1.2. The homogeneous underground
7
The homogeneous underground represents an electrical conductive half-
space with the resistivity ρ. It is limited by the earth's surface (insulator).
A current electrode A will be placed at the earth's surface. A second elec-
trode B is moved to infinity (this is necessary in order to complete the
current circle). If the electrodes are supplied by a direct voltage, then a
direct current I will flow through the earth. At the electrode in point A the
current spreads radially (Fig.2). Now we consider in the subsurface the
skin of a hemisphere with radius r and thickness dr, which has its central
point in A. We then determine the resistance of this hemispherical skin
with respect to the current running radially from the "point source" in A.
We apply Ohm's law for the wire (equ.1)
R =
a
ρ
q
The length a corresponds to the thickness dr of the spherical skin. The
cross-section q corresponds to the surface of the hemisphere with radius
r, which is 2πr 2 , because the surface of the whole sphere is 4πr 2 . The re-
sistivity is that of the homogeneous earth. Therefore the resistance dR of
this thin skin is:
ρ
π 2
dr
dR = (4)
2 r
The next step will be to determine the resistance of a thick hemispherical
body with an inner radius r1 and an outer radius r2 (Fig.3). We do this by
summing up hemispherical skins the radius of which arises from r1 up to
r2.
This is a simple integration of the skin-resistances dR from r1 to r2. Those
who are not familiar with solving simple integrals will find the used for-
mula in elementary mathematical tables given e.g. in any technical man-
ual. The resistance R1,2 of our hemispherical body is, using equation (4),
R
1,
2
r
2
2
ρ 1
= ∫ dR =
2 ∫ dr
π r
r1
r
r1
=
ρ ⎡ 1⎤
⎢
−
2π
⎣ r ⎥
⎦
r2
r1
=
ρ ⎛ 1 1
⎜ −
2π
⎝ r1
r2
⎟ ⎞
⎠
(5)