Problems
Application 6.1 A Uniformly Polarized Sphere
6.5 Simple Dielectric Matter
inject quantum mechanical and/or statistical mechanical information into classical electrodynamics.
For example, aferroelectricis a form of matter wherePcan exist in the absence ofE. We will not pause to treat this case because it is rather uncommon. Instead, we focus on the vast majority of systems which are unpolarized in the absence of a field but which acquire a uniform macroscopic polarization in the presence of a uniform external electric field. The general rule revealed by experiment is
Pi=0χijEj +0χij k(2)EjEk+ · · ·. (6.34) The tensor character of the constantsχij andχij k(2)allows for the possibility thatPis not parallel to E. This is realized in spatially anisotropic matter. The second-order and higher-order terms in (6.34) allow for the possibility that the polarization depends non-linearly on the field. This is realized in all matter when the electric field strength is large enough.
6.5 Simple Dielectric Matter
The first term on the right-hand side of (6.34) is sufficient to describe the polarization of alinear dielectric. In this book, a dielectric that is both linear and spatially isotropic will be calledsimple. A simple dielectric obeys the constitutive relation
P=0χE. (6.35)
The constant χ is called the electric susceptibility. We also introduce the permittivity and the dimensionlessdielectric constantκthrough
P=(−0)E=0(κ−1)E. (6.36)
Statistical mechanical arguments show thatχ ≥0.8 Therefore, ≥0 andκ≥1 as well. This is plausible because a dielectric incompletely screens an external field. In a simple medium, the auxiliary field defined in (6.25) becomes
D=E=κ0E=0(1+χ)E. (6.37)
Experiments show that (6.35), (6.36), and (6.37) apply equally well to macroscopic fields that vary with position in a simple medium.
Example 6.1 A point dipolep0is embedded at the center of a dielectric sphere with volumeVand dielectric constantκ. Find the total dipole momentptotof the entire system.
Solution: From (6.9) and (6.36), the dipole moment of the polarizable dielectric is p=
V
d3rP=0(κ−1)
V
d3rE.
On the other hand, we proved in Example 4.1 that
V
d3rE= −ptot
30
. Therefore,
ptot=p0+p=p0−13(κ−1)ptot,
8 See, for example, T.M. Sanders, Jr., “On the sign of the static susceptibility”,American Journal of Physics56, 448 (1988).
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168 DIELECTRIC MATTER: POLARIZATION AND ITS CONSEQUENCES
or
ptot= 3 2+κp0.
Sinceκ≥1, a dielectric medium generally screens (reduces the magnitude of) the embedded dipole.
6.5.1 Fields and Sources in Simple Dielectric Matter
We begin our study of simple dielectrics by inserting (6.37) into the Maxwell equation (6.27) to get
∇ ·E+E· ∇=ρf. (6.38)
The general problem posed by (6.38) is difficult, particularly when the dielectric “constant” varies smoothly with position. In this book, we restrict ourselves to situations no more complicated than Figure 6.7 in Section 6.5.3 below, where the dielectric constant takes (different) constant values in distinct regions of space separated by sharp boundaries. In each region, (6.38) simplifies to
∇ ·E=ρf. (6.39)
The global electric field is constructed from the fields calculated in each region by enforcing the matching conditions (6.31) and (6.33) at each sharp boundary. Alternately, because∇ ×E=0 is always true in electrostatics, we can insertE= −∇ϕinto (6.39) and solve a Poisson equation,
∇2ϕ= −ρf, (6.40)
in each region where the permittivity is constant. Section 6.5.5 explores this potential-theory approach to simple dielectric matter.
Inserting (6.35) and (6.37) into the Maxwell equation (6.28) gives zero on both sides because
∇ ×E=0. On the other hand, writing (6.28) in the form
∇ ×P=0 (6.41)
seems to imply that the second integral in (6.30) vanishes, from which we might conclude that free charge is the only source of the fieldD(r). This is not true (in general) because a contribution to the polarization integral survives from every surface Sk where the dielectric constant changes discontinuously. We leave it as an exercise for the reader to show that (6.30) reduces in this case to
D(r)= −∇
d3r ρf(r)
4π|r−r|+
k
∇ ×
dSk P(r)×n(rˆ )
4π|r−r| . (6.42) Finally, we have learned that polarization charge is the fundamental source of the field produced by a polarizationP(r). Using (6.36) and (6.39), the volume and surface polarization charge densities produced by a simple dielectric with permittivity0κare
ρP= −∇ ·P= 1
κ −1
ρf (6.43)
and
σP=P·nˆ|S =0(κ−1)E|S. (6.44) Typically, there is a contribution to (6.44) from both sides of an interface between two simple dielectrics.
6.5 Simple Dielectric Matter 169
d
f
E = E0 –σf
Figure 6.6: A capacitor with fixed plate charge densityσf. Plus and minus signs denote polarization charge at the surface of the dielectric. A few lines of the electric fieldE=Eˆzare indicated.
6.5.2 Simple Dielectric Response to Free Charge
A point charge embedded in a medium with dielectric constantκis an example of a volume distribution of free chargeρf(r). According to (6.43), the point charge induces a polarization charge which occupies the same point in space as the point charge itself. This macroscopic statement is a consequence of Lorentz averaging (Section 2.3). In microscopic reality, the point charge polarizes nearby matter and partially neutralizes itself by attracting nearby charges in the dielectric (of opposite sign) to itself.
Becauseκ≥1, this partial neutralization appears on the macroscopic level when we use (6.43) to calculate the total volume charge density,
ρ(r)=ρP(r)+ρf(r)= ρf(r)
κ . (6.45)
The screening physics of (6.45) is built in when we solve Gauss’ law (6.39) for a point chargeq0
embedded atr0in a simple dielectric medium:
E(r)= q0
4π r−r0
|r−r0|3 =Evac(r)
κ . (6.46)
The corresponding solution of the Poisson equation (6.40) for the electrostatic potential is ϕ(r)= q0
4π 1
|r−r0| = ϕvac(r)
κ . (6.47)
It is not always appreciated that the field and potential produced by the embedded point charge are screened (reduced) by a factor ofκcompared to the case of a point charge in vacuum, whether the observation point lies within the dielectric medium or not.9This is a consequence of the spatiallylocal character of (6.45).
A Parallel-Plate Capacitor with Fixed Charge
Figure 6.6 shows a parallel-plate capacitor with vacuum capacitance C0=0A/d. What happens when a dielectric is inserted to fill the space between the plates and the system is isolated so the plate chargeQremains fixed? The charge densityσf =Qf/Ais fixed also, so Gauss’ law (6.29) and a pillbox-shaped Gaussian surface with one face inside the lower conducting plate and one face inside the dielectric tell us thatD=σfz.ˆ
Therefore,
E= D = E0
κ . (6.48)
The electric field is reduced (screened) compared to its valueE0=zσ/ˆ 0 in the absence of the dielectric. This is consistent with (6.46) because the free charge at the surface of each metal plate is
9 If the observation point lies outside the medium, there are generally other contributions to the total field besides the field from the screened point charge itself. See Section 6.5.3.
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1 2 q
Figure 6.7: A point chargeqembedded at the center of a sphere with dielectric constantκ1. The sphere is itself embedded in an infinite volume with dielectric constantκ2.
partially neutralized by polarization charge that appears at the dielectric surfaces. At the lower surface, say, the definition (6.5) ofσPgives
σP= −P·zˆ= −0(κ−1)E·zˆ=1−κ
κ σf. (6.49)
This gives the total charge density at the metal/dielectric interface (which is the source ofE) as a special case of (6.45):
σ=σf+σP=σf/κ. (6.50)
The potential difference between the plates is ϕ=
d 0
d·E= 1 κ d
0
d·E0. (6.51)
We have assumed that the chargeQof the plate does not change. Therefore, the capacitance of the dielectric-filled capacitor is
C= Q
ϕ =κC0. (6.52)
6.5.3 Polarization Charge at a Simple Interface
The polarization charge induced at the common interface between two polarized dielectrics is the source of an electric field at every point in space. This observation resolves the following “paradox”.
Figure 6.7 shows a point chargeqembedded in a sphere of radiusRand dielectric constantκ1. The sphere is itself embedded in an infinite medium with dielectric constantκ2. We begin with the spherical symmetry of the problem and use Gauss’ law in the form (6.29) to get
D(r)= q 4π
ˆ r
r2. (6.53)
Since D=E, we conclude that E1(r)=qr/4π ˆ 1r2 in medium 1 and E2(r)=qr/4π ˆ 2r2 in medium 2. On the other hand, (6.46) tells us that the electric field produced by the embedded point chargeat every point in spaceis
Eq(r)= q 4π 1
ˆ r
r2. (6.54)
If (6.54) holds true in medium 2, how does the Gauss’ law field quoted just above forE2(r) arise?
There is no true paradox here becauseE2(r) is the sum of two terms: the field (6.54) produced by q and a fieldEσ(r) produced by a uniform density of surface polarization chargeσPat ther=R
6.5 Simple Dielectric Matter 171
dielectric interface. The spherical symmetry of the boundary guarantees that there is no comparable contribution toE1(r). By Gauss’ law,
Eσ(r)= σPR2 0
ˆ r
r2 r > R. (6.55)
We getσPitself from (6.5), noting that both dielectrics make a contribution:
σP(r)=rˆ·(P1−P2)|r=R=rˆ·[0(κ1−1)E1−0(κ2−1)E2]r=R. (6.56) However,D1(R)=D2(R) because there is no free charge at the interface. Therefore
σP=rˆ·(D1−D2)r=R−0ˆr·(E1−E2)r=R= −0rˆ·D(R) 1
1
− 1 2
. (6.57)
InsertingD(R) from (6.53) into (6.57) gives Eσ(r)= q
4π 0
ˆ r r2
1 κ2
− 1 κ1
r > R. (6.58)
Combining (6.54) with (6.55) produces the elementary Gauss’ law result, E2(r)=Eq(r)+Eσ(r)= q
4π 2 ˆ r
r2 r > R. (6.59)
Example 6.2 A point chargeq sits atz= −d on thez-axis (Figure 6.8). Find the polarization charge density σP(x, y) induced on the plane z=0 when the half-space z >0 is filled with a medium with dielectric constantκRand the half-spacez <0 is filled with a medium with dielectric constantκL.
L R
q d
r
z
0 ( , )xy r
Figure 6.8: Two semi-infinite dielectrics meet at a planar interface. A point chargeqsits a distancedto the left of the interface. The pointr0=(x, y) lies in thez=0 plane.
Solution: This problem is typically solved using the method of images (Section 8.3.3). The alternative method offered here is less elegant, but will deepen the reader’s understanding of the electrostatics of dielectrics.
LetEL andER be the electric fields at points infinitely close to, but on opposite sides of, an interface pointr0=(x, y). The surface polarization charge density atr0is
σP(r0)=zˆ·(PL−PR)=0(χLEL−χRER)·z.ˆ
The sources ofELandERare the screened point chargeq/κLand the surface polarization charge σP(r0) we are trying to find. Following Section 3.4.2, the field produced by the surface charge on
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either side of (but very near) the surface itself is the sum of two terms: a contribution±zσˆ P(r0)/20
from the charge infinitesimally near tor0and a contributionESfrom the surface charge away from r0. ButEShas no component along ˆz. Therefore, using the trigonometric factord/rto project out thez-component of the point charge field,
0EL·zˆ= 1 4π κL
q r2
d
r − σP(r0)
2 0ER·zˆ= 1
4π κL q r2
d
r +σP(r0) 2 . There is no free charge at the interface, so the matching condition (6.31) is
κLEL·zˆ=κRER·z.ˆ
Combining all these equations gives the polarization surface charge density as σP(x, y)= 1
2π κL
κL−κR
κL+κR
qd (x2+y2+d2)3/2.
We will use this result in Example 6.7 to find the force exerted onqby the dielectric.
6.5.4 Simple Dielectric Response to Fixed Fields
In the absence of free charge, the Maxwell equations (6.26) and (6.27) simplify to
∇ ×E=0 and ∇ ·D=0. (6.60)
The most interesting aspect of these equations is the matching conditions they imply, namely, ˆ
n×[E1−E2]S=0 and nˆ·[D1−D2]S =0. (6.61) The tangential component ofEand the normal component ofDare continuous at a dielectric interface with no free charge.
A Parallel-Plate Capacitor with Fixed Potential
Figure 6.9 shows a parallel-plate capacitor with vacuum capacitanceC0=0A/d. What happens when a dielectric is inserted to fill the space between the plates and the potential differenceV between the plates is held constant? Because the left side of
V = d 0
d·E (6.62)
is fixed, the field between the plates is
E=E0= V
dz.ˆ (6.63)
In contrast to (6.48), the electric field isnotscreened by the dielectric.
The polarization charge density induced at the lower surface of the dielectric is
σP= −P·zˆ= −0(κ−1)V /d. (6.64)
On the other hand,D=0κE, so the charge density induced on the lower metal plate is σf =D·ˆz=0κV
d. (6.65)
This shows that the capacitanceCwith fixed potential is the same as the capacitance (6.52) we found earlier for fixed charge:
C= σfA
V =κC0. (6.66)
6.5 Simple Dielectric Matter 173
d
V 0
E = E0
Figure 6.9: A capacitor with fixed plate potentials. Plus and minus signs denote polarization charge at the surface of the dielectric. A few lines of the electric fieldE=Ezˆare indicated.
The sum of (6.65) and (6.64) is exactly the free charge densityσ0=0V /dthat produces the field E0when the dielectric is absent. In other words,E=E0when the dielectric is present because charge flows from the battery (or whatever maintains the plates at fixed potential) to the surface of the plates to exactly cancel the polarization charge on the adjacent dielectric surfaces. Faraday exploited (6.65) to determineκ for many materials. His method was to measure the change in the amount of charge drawn onto the metal plates of a (spherical) capacitor when a dielectric was interposed between them.
The SI unit of capacitance is called thefaradto honor this aspect of Faraday’s work.
Macroscopically, the charge densities (6.65) and (6.64) are spatially coincident. This is not so microscopically, and (6.65) implies that the local electric field immediately adjacent to both metal plates islargerin magnitude than the electric field produced by the vacuum capacitor. If so, there must also be regions inside the dielectric where the electric fields is smaller than (or oppositely directed from) the field of the vacuum capacitor. We will return to this point at the end of Section 6.6.2.