# Note ```Potential Fields in Electrical Methods
An important class of methods used for exploration rely on measuring the
response of the Earth to applied potential differences (and hence injection
of current into the ground).
We will see that this subject is intimately related to the study of potential
fields.
Ohm’s Law
We begin by reviewing electrical resistance, resistivity and conductivity.
Ohm’s law states
V = IR
(1)
where V is the applied potential difference, R is the resistance and I the
current that flows. R is an extrinsic variable, for a continuous medium we
need an intrinsic variable — this is ρ, the resistivity, the resistance of a 1m3
block of material. In general,
ρL
A
R=
(2)
We write V = IR in continuous form:
V =
⇒E=
ρL
I
A
V
ρI
=
= ρJ
L
A
where
• E=Electric field strength (potential gradient)
• J=Current density (Am−1 )
• ρ is resistivity in Ωm.
• σ = ρ−1 is conductivity in Ω−1 m−1 or Siemens/m (SI)
1
(3)
(4)
So we can also write J = σE, a convenient form of Ohm’s Law.
We can also record the relationship between E and the potential in vector
form:
E = −∇V
(5)
where the minus signs signifies that Electric fields, and the associated currents, point in the direction of decreasing potential, whereas ∇V gives the
direction of increasing potential.
Point electrode
We now consider current I injected into a uniform half-space, of electrical
conductivity σ or resistivity ρ.
It is clear that the current spreads out in all directions beneath the
surface (and none flows across the surface). Hence we can write that the
current density J as a function of radius r is
Jr =
I
2πr 2
(6)
1
Now E = −∇V = − ∂V
∂r and J = ρ E so
Jr = −
1 ∂V
ρ ∂r
⇒
∂V
ρI
=−
∂r
2πr 2
(7)
and thus by integrating we find the potential V is
ρI
for point source at the surface of a half-space
(8)
2πr
Apart from the factor of 2π rather than 4π, this looks like the fundamental potential that we have seen before in a gravitational context, and
the reason is that indeed the two are linked by potential theory. Note that
the factor of 2 (rather than 4) comes from the fact that in a half-space no
current can flow above z = 0, and there is a doubling of the current and the
potential below the surface. If one were to embed a point source of current
deep within an infinite medium, one would find that the current flows in all
directions, and that the potential would be
V =
V =
ρI
4πr
for point source in an infinite medium
2
(9)
Resistivity surveying
The principle of superposition
To use a single electrode we would need to apply voltage V relative
to zero, which is at infinity and thus inconvenient. It is easier to apply a
potential difference between two closely spaced electrodes supplying current
Iin and Iout .
Potentials due to the electrodes at A and B are
VA =
Iρ
;
2πr1
VB = −
Iρ
V = VA + VB =
2π
Iρ
2πr2
1
1
−
r1 r2
(10)
(11)
One might imagine that an experiment could be set up with just these
two electrodes, and that V and I could be measured and the resistivity
ρ deduced. This is not a good idea!. This is because there is typically a
huge contact resistance (or stake resistance) between the electrodes and the
ground. The electrode is typically a steel stake, and the amount of current
entering depends greatly on the contact between the stake and the individual
soil/sand/ice particles.
A better way is to use 4 electrodes — 2 current electrodes (usually called
A and B) and 2 voltage electrodes (usually called M and N).
We measure the voltage between M and N with a high impedance voltmeter — thus no current is drawn, and thus the stake resistance at M and
N is irrelevant. Similarly, we know exactly that the current entering the
ground at A and B is I and −I respectively, and the stake resistances here
are irrelevant.
3
Consider uniform Earth
Measure I between A &amp; B ad potential difference ∆V between M &amp; N.
Hence we calculate the resistance R = ∆V /I.
The potentials at the electrodes are
VM =
Iρ
2π
1
1
−
AM
BM
VN =
Iρ
2π
1
1
−
AN
BN
So
∆V = VM − VN =
Iρ
2π
(12)
(13)
1
1
1
1
−
−
+
AM
BM
AN
BN
(14)
which means we can deduce the true resistivity
ρ= 2πR
1
AM
−
1
BM
−
1
AN
+
1
BN
= GR
(15)
where G is the geometrical factor which is simply a function of the electrode
configuration.
Non-uniform earth
We still continue to compute the above expression — we call it the “apparent
resitivity” ρa , and it is plotted as a function of electrode separation.
In this case it is a complicated function of all the rocks through which
the current has passed. Nevertheless, in the case that the Earth is uniform
the apparent resistivity curve will be flat and equal to the true resistivity.
In other cases it can tend to the values of one or more of the layers.
One of the simplest techniques for interpreting apparent resitivity curves
is to compare them with synthetic curves which have been computed for
different scenarios — in other words, solving the “forward problem”. The
inverse problem of deducing the conductivity as a function of depth from a
curve of ρa versus electrode spacing is much harder.
In a heterogeneous earth composed of layers, in order to compute the
response we make use of the fact that the potential must always satisfy
∇ &middot; (σ∇V ) = 0.
(16)
Therefore in each layer, in which σ is constant, we have ∇2 V = 0. Hence
Laplace’s equation is satisfied in each layer, and we construct the total solution from the individual solutions in each layer, but ensuring that the
matching conditions at each interface are satisfied.
4
```