# 4.6 Conductors

The electromagnetic constitutive parameters of a material medium are its electrical permittivity ε, magnetic permeability μ, and conductivity σ. A material is said to be homogeneous if its constitutive parameters do not vary from point to point, and isotropic if they are independent of direction. Most materials are isotropic, but some crystals are not. Throughout this book, all materials are assumed to be homogeneous and isotropic. This section is concerned with σ, Section 4-7 examines ε, and discussion of μ is deferred to Chapter 5.

Materials are classified as conductors (metals) or dielectrics (insulators) according to the magnitudes of their conductivities. A conductor has a large number of loosely attached electrons in the outermost shells of its atoms. In the absence of an external electric field, these free electrons move in random directions and with varying speeds. Their random motion produces zero average current through the conductor. Upon applying an external electric field, however, the electrons migrate from one atom to the next in the direction opposite that of the external field. Their movement gives rise to a conduction current density

where σ is the material’s conductivity with units of siemen per meter (S/m).

In yet other materials, called dielectrics, the electrons are tightly bound to the atoms—so much so that it is very difficult to detach them under the influence of an electric field. Consequently, no significant conduction current can flow through them.

The conductivity σ of most metals is in the range from 10 6 to 10 7 S/m when compared with 10−10 to 10−17 S/m for good insulators (Table 4-1). A class of materials called semiconductors allow for conduction currents even though their conductivities are much smaller than those of metals. The conductivity of pure germanium, for example, is 2.2 S/m. Tabulated values of σ at room temperature (20 °C) are given in Appendix B for some common materials, and a subset is reproduced in Table 4-1.

# Table 4-1 Conductivity of some common materials at 20 °C.

Table 4 1 Full Alternative Text

Concept Question 4-14: What are the electromagnetic constitutive parameters of a material?

Concept Question 4-15: What classifies a material as a conductor, a semiconductor, or a dielectric? What is a superconductor?

Concept Question 4-16: What is the conductivity of a perfect dielectric?

# 4.6.1 Drift Velocity

The drift velocity u e of electrons in a conducting material is related to the externally applied electric field E through

u e=−μ e E(m/s), (4.74a)

where μ e is a material property call the electron mobility with units of (m 2/V⋅s). In a semiconductor, current flow is due to the movement of both electrons and holes, and since holes are positive-charge carriers, the hole drift velocity u h is in the same direction as E,

u h=μ h E(m/s), (4.74b)

where μ h is the hole mobility. The mobility accounts for the effective mass of a charged particle and the average distance over which the applied electric field can accelerate it before it is stopped by colliding with an atom and then starts accelerating all over again. From Eq. (4.11), the current density in a medium containing a volume density ρ v of charges moving with velocity u is J=ρ v u. In the most general case, the current density consists of a component J e due to electrons and a component J h due to holes. Thus, the total conduction current density is

J=J e+J h=ρ v e u e+ρ v h u h(A/m 2),(4.75)

where ρ v e=−N e e and ρ v h=N h e with N e and N h being the number of free electrons and the number of free holes per unit volume and e=1.6×10−19 C is the absolute charge of a single hole or electron. Use of Eqs. (4.74a) and (4.74b) gives

J=(−ρ v e μ e+ρ v h μ h)E=σ E,(4.76)

where the quantity inside the parentheses is defined as the conductivity of the material, σ. Thus,

and its unit is siemens per meter (S/m). For a good conductor, N h μ h≪N e μ e, and Eq. (4.77a) reduces to

That is,

Perfect dielectric:J=0,Perfect conductor:E=0.

Because σ is on the order of 10 6 S/m for most metals, such as silver, copper, gold, and aluminum (Table 4-1), it is common practice to treat them as perfect conductors and to set E=0 inside them.

A perfect conductor is an equipotential medium, meaning that the electric potential is the same at every point in the conductor. This property follows from the fact that V 21, which is the voltage difference between two points in the conductor, equals the line integral of E between them, as indicated by Eq. (4.39). Since E=0 everywhere in the perfect conductor, the voltage difference V 21=0. The fact that the conductor is an equipotential medium, however, does not necessarily imply that the potential difference between the conductor and some other conductor is zero. Each conductor is an equipotential medium, but the presence of different distributions of charges on their two surfaces can generate a potential difference between them.

# Example 4-10: Conduction Current in a Copper Wire

A 2-mm diameter copper wire with conductivity of 5.8×10 7 S/m and electron mobility of 0.0032 (m 2/V⋅s) is subjected to an electric field of 20 (mV/m). Find (a) the volume charge density of the free electrons, (b) the current density, (c) the current flowing in the wire, (d) the electron drift velocity, and (e) the volume density of the free electrons.

Solution: (a)

ρ v e=−σ μ e=−5.8×10 7 0.0032=−1.81×10 10 (C/m 3).

(b)

J=σ E=5.8×10 7×20×10−3=1.16×10 6 (A/m 2).

(c)

I=J A=J π d 2 4=1.16×10 6 π×4×10−6 4=3.64 A.

(d)

u e=−μ e E=−0.0032×20×10−3=−6.4×10−5 m/s.

The minus sign indicates that u e is in the opposite direction of E.

(e)

N e=−ρ v e e=1.81×10 10 1.6×10−19=1.13×10 29 electrons/m 3.

# 4.6.2 Resistance

To demonstrate the utility of the point form of Ohm’s law, we apply it to derive an expression for the resistance R of a conductor of length l and uniform cross section A, as shown in Fig. 4-16. The conductor axis is along the x direction and extends between points x 1 and x 2, with l=x 2−x 1. A voltage V applied across the conductor terminals establishes an electric field E=x ˆ E x; the direction of E is from the point with higher potential (point 1 in Fig. 4-16) to the point with lower potential (point 2). The relation between V and E x is obtained by applying Eq. (4.39):

V=V 1−V 2=−∫x 2 x 1 E⋅d l=−∫x 2 x 1 x ˆ E x⋅x ˆ d l=E x l(V).(4.78)

# Figure 4-16 Linear resistor of cross section A and length l connected to a dc voltage source V.

Figure 4 16 Full Alternative Text

Using Eq. (4.73), the current flowing through the cross section A at x 2 is

I=∫A J⋅d s=∫A σ E⋅d s=σ E x A(A).(4.79)

From R=V/I, the ratio of Eq. (4.78) to Eq. (4.79) gives

R=l σ A(Ω).(4.80)

We now generalize our result for R to any resistor of arbitrary shape by noting that the voltage V across the resistor is equal to the line integral of E over a path l between two specified points and the current I is equal to the flux of J through the surface S of the resistor. Thus,

The reciprocal of R is called the conductance G, and the unit of G is (Ω−1) or siemens (S). For the linear resistor,

G=1 R=σ A l(S).(4.82)

# Example 4-11: Conductance of Coaxial Cable

The radii of the inner and outer conductors of a coaxial cable of length l are a and b, respectively (Fig. 4-17). The insulation material has conductivity σ. Obtain an expression for G′, which is the conductance per unit length of the insulation layer.

# Figure 4-17 Coaxial cable of Example 4-11.

Figure 4 17 Full Alternative Text

Solution: Let I be the total current flowing radially (along r ˆ) from the inner conductor to the outer conductor through the insulation material. At any radial distance r from the axis of the center conductor, the area through which the current flows is A=2 π r l. Hence,

J=r ˆ I A=r ˆ I 2 π r l,(4.83)

and from J=σ E,

E=r ˆ I 2 π σ r l.(4.84)

In a resistor, the current flows from higher electric potential to lower potential. Hence, if J is in the r ˆ direction, the inner conductor must be at a potential higher than that at the outer conductor. Accordingly, the voltage difference between the conductors is

V a b=−∫b a E⋅d l=−∫b a I 2 π σ l r ˆ⋅r ˆ d r r=I 2 π σ l ln b a.(4.85)

The conductance per unit length is then

G′=G l=1 R l=I V a b l=2 π σ ln(b/a)(S/m).(4.86)

# 4.6.3 Joule’s Law

We now consider the power dissipated in a conducting medium in the presence of an electrostatic field E. The medium contains free electrons and holes with volume charge densities ρ v e and ρ v h, respectively. The electron and hole charge contained in an elemental volume Δ υ is q e=ρ v e Δ υ and q h=ρ v h Δ υ, respectively. The electric forces acting on q e and q h are F e=q e E=ρ v e E Δ υ and F h=q h E=ρ v h E Δ υ. The work (energy) expended by the electric field in moving q e a differential distance Δ l e and moving q h a distance Δ l h is

Δ W=F e⋅Δ l e+F h⋅Δ l h.(4.87)

Power P is measured in watts (W) and is defined as the time rate of change of energy. The power corresponding to Δ W is

Δ P=Δ W Δ t=F e⋅Δ l e Δ t+F h⋅Δ l h Δ t=F e⋅u e+F h⋅u h=(ρ v e E⋅u e+ρ v h E⋅u h)Δ υ=E⋅J Δ υ,(4.88)

where u e=Δ l e/Δ t and u h=Δ l h/Δ t are the electron and hole drift velocities, respectively. Equation (4.75) was used in the last step of the derivation leading to Eq. (4.88). For a volume υ, the total dissipated power is

and in view of Eq. (4.73),

P=∫υ σ|E|2 d υ(W).(4.90)

Equation (4.89) is a mathematical statement of Joule’s law. For the resistor example considered earlier, |E|=E x and its volume is υ=l A. Separating the volume integral in Eq. (4.90) into a product of a surface integral over A and a line integral over l, we have

P=∫υ σ|E|2 d υ=∫A σ E x d s∫l E x d l=(σ E x A)(E x l)=I V(W),(4.91)

where use was made of Eq. (4.78) for the voltage V and Eq. (4.79) for the current I. With V=I R, we obtain the familiar expression

P=I 2 R(W).(4.92)

Concept Question 4-17: What is the fundamental difference between an insulator, a semiconductor, and a conductor?

Concept Question 4-18: Show that the power dissipated in the coaxial cable of Fig. 4-17 is

P=I 2 ln(b/a)2 π σ l.