Static Electromagnetic Fields in Matter
7.4 Boundary Conditions for the Static Fields
accelerate briefly in the direction of (or opposite to) the field before being scattered by other relatively immobile components. The random component of the carriers’
velocity yields no net current, but the short, directed segments yield a drift velocity along the field (for isotropic conductors). This leads us to postulate that
J=g(E)E (7–51)
For many materials,g(E) is almost independent ofE, in which case the material is labelled ohmic with constitutive relation J =g E. The constant g is called the conductivity of the material and is generally a function of temperature as well as dislocations in the material. (Many authors useσto denote the conductivity.) The resistivityη ≡1/g is also frequently employed.
A rough microscopic description can be given in terms of the carriers’ mean time between collisions, τ, since v 12aτ = 12q Eτ /m. The quantity qτ /m is commonly called the carrier mobility. Computing the net current density as J = nqv= 12(nq2τ /m)E wheren is the carrier number density, leads us to write the conductivity as
g=nq2τ
2m (7–52)
When a magnetic field is present, we expect the current to be influenced by the magnetic force on the charge carriers. The modified law of conduction should read
J=g(E +v ×B) (7–53)
This form of the conduction law governs the decay of magnetic fields in conduc- tors. Substituting (7-53) into Amp`ere’s law and assuming, that inside the conductor
∂ E/∂tis sufficiently small to ignore, we have ∇ × B =µ J. Taking the curl once more we obtain
∇2B =gµ
∂ B
∂t −∇ × (v ×B)
(7–54) If the conductor is stationary, the equation above reduces to a diffusion equation.
The time-independent simplified Maxwell equations (3–27) are modified in the pres- ence of matter to read
∇ · D =ρ ∇ × E = 0
∇ · B = 0 ∇ × H =J
(7–55)
whereD =ε0E+P andB =µ0(H +M). The first and the last of these equations may be integrated to give, respectively, Gauss’ law
S
D ·dS=
τ
ρd3r (7–56)
Figure 7.9: The Gaussian pillbox straddles the interface between two di- electrics.
and Amp`ere’s law
Γ
H ·d=
S
J·dS (7–57)
in the presence of matter. To determine the behavior of each of the fields at the interface between differing materials, we integrate each of the equations (7–55). For ease of referring to various direction, we take, without loss of generality, the x-y plane to be tangential to the dielectric interface.
∇ · D =ρ
Consider ∇ · D inside a thin pillbox of width whose flat sides lie on opposite sides of a dielectric interface, as shown in Figure 7.9. We let the charge density be described byρ=ρv(x, y, z) +σ(x, y)δ(z), whereρv is a volume charge density and σa surface charge density confined to the interface. Integrating∇ · D =ρover the volume of the pillbox, we get
τ
∇ · D d 3r=
τ
ρ d3r=
τ
ρv(x, y, z)d3r+
S
σ(x, y)dxdy (7–58) which, with the aid of the mean value theorem and the divergence theorem (20), becomes
S
D ·dS= ¯ρv·τ+
S
σ(x, y)dxdy (7–59) We break the surface integral of the electric displacement into integrals over each of the three surfaces of the pillbox to write
S
DIz(x, y, /2)dxdy−
S
DIIz (x, y,−/2)dxdy +
curved side
D ·dS = ¯ρvS+
S
σ(x, y)dxdy (7–60) In the limit of vanishing this becomes
S
(DIz−DIIz )dxdy=
S
σ(x, y)dxdy (7–61) which can hold true for arbitraryS only ifDIz−DzII =σ, or to make our conclusion coordinate independent,
DI−DII
·nˆ =σ (7–62)
Figure 7.10:The rectangular loop in (a) lies in thex-zplane, andy points into the page. In (b), the loop is placed in they-z plane leavingx to point out of the page.
where ˆn is a unit normal to the interface pointing from region II to region I. In conclusion,the perpendicular component ofD is discontinuous byσ.
∇ · B = 0
The same argument as above, may be applied to∇ · B = 0 and results in BI−BII
·ˆn= 0 (7–63)
In other words,the perpendicular component ofB is continuousacross the interface.
∇ × E = 0
We obtain the boundary condition on the tangential component ofE by integrating
∇ × E over the area of the thin loop whose two long sides lie on opposite sides of the interface illustrated in Figure 7.10a.
0 =
S
(∇ × E) ·dS= E ·d
= x0+L
x0
ExII−ExI
dx+W!E¯z(x0+L)−E¯z(x0)"
(7–64) Taking the limit asW →0, we require
x0+L
x0
ExII −ExI
dx= 0 (7–65)
which can hold for allLonly ifExII =ExI. Generally, then,the tangential component of E is continuous across the interface.
∇ × H =J
Consider∇× H inside the loop of Figure 7.10a, and let the current density consist of a body current densityJ(x, y, z) and a surface current densityj(x, y)δ(z) confined
to the interface. (Surface currents are, strictly speaking, possible only with perfect conductors, but they often constitute a good approximation to the currents on metallic interfaces.) Asj lies in thex-y plane,jz = 0. Integrating∇ × H over the surface included by the loop of figure 7.10a gives
(∇ × H)·dS=
−JydS+
−jyδ(z)dxdz
H ·d=−JyS− x0+L
x0
jydx (7–66)
Again, taking the limit asW →0, we get x0+L
x0
HxII −HxI
dx=− x0+L
x0
jydx (7–67)
independent ofx0orL. Equating the integrands, givesHxI−HxII =jy. Next, placing the loop in the z-y plane as illustrated in Figure 7.10b, we obtain a similar result for Hy: HyI−HyII =−jx. (The change in sign occurs because the x axis points out of the page, meaning thatdS = +ˆıdx dy.) We can generalize these results by writing
ˆ n×
HI−HII
=j (7–68)
In words, theparallel component ofH is discontinuous by j.
We summarize these boundary conditions below, labelling the components per- pendicular and parallel to the surface by⊥and respectively.
(D) ⊥ is discontinuous byσ (E) is continuous
(B) ⊥ is continuous (H) is discontinuous byj
(7–69)
Example 7.5: A large slab of uniform dielectric is placed in a uniform electric fieldE with its parallel faces making angleθwith the field. Determine the angle that the internal electric field makes with the faces. (Note that this is not Snell’s law, which arises from interference of waves.)
Solution: The following are the boundary conditions for the electric field: D⊥ is continuous andE is continuous. Labelling the vacuum fields with the subscriptv and those in the dielectric with subscriptd, we have in vacuumDv⊥ =ε0Evcosθv
andE=Evsinθv. The boundary conditions then translate to
ε0Evcosθv =εdEdcosθd and Evsinθv=Edsinθd (Ex 7.5.1) Dividing one equation by the other, we have tanθd= (εd/ε0) tanθv. In other words, the electric field bendsaway from the normal on entering the dielectric.
A magnetic induction field, in exactly the same fashion, also bends away from the normal when entering a medium of high permeability.
As a simple illustration of using the boundary conditions at the interface with an anisotropic medium, we reconsider the preceding example with a hypothetical dielectric having two different permittivities.
Example 7.6: A crystal whose two equivalent principal axes lie in the x-y plane while the third lies along thez axis has two distinct permittivities,ε0(1 +χxx) = ε0(1 +χyy) =εxx= 2ε0andε0(1 +χzz) =εzz = 3ε0. A large slab of this material is placed in a uniform electric field in vacuum, making angleθ with the (normal)z axis of the slab. Determine the directions ofE andD in the dielectric.
Solution: In the medium, we have, denoting the dielectric values with subscriptd Dd,x
Dd,z
=
2ε0 0 0 3ε0
Ed,x
Ed,z
=
2ε0EdsinθE,d
3ε0EdcosθE,d
(Ex 7.6.1) which relates the displacement field to the electric field. We denote the vacuum field and angles with a subscriptv. The boundary conditions
E is continuous ⇒ Evsinθv =EdsinθE,d (Ex 7.6.2) andD⊥ is continuous ⇒ ε0Evcosθv = 3ε0EdcosθE,d (Ex 7.6.3) Dividing (Ex 7.6.2) by (Ex 7.6.3) gives tanθE,d= 3 tanθv. Further, from (Ex 7.6.1) tanθD,d= 23tanθE,d. The angle of the electric displacement field is now obtained from tanθD,d = 23tanθE,d= 2 tanθv.
The electrostatic field E satisfies ∇ × E = 0, implying that E may be written E = −∇V. Furthermore, ∇ · D = ρ becomes for a linear, isotropic dielectric,
∇ · (ε E) =∇ · (−ε∇V) =ρ. Ifεis piecewise constant, then the potential,V, must satisfy Poisson’s equation,∇2V =−ρ/ε. In a region of space devoid of free charges,
∇2V = 0. The same methods as those used for boundary condition problems in vacuum may now be applied when dielectrics are involved, as illustrated below.
Example 7.7: Find the potential in the neighborhood of a sphere of radiusR, made of a linear isotropic dielectric placed in an initially uniform fieldE0.
Solution: Solving the Laplace equation in spherical polar coordinates with the z axis chosen to lie alongE0 and assuming no azimuthal angle dependence, we have
V(r < R) =
=0
ArP(cosθ) (Ex 7.7.1) We eliminate all terms that grow faster thanr1 and separate the= 1 term from the rest of the sum
V(r > R) = ∞
=0
B
r+1P(cosθ)−E0rcosθ