Static Electromagnetic Fields in Matter
7.1 The Electric Field Due to a Polarized Dielectric
Chapter 7
Figure 7.1:The divergent polarization leads to a concentration of negative charge that acts as a source of electric field.
=− 1
4πε0 ρ(r)∇2 1
|r−r|+P(r)·∇
∇2 1
|r−r|
d3r (7–3) Using the relation (26),∇2
1/|r−r|
=−4πδ(r−r), we have
∇ · E = 1 ε0
$
ρ(r)δ(r−r) +P( r)·∇δ(r−r)
%
d3r (7–4) The first term of the integral in (7–4) integrates without difficulty, and the second part can be expanded in Cartesian coordinates
P(r)·∇δ(r−r) =
Px(r) ∂
∂x[δ(x−x)δ(y−y)δ(z−z)]dxdydz +
Py(r) ∂
∂yδ(r−r)d3r+
Pz(r) ∂
∂zδ(r−r)d3r(7–5) With the help of∞
−∞f(x)δ(x−a)dx=−f(a) we integrate each term. Focusing our attention on the first integral on the right hand side of (7–5) we obtain
Px(r) ∂
∂xδ(x−x)δ(y−y)δ(z−z)dxdydz=
Px(x, y, z) ∂
∂xδ(x−x)dx
=−∂Px(x, y, z)
∂x =−∂Px(r)
∂x (7–6) In exactly the same fashion, the second and third integrals of (7–5) integrate re- spectively to
−∂Py(r)
∂y and −∂Pz(r)
∂z (7–7)
to give
∇ · E = ρ
ε0 −∇ · P( r)
ε0 (7–8)
Figure 7.2: At the boundary of a polarized dielectric the exposed ends of the dipoles generate a bound surface charge equal toP ·ˆn, where ˆnis the outward facing normal.
In other words,−∇ · P acts as a source of electric field and is often given the name bound charge. Its origin is physically obvious if we consider a collection of dipoles having a nonzero divergence. The divergence ofP in Figure 7.1 is positive, resulting in an accumulation of negative charge in the center.
At discontinuities inP, as at the boundaries of dielectrics, we get an accumu- lation of bound surface charge even for uniform polarizations (Figure 7.2). This observation is readily verified from equation (7–8). Setting the free charge to zero, we write
∇ · E =−∇ · P
ε0 (7–9)
Applying Gauss’ law to a small flat pillbox of Figure 7.3 with top and bottom surface parallel to the dielectric surface, we find
τ
∇ · E d 3r=−
τ
∇ · P
ε0 d3r (7–10)
whence,
S
E ·dS=−1
ε0 SP ·dS (7–11)
Noting that the surface of the pillbox inside the dielectric is directed opposite to the outward-facing normal ˆn, we find
(Ee−Ei)·Se=−1
ε0P ·Si= P ·Se
ε0 (7–12)
Figure 7.3:The Gaussian pillbox has ‘bottom’ surfaceSiwith normal facing into the dielectric and ‘top’ surfaceSe with normal facing out.
requiring that
(Ee−Ei)·nˆ= 1
ε0P ·ˆn (7–13)
where ˆnis a normal facing outward from the dielectric (Figure 7.3). This discon- tinuity in the field is exactly the effect that a surface charge, σ = P ·n, at theˆ surface of the dielectric would have. In total then, the electric field produced by a polarized dielectric medium is identical to that produced by a bound charge density ρb =−∇ · P and a bound surface charge densityσb = P ·nˆ on the surface. The resulting electric field may be written
E( r) = 1 4πε0
τ
$−∇·P(r)
%
(r−r)
|r−r|3 d3r+ 1 4πε0 S
(P ·ˆn)(r−r)
|r−r|3 dS (7–14) Example 7.1: Find the electric field along thez-axis of a dielectric cylinder of length Land radiusawhose axis coincides with thez axis when the cylinder is uniformly polarized along its axis.
Solution: For convenience we take one face, say the bottom, in the x-y plane.
According to (7–14) the electric field in the cylinder is given by E(z) = 1
4πε0
(P ·ˆn)(zˆk−r)
|r−r|3 dS
= 1
4πε0 a
0
2π
0
−P(zˆk−rˆr)rdrdϕ (z2+r2)3/2 + 1
4πε0 a
0
2π
0
P(zkˆ−Lkˆ−rˆr)r drdϕ [(z−L)2+r2]3/2
= Pˆk 2ε0
z
(z2+r2)1/2− (z−L) [(z−L)2+r2]1/2
0
a
0
= Pˆk 2ε0
z
(z2+a2)1/2− z−L
[(z−L)2+a2]1/2 −2 0
(Ex 7.1.1) Inside the cylinder, first two terms are both positive and less than one, meaning the electric field points in the negative z direction. If a L, the electric field becomes simplyE =Pk/εˆ 0, the field between two parallel plates carrying charge densityσ=P.
7.1.1 Empirical Description of Dielectrics
The molecules of a dielectric may be classed as either polar or nonpolar. We consider first the case of nonpolar molecules. When such molecules (or atoms) are placed in an electric field, the positive charges will move slightly in the direction of the field while the negative charges move slightly in the opposite direction, creating a polarization of the medium.
If, on the other hand, the molecules have intrinsic (permanent) dipole moments that in the absence of an electric field are randomly oriented, they will attempt to
align with the electric field, and their non-random alignment will lead to a polar- ization of the medium. In either case, the resulting polarization will be a function of the local electric field (including that of neighboring dipoles). We may write the empirical relation
Pi=P0i+χijε0Ej+χi(2)jk ε20EjEk+χi(3)jkε30EjEkE+· · · (7–15) (summation over repeated indices is implied). For isotropic materials, only the diagonal terms of thedielectric susceptibility tensorχsurvive, and (7–15) becomes merely a power series expansion for the polarization P.
Materials exhibiting large spontaneous polarization are known as ferroelectrics (clearly there must be polar molecules involved). In analogy to magnets, ferro- electric objects are known as electrets. The best known example of a ferroelectric crystal is BaTiO3. Mechanical distortions of the crystal may result in large changes of the polarization, giving rise topiezoelectricity. Similarly, changes in temperature give rise topyroelectricity.
A number of crystals have sufficiently large second or third order susceptibility that optical radiation traversing the crystal may excite a polarization with cos2ωt or cos3ωtdependence giving rise to the generation of frequency doubled or tripled light. The efficiency of such doubling or tripling would be expected to increase—
linearly for doubling or quadratically for tripling—with incident field strength.
In sufficiently small electric fields, the relationship betweenE andP for isotropic materials becomes simply
P =χε0E (7–16)
The constantχis called thelinear dielectric susceptibility of the dielectric. The ratio between the induced molecular dipole and ε0E, the polarizing field, is known as thepolarizability,α. Thus
p=αε0E (7–17)
7.1.2 Electric Displacement Field
The calculation of the microscopic fieldE arising from charges and molecular dipoles of the medium requires considerable care. It is frequently useful to think of the polarization of the medium as merely a property of the medium rather than as a source of field. To do so requires the definition of the electric displacement field.
The differential equation (7–8),
∇ · E = ρ
ε0 −∇ · P ε0 is more conveniently written
∇ · (ε0E +P) =ρ (7–18)
The quantity D ≡ε0E +P is the electric displacement field. In terms of D, Maxwell’s first equation becomes
∇ · D =ρ (7–19) The dipoles of the medium are not a source forD; only the so-called free charges act as sources.
WhenP can be adequately approximated byP =χε0E we find that
D = ε0E +χε0E =ε0(1 +χ)E =ε E (7–20) The constantεis called the permittivity of the dielectric. Thedielectric constant, κ, is defined by
κ≡ ε
ε0 = 1 +χ (7–21)
In general, because it takes time for dipoles to respond to the applied field, all three constants–χ,ε, andκ–are frequency dependent.
Anticipating the boundary condition implied by (7–19), namely thatD⊥ is con- tinuous across a dielectric interface bearing no free charge, we illustrate these ideas with a simple example.
Example 7.2: A large parallel plate capacitor has a potential V applied across its plates. A slab of dielectric with dielectric constantκfills 9/10 of the gap between its plates with air (κair = 1) filling the remaining space. Find the resulting electric field in the air and in the dielectric between the capacitor plates as well as the capacitance. Assume a separationt between the plates.
Solution: We take the lower capacitor plate to lie in thex-yplane and for simplicity assume the air layer is the top 10%. In terms of the electric fieldEa in air and the electric fieldEd in the dielectric, the potential differenceV between the plates is
V = 0.1tEa+ 0.9tEd (Ex 7.2.1)
The continuity of the perpendicular components ofD givesε0Ea =εEd, orEa = κEd. With this substitution, we solve the equation above to get
Ed = V
(0.9 + 0.1κ)t and Ea=
κ 0.9 + 0.1κ
V
t (Ex 7.2.2)
Note that the electric field in the air space isκtimes as large as that in the dielectric.
The stored charge is most easily found from the air value of the electric field sinceσ=ε0E⊥. We have then
Q=
ε0κA 0.9 + 0.1κ
V
t (Ex 7.2.3)
leading to capacitance
C= Q
V = ε0κA
(0.9 + 0.1κ)t (Ex 7.2.4)
As the fraction of air layer decreases, the capacitance tends toκtimes that of the air spaced capacitor. The electric field in the dielectric is evidently smaller than
it is in air. The diminution may be attributed to shielding produced by the aligned dipoles at the air-dielectric interface. The thin air layer, on the other hand, has a significantly increased electric field due to the dielectric, and is a major contributor to the voltage difference between the plates.