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Potential Fields in Electrical Methods

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Potential Fields in Electrical Methods 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 R = (2) A We write V = IR in continuous form: ρL V = I (3) A V ρI ⇒ E = = = ρJ (4) 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−1m−1 or Siemens/m (SI) 1 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 cur- rents, 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 I J = (6) r 2πr2 E −∇ − ∂V J 1 E Now = V = ∂r and = ρ so 1 ∂V ∂V ρI J = − ⇒ = − (7) r ρ ∂r ∂r 2πr2 and thus by integrating we find the potential V is ρI V = 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 funda- mental 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 ρI V = for point source in an infinite medium (9) 4πr 2 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 Iρ Iρ VA = ; VB = − (10) 2πr1 2πr2 Iρ 1 1 V = VA + VB = − (11) 2π r1 r2 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 volt- meter — 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 & B ad potential difference ∆V between M & N. Hence we calculate the resistance R =∆V/I. The potentials at the electrodes are Iρ 1 1 V = − (12) M 2π AM BM Iρ 1 1 V = − (13) N 2π AN BN So Iρ 1 1 1 1 ∆V = V − V = − − + (14) M N 2π AM BM AN BN which means we can deduce the true resistivity 2πR ρ = = GR (15) 1 − 1 − 1 1 AM BM AN + BN 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 ∇ · (σ∇V ) = 0. (16) Therefore in each layer, in which σ is constant, we have ∇2V = 0. Hence Laplace’s equation is satisfied in each layer, and we construct the total so- lution from the individual solutions in each layer, but ensuring that the matching conditions at each interface are satisfied. 4.
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