the creation of a collimated beam of charge, as occurs in an electron tube where a series of permanent magnets can be used to create a beam of steady current More typically, steady currents are created using wire conductors to guide the moving charge. When an external force, such as the electric field created by a battery, is applie to an uncharged conductor, the free electrons will begin to move through the positive lattice, forming a current. Each electron moves only a short distance before colliding with the positive lattice, and if the wire is bent into a loop the resulting macroscopic current will be steady in the sense that the temporally and spatially averaged microscopic current will obey V.J=0. We note from the examples above that any charges attempting to leave the surface of the wire are drawn back by the electrostatic force produced by th esulting imbalance in electrical charge. For conductors, the drift "velocity associated with the moving electrons is proportional to the applied field ud=-HeE where ue is the electron mobility. The mobility of copper (3.2 x 10-'m2/V. s) is such that an applied field of 1 V/m results in a drift velocity of only a third of a centimeter per second Integral properties of a steady current. Steady currents obey several useful inte- gral properties. To develop these properties we need an integral identity. Let f(r) and g(r)be scalar functions, continuous and with continuous derivatives in a volume region V. Let J represent a steady current field of finite extent, completely contained within V. We begin by using(B 42) to expand v·(fgJ=fg(V·J)+J.V(fg) Noting that V.J=0 and using(B 41), we get V·(fgJ)=(fJ)·Vg+(gJ)·Vf Now let us integrate over V and employ the divergence theorem fg)J·d (fJ)·Vg+(gJ)·VfdV. Since J is contained entirely within S, we must have f J=0 everywhere on S. Hence (fJ)·Vg+(gJ·Vfdv=0. We can obtain a useful relation by letting f =l and g=x; in(3. 25), where(x, y, 2)= Ji(rdV=0, (326) where Ji=Jx and so on. Hence the volume integral of any rectangular component of J is zero. Similarly, letting f=g=x; we find that Ji(r)dV=0 (327) With f = xi and g=xj we obtain Lx J; (r)+xj Ji (r)dv=0 ②2001 by CRC Press LLC
the creation of a collimated beam of charge, as occurs in an electron tube where a series of permanent magnets can be used to create a beam of steady current. More typically, steady currents are created using wire conductors to guide the moving charge. When an external force, such as the electric field created by a battery, is applied to an uncharged conductor, the free electrons will begin to move through the positive lattice, forming a current. Each electron moves only a short distance before colliding with the positive lattice, and if the wire is bent into a loop the resulting macroscopic current will be steady in the sense that the temporally and spatially averaged microscopic current will obey ∇ · J = 0. We note from the examples above that any charges attempting to leave the surface of the wire are drawn back by the electrostatic force produced by the resulting imbalance in electrical charge. For conductors, the “drift” velocity associated with the moving electrons is proportional to the applied field: ud = −µeE where µe is the electron mobility. The mobility of copper (3.2 × 10−3m2/V · s)is such that an applied field of 1 V/m results in a drift velocity of only a third of a centimeter per second. Integral properties of a steady current. Steady currents obey several useful integral properties. To develop these properties we need an integral identity. Let f (r) and g(r) be scalar functions, continuous and with continuous derivatives in a volume region V. Let J represent a steady current field of finite extent, completely contained within V. We begin by using (B.42)to expand ∇ · ( f gJ) = f g(∇ · J) + J · ∇( f g). Noting that ∇ · J = 0 and using (B.41), we get ∇ · ( f gJ) = ( f J) · ∇g + (gJ) · ∇ f. Now let us integrate over V and employ the divergence theorem: S ( f g)J · dS = V [( f J) · ∇g + (gJ) · ∇ f ] dV. Since J is contained entirely within S, we must have nˆ · J = 0 everywhere on S. Hence V [( f J) · ∇g + (gJ) · ∇ f ] dV = 0. (3.25) We can obtain a useful relation by letting f = 1 and g = xi in (3.25), where (x, y,z) = (x1, x2, x3). This gives V Ji(r) dV = 0, (3.26) where J1 = Jx and so on. Hence the volume integral of any rectangular component of J is zero. Similarly, letting f = g = xi we find that V xi Ji(r) dV = 0. (3.27) With f = xi and g = x j we obtain V xiJj(r) + x jJi(r) dV = 0. (3.28)
3.2 Electrostatics 3.2.1 The electrostatic potential and work The equation Ed=0 satisfied by the electrostatic field E(r)is particularly interesting. A field with zero circulation is said to be conservative. To see why, let us examine the work required to move a particle of charge Q around a closed path in the presence of e(r). Since work is the line integral of force and b=0, the work expended by the external system moving the charge against the lorentz for rce Is W=-∮(QE+QvxB)·d=-Q∮E·d=0 This property is analogous to the conservation property for a classical gravitational field: ly potential energy gained by raising a point mass is lost when the mass is lowered Direct experimental verification of the electrostatic conservative property is difficult aside from the fact that the motion of Q may alter e by interacting with the sources of E. By moving Q with nonuniform velocity (i.e, with acceleration at the beginning of the loop, direction changes in transit, and deceleration at the end) we observe a radiative loss of energy, and this energy cannot be regained by the mechanical system providing the motion. To avoid this problem we may assume that the charge is moved so slowly, or in such small increments, that it does not radiate. We shall use this concept later te determine the"assembly energy"in a charge distribution The electrostatic potential. By the point form of (3.29) V×E(r)=0 we can introduce a scalar field p= (r) such that E(r)=-VΦ(r) (330) The function p carries units of volts and is known as the electrostatic potential. Let us consider the work expended by an external agent in moving a charge between points Pl at ri and P, at r Q r):d=Q/d(r)=Q(r)-(r) The work W2l is clearly independent of the path taken between Pi and P2; the quantit d(r2) m)= (331) called the potential difference, has an obvious physical meaning as work per unit charge required to move a particle against an electric field between two point ②2001 by CRC Press LLC
3.2 Electrostatics 3.2.1 The electrostatic potential and work The equation E · dl = 0 (3.29) satisfied by the electrostatic field E(r) is particularly interesting. A field with zero circulation is said to be conservative. To see why, let us examine the work required to move a particle of charge Q around a closed path in the presence of E(r). Since work is the line integral of force and B = 0, the work expended by the external system moving the charge against the Lorentz force is W = − (QE + Qv × B) · dl = −Q E · dl = 0. This property is analogous to the conservation property for a classical gravitational field: any potential energy gained by raising a point mass is lost when the mass is lowered. Direct experimental verification of the electrostatic conservative property is difficult, aside from the fact that the motion of Q may alter E by interacting with the sources of E. By moving Q with nonuniform velocity (i.e., with acceleration at the beginning of the loop, direction changes in transit, and deceleration at the end)we observe a radiative loss of energy, and this energy cannot be regained by the mechanical system providing the motion. To avoid this problem we may assume that the charge is moved so slowly, or in such small increments, that it does not radiate. We shall use this concept later to determine the “assembly energy” in a charge distribution. The electrostatic potential. By the point form of (3.29), ∇ × E(r) = 0, we can introduce a scalar field = (r) such that E(r) = −∇(r). (3.30) The function carries units of volts and is known as the electrostatic potential. Let us consider the work expended by an external agent in moving a charge between points P1 at r1 and P2 at r2: W21 = −Q P2 P1 −∇(r) · dl = Q P2 P1 d(r) = Q [(r2) − (r1)] . The work W21 is clearly independent of the path taken between P1 and P2; the quantity V21 = W21 Q = (r2) − (r1) = − P2 P1 E · dl, (3.31) called the potential difference, has an obvious physical meaning as work per unit charge required to move a particle against an electric field between two points.
T1 Figure 3.3: Demonstration of path independence of the electric field line integral Of course, the large-scale form(3.29 )also implies the path-independence of work in the electrostatic field. Indeed, we may pass an arbitrary closed contour r through Pr and P2 and then split it into two pieces TI and T2 as shown in Figure 3.3. Since EdI=-Q/EdI+0/EdI=0. 1-r2 we hav eE.dI=-Q/E We sometimes refer to p(r) as the absolute electrostatic potential. Choosing a suitable reference point Po at location ro and writing the potential difference as V21=[(r2)-Φ(ro)-[Φ(ri)-中(ro)] we can justify calling p(r) the absolute potential referred to Po. Note that Po might describe a locus of points, rather than a single point, since many points can be at the same value of E found from( 3.30), for simplicity we often choose ro such that (ro)=0. potential. Although we can choose any reference point without changing the result Several properties of the electrostatic potential make lectric fields. We know that, at equilibrium, the electrostatic field within a conducting body must vanish. By(3. 30) the potential at all points within the body must therefore have the same constant value. It follows that the surface of a conductor is an equipotential surface: a surface for which p(r) is constant As an infinite reservoir of charge that can be tapped through a filamentary conductor the entity we call"ground"must also be an equipotential object. If we connect a con- ductor to ground, we have seen that charge may flow freely onto the conductor. Since no work is expended, "grounding a conductor obviously places the conductor at the same absolute potential as ground. For this reason, ground is often assigned the role as the potential reference with an absolute potential of zero volts. Later we shall see that for sources of finite extent ground must be located at infinit ②2001 by CRC Press LLC
Figure 3.3: Demonstration of path independence of the electric field line integral. Of course, the large-scale form (3.29)also implies the path-independence of work in the electrostatic field. Indeed, we may pass an arbitrary closed contour through P1 and P2 and then split it into two pieces 1 and 2 as shown in Figure 3.3. Since −Q 1−2 E · dl = −Q 1 E · dl + Q 2 E · dl = 0, we have −Q 1 E · dl = −Q 2 E · dl as desired. We sometimes refer to (r) as the absolute electrostatic potential. Choosing a suitable reference point P0 at location r0 and writing the potential difference as V21 = [(r2) − (r0)] − [(r1) − (r0)], we can justify calling (r) the absolute potential referred to P0. Note that P0 might describe a locus of points, rather than a single point, since many points can be at the same potential. Although we can choose any reference point without changing the resulting value of E found from (3.30), for simplicity we often choose r0 such that (r0) = 0. Several properties of the electrostatic potential make it convenient for describing static electric fields. We know that, at equilibrium, the electrostatic field within a conducting body must vanish. By (3.30)the potential at all points within the body must therefore have the same constant value. It follows that the surface of a conductor is an equipotential surface: a surface for which (r) is constant. As an infinite reservoir of charge that can be tapped through a filamentary conductor, the entity we call “ground” must also be an equipotential object. If we connect a conductor to ground, we have seen that charge may flow freely onto the conductor. Since no work is expended, “grounding” a conductor obviously places the conductor at the same absolute potential as ground. For this reason, ground is often assigned the role as the potential reference with an absolute potential of zero volts. Later we shall see that for sources of finite extent ground must be located at infinity.
3.2.2 Boundary conditions Boundary conditions for the electrostatic field. The boundary conditions found for the dynamic electric field remain valid in the electrostatic case. Thus n12×(E1-E2) (3.32) Here f12 points into region 1 from region 2. Because the static curl and divergence equations are independent, so are the boundary conditions(3. 32)and(3. 33) For a linear and isotropic dielectric where D=EE, equation(3. 33)becomes (∈E1-∈2E2)=p (334) Alternatively, using D=EoE+P we can write (3. 33)as (Ps pps +p (335) n. P is the polarization surface charge with f pointing outward from the material body We can also write the boundary conditions in terms of the electrostatic potential. with E=-Vq, equation(3. 32) becomes d1(r)=Φ2(r) for all points r on the surface. Actually p and p2 may differ by a constant; because this constant is eliminated when the gradient is taken to find E, it is generally ignored We can write (3. 35)as dd1a中2 =-Ps-pPsl-PPs2 where the normal derivative is taken in the f12 direction. For a linear, isotropic dielectric (337) Again, we note that (3.36)and (3. 37) are independent Boundary conditions for steady electric current. The boundary condition on the normal component of current found in 8 2.8.2 remains valid in the steady current case. Assume that the boundary exists between two linear, isotropic conducting regions having onstitutive parameters(E1, 01) and(E2, 2), respectively. By(2.198)we have n12·(J1-J2)=-Vx·J (338) where f12 points into region 1 from region 2. A surface current will not appear on the boundary between two regions having finite conductivity, although a surface charge may cumulate there during the transient period when the currents are established 31]. If is influenced to move from the surface, it will move into the adjacent regions, ②2001 by CRC Press LLC
3.2.2 Boundary conditions Boundary conditions for the electrostatic field. The boundary conditions found for the dynamic electric field remain valid in the electrostatic case. Thus nˆ 12 × (E1 − E2) = 0 (3.32) and nˆ 12 · (D1 − D2) = ρs. (3.33) Here nˆ 12 points into region 1 from region 2. Because the static curl and divergence equations are independent, so are the boundary conditions (3.32)and (3.33). For a linear and isotropic dielectric where D = E, equation (3.33)becomes nˆ 12 · ( 1E1 − 2E2) = ρs. (3.34) Alternatively, using D = 0E + P we can write (3.33)as nˆ 12 · (E1 − E2) = 1 0 (ρs + ρPs1 + ρPs2) (3.35) where ρPs = nˆ · P is the polarization surface charge with nˆ pointing outward from the material body. We can also write the boundary conditions in terms of the electrostatic potential. With E = −∇, equation (3.32)becomes 1(r) = 2(r) (3.36) for all points r on the surface. Actually 1 and 2 may differ by a constant; because this constant is eliminated when the gradient is taken to find E, it is generally ignored. We can write (3.35)as 0 ∂1 ∂n − ∂2 ∂n = −ρs − ρPs1 − ρPs2 where the normal derivative is taken in the nˆ 12 direction. For a linear, isotropic dielectric (3.33)becomes 1 ∂1 ∂n − 2 ∂2 ∂n = −ρs. (3.37) Again, we note that (3.36)and (3.37)are independent. Boundary conditions for steady electric current. The boundary condition on the normal component of current found in § 2.8.2 remains valid in the steady current case. Assume that the boundary exists between two linear, isotropic conducting regions having constitutive parameters ( 1,σ1) and ( 2,σ2), respectively. By (2.198) we have nˆ 12 · (J1 − J2) = −∇s · Js (3.38) where nˆ 12 points into region 1 from region 2. A surface current will not appear on the boundary between two regions having finite conductivity, although a surface charge may accumulate there during the transient period when the currents are established [31]. If charge is influenced to move from the surface, it will move into the adjacent regions
(02,E2) Figure 3.4: Refraction of steady current at a material interface ather than along the surface, and a new charge will replace it, supplied by the current for finite conducting regions (3. 38)becomes n12·(J1-J2)=0 (339) a boundary condition on the tangential component of current can also be found Substituting E=J/o into(3. 32) we have JI J2 = We can also write this as Ju J2e (3.40) he We may combine the boundary conditions for the normal components of current electric field to better understand the behavior of current at a material boundary stituting E=J/o into ( 3. 34)we hay EIJun-EJ2n=p (341) where Jin= f12 JI and J2n =112. J2. Combining (3.41)with(3. 39), we have E1n(∈1 where Unless E102-01E2=0, a surface charge will exist on the interface between dissimilar urrent-carrying conductor We may also combine the vector components of current on each side of the boundary to determine the effects of the boundary on current direction(Figure 3. 4). Let 81.2 denote en J1.2 and f12 so that Jin = Ji cos 01, Jir= Ju sin 81 J2= J2 sin g. ②2001 by CRC Press LLC
Figure 3.4: Refraction of steady current at a material interface. rather than along the surface, and a new charge will replace it, supplied by the current. Thus, for finite conducting regions (3.38)becomes nˆ 12 · (J1 − J2) = 0. (3.39) A boundary condition on the tangential component of current can also be found. Substituting E = J/σ into (3.32)we have nˆ 12 × J1 σ1 − J2 σ2 = 0. We can also write this as J1t σ1 = J2t σ2 (3.40) where J1t = nˆ 12 × J1, J2t = nˆ 12 × J2. We may combine the boundary conditions for the normal components of current and electric field to better understand the behavior of current at a material boundary. Substituting E = J/σ into (3.34)we have 1 σ1 J1n − 2 σ2 J2n = ρs (3.41) where J1n = nˆ 12 · J1 and J2n = nˆ 12 · J2. Combining (3.41)with (3.39), we have ρs = J1n 1 σ1 − 2 σ2 = E1n 1 − σ1 σ2 2 = J2n 1 σ1 − 2 σ2 = E2n 1 σ2 σ1 − 2 where E1n = nˆ 12 · E1, E2n = nˆ 12 · E2. Unless 1σ2 − σ1 2 = 0, a surface charge will exist on the interface between dissimilar current-carrying conductors. We may also combine the vector components of current on each side of the boundary to determine the effects of the boundary on current direction (Figure 3.4). Let θ1,2 denote the angle between J1,2 and nˆ 12 so that J1n = J1 cos θ1, J1t = J1 sin θ1 J2n = J2 cos θ2, J2t = J2 sin θ2