30
PHYS 2012 Physics 2B PHYS 2912 Physics 2B (Advanced) Electromagnetic Properties of Matter Module 2009 Part 1 - Dielectric Materials Martijn de Sterke and B. James School of Physics, University of Sydney Prologue In teaching physics we need to balance width and depth. In keeping with physics departments everywhere, the School of Physics addresses this by teaching the key topic areas a number of different times at increasing levels of sophistication. For example, our honours level course on Relativistic Quantum Mechanics is significantly more challenging than the first year Quantum Mechanics module you did last year. Of course this can only work because, you, the students, get more sophisticated over the years as well! After all you take more physics, maths and other courses, and so you think more about physics, and can start to make connections between different areas. The necessity for this is clear–by the time students do honours in 4th year, they need to carry out a research project, which typically involves acquiring new knowledge. In order to get to this level, you are exposed over the years to increasingly more fundamental descriptions of nature. The reason I am telling you this is that this semester is the next step in this increased level of difficulty and sophistication. The School of Physics considers the first three semesters of its program to be a broad-brush overview of a wide variety of subjects within physics. The treatment in this phase tends to be predominantly qualitative, with some numerical calculations. The second semester of second year is where we return to some of the topics and teach them at a deeper level, and this module on the Electromagnetic Properties of Matter is a prime example; after all, you already did electromagnetism last year. One of the key differences with the previous treatment is that we will be less qualitative. Mathematics is the language of physics, and in order to understand nature at an increasingly deep level we need to use it. We simply cannot solve our problems by staring at it long enough and coming up with the answer in that way. This was well summarized by E. Wigner in 1960 when he wrote about “The Unreasonable Effectiveness of Mathematics in the Natural Sciences.” In summary, these 19 lectures will be different than the previous lectures you have had–they are more quantitative and more mathematical, since ultimately that is how nature needs to be described. 1

PHYS 2012 Physics 2B PHYS 2912 Physics 2B (Advanced ...€¦ · PHYS 2012 Physics 2B PHYS 2912 Physics 2B (Advanced) Electromagnetic Properties of Matter Module 2009 Part 1 - Dielectric

  • Upload
    others

  • View
    28

  • Download
    0

Embed Size (px)

Citation preview

PHYS 2012 Physics 2B

PHYS 2912 Physics 2B (Advanced)

Electromagnetic Properties of Matter Module 2009

Part 1 - Dielectric Materials

Martijn de Sterke and B. JamesSchool of Physics, University of Sydney

Prologue

In teaching physics we need to balance width and depth. In keeping with physics departmentseverywhere, the School of Physics addresses this by teaching the key topic areas a number ofdifferent times at increasing levels of sophistication. For example, our honours level course onRelativistic Quantum Mechanics is significantly more challenging than the first year QuantumMechanics module you did last year. Of course this can only work because, you, the students,get more sophisticated over the years as well! After all you take more physics, maths andother courses, and so you think more about physics, and can start to make connections betweendifferent areas. The necessity for this is clear–by the time students do honours in 4th year, theyneed to carry out a research project, which typically involves acquiring new knowledge. In orderto get to this level, you are exposed over the years to increasingly more fundamental descriptionsof nature.The reason I am telling you this is that this semester is the next step in this increased levelof difficulty and sophistication. The School of Physics considers the first three semesters ofits program to be a broad-brush overview of a wide variety of subjects within physics. Thetreatment in this phase tends to be predominantly qualitative, with some numerical calculations.The second semester of second year is where we return to some of the topics and teach them ata deeper level, and this module on the Electromagnetic Properties of Matter is a prime example;after all, you already did electromagnetism last year. One of the key differences with the previoustreatment is that we will be less qualitative. Mathematics is the language of physics, and in orderto understand nature at an increasingly deep level we need to use it. We simply cannot solveour problems by staring at it long enough and coming up with the answer in that way. This waswell summarized by E. Wigner in 1960 when he wrote about “The Unreasonable Effectivenessof Mathematics in the Natural Sciences.”In summary, these 19 lectures will be different than the previous lectures you have had–theyare more quantitative and more mathematical, since ultimately that is how nature needs to bedescribed.

1

1 Introduction

It is assumed that you are familiar with all the electromagnetic topics covered in first yearphysics. Strictly speaking these deal with electromagnetism in free space, but the presence ofair makes minimal difference. In this module we will consider the subject of electromagnetismin the presence of materials, usually solids, the properties of which differ significantly from thoseof air.It is worth recalling two topics from last year: Gauss’s Law and the relationship between electricfield and electric potential.

1.1 Gauss’s Law

Gauss’s Law is in the form of an integral. It tells us that the flux of electric field through a closedsurface is proportional to the net charge inside the surface, with a constant of proportionality of1/ε0 ∮

E · dA =qenclosed

ε0=

1ε0

∫dr ρ(r), (1)

where ρ(r) is the charge density and the integration includes all space. The second equalityexpresses the fact that the integral over the charge density in the enclosed volume gives the totalcharge. This law is true only because the Coulomb force has an inverse square dependence. Itapplies to any closed surface, and to any distribution of charge. It can be used to determineexpressions for the electric field produced by sufficiently symmetric charge distributions, such asthe electric field due to:

• a point charge, or any spherically symmetric charge distribution (charged spherical shell,uniformly charged sphere, etc.)

• a line charge, or cylindrically symmetric charge distribution

• an infinite sheet of charge

Note that in the form in which it was given here, Gauss’ law can only deal with vacuum. Wewill see in Section 9 how this is generalized to dielectric media and how to deal with dielectricmedia more generally.

1.2 Electric field and electric potential

The effect of electric charge on the surrounding space can be described in terms of a vector field(electric field, E) or a scalar field (electric potential, V ). Obviously one must be related to theother. The potential difference between two points is given in terms of the electric field by

Vb − Va = −∫ b

a

E · dl (2)

where the integral is independent of the path taken from point a to point b. Another way ofstating the latter fact is that the line integral of the electric field around a closed loop is zero.The electrostatic field is therefore a conservative field. Of course if the integral did depend onthe path taken then there would be little defining V in the first place.

2

The inverse relation gives us the component of the electric field in a particular direction. Forexample, the component of electric field in a direction specified by the coordinate l is

El = −∂V

∂l(3)

where, for example, l could be the coordinates x, y, or z. We can write this as a single vectorequation (in cartesian coordinates):

E = (Ex, Ey, Ez) = −(

∂V

∂x,∂V

∂y.∂V

∂z

)= −~∇V (4)

where ~∇ is the vector differential operator (called del or nabla):

~∇ ≡ (∂

∂x,

∂y,

∂z). (5)

This operator plays a central role in the differential formulation of the Maxwell equations. How-ever, in this set of lectures we mostly, though not exclusively, deal with the integral formulation.

1.3 Differential form of Gauss’s law

Consider a volume element δxδyδz in a region of space with a volume charge density ρ. We applyGauss’s law to this volume. Consider first the flux through the sides of the cube in the y − zplane, which is due to the electric field component Ex (see figure 1). If at x, the component isEx then at x + δx it is Ex + (∂Ex/∂x)δx. The net outwards flux through the y − z sides of thevolume element is therefore

−Exδyδz +(

Ex +∂Ex

∂xδx

)δyδz =

∂Ex

∂xδxδyδz (6)

x

y

z

Ex Ex + ∂Ex

∂xδx

δxδy

δz E

Figure 1: Electric field flux through a volume element

Taking into account the other components, the total flux outwards through the volume elementis (

∂Ex

∂x+

∂Ey

∂y+

∂Ez

∂z

)δxδyδz (7)

By Gauss’s law this equals the enclosed charge, which is ρδxδyδz, divided by ε0. It follows(

∂Ex

∂x+

∂Ey

∂y+

∂Ez

∂z

)δxδyδz =

ρ

ε0δxδyδz (8)

3

Thus,(

∂Ex

∂x+

∂Ey

∂y+

∂Ez

∂z

)=

ρ

ε0or (9)

~∇ ·E =ρ

ε0(10)

This is the differential form of Gauss’s law - it applies at any point in space. The quantity ~∇ ·Eis called the divergence of the electric field. The differential equation embodies that fact that ifthere is positive (negative) charge density in a region there must be a net flux of field lines from(into) this region.

Interlude

Before diving into the new material it is probably good to outline the main lines of argument inthis part of the unit. Some of the items below may make little sense at this time, so I recommendthat you look at these points while we progress through the notes.

• Section 2: properties of dipoles.

• Section 3: from a large distance, the field of an electrically neutral charge distribution isapproximately that of a dipole.

• Sections 4 and 5: Since neutral atoms can evidently be considered to be dipole-like, weconsider the properties of a distribution of a large number of dipoles as a model for adielectric medium (a medium without free charges).

Having established the properties of a large collection of dipoles, we next need to work themagnitude of each of the individual dipoles. We do so in two steps.

• Section 6: we establish the relation between the applied electric field and the dipole momentof an atom.

• Section 7: the previous result lets us introduce a new field, the electric displacement, whichis convenient for calculations.

• Section 8: the last piece of the puzzle, namely the relation between the applied macroscopicfield, and the field which an individual dipole notices. The two differ because inside themedium the field is affected by the presence of all the other dipoles.

With the exception of Section 10, in which the energy associated with an electric field is calcu-lated, the remaining chapters are applications and special cases of the general theory outlinedabove.

2 The electric dipole

2.1 Electric field of a dipole

An electric dipole consists of two charges of equal magnitude but opposite sign (±q), separatedby a distance d. The quantity p = qd is called the dipole moment. The dipole moment has a

4

vector representation p = qd, where the direction is from the negative to the positive charge, asshown in figure 2. Dipoles are key elements of electromagnetism, and are particularly importantin the limit in which the observation point is much larger than the distance d. In this limitthe electric fields, or the potential for that matter, of the positive and negative charges almostcancel. In fact the cancelation becomes better when the observation points moves further away.We will see the effect of this below. The reason we are interested in dipoles here is that they areat the basis of the model for dielectric materials that we use.

-q +q

p = qd

P

θ

rr- r+

x

y

Figure 2: Electric dipole

At an arbitrary point, P, shown in figure 2, the electric field can be written as

E =1

4πε0

(q

r3+

r+ − q

r3−r−

)(11)

where r± = r∓ d/2. We first consider two special cases:

Case 1: θ = 0

Ex =1

4πε0

(q

(r − d/2)2− q

(r + d/2)2

)(12)

=q

4πε0

((r + d/2)2 − (r − d/2)2

(r2 − d2/4)2

)(13)

≈ 14πε0

2p

r3(14)

for r À d/2. Clearly Ey = 0.

5

Case 2: θ = π/2

Ex = − 14πε0

(2q cosφ

r2 + d2/4

)(15)

where

cos φ =d/2

(r2 + d2/4)1/2(16)

For r À d/2 this reduces to

Ex = − 14πε0

p

r3(17)

Again, clearly Ey = 0.The electric field pattern for an electric dipole, showing field lines and surface of equal potential,is shown in Fig. 3.

Figure 3: Electric field lines and cross-sectional view of equipotential surfaces for an electricdipole

2.2 Electric potential of a dipole

As we know from first year, getting the electric field, which is a vector quantity, is somewhattedious. It is more convenient to calculate the electric potential, and the field through Eq. (4)The electric potential for an electric dipole is given by

V =1

4πε0

(q

r+− q

r−

)(18)

6

where, if r À d/2,

r+ ≈ r − d cos θ/2 (19)r− ≈ r + d cos θ/2 (20)

Using (r ± d cos θ/2)−1 = r−1(1± d cos θ/2r)−1 ≈ r−1(1∓ d cos θ/2r), we find

V =1

4πε0

p cos θ

r2, (21)

which in terms of vectors can be written as

V =1

4πε0

p · rr2

, (22)

where r is the unit vector pointing in the direction of r. This confirms our earlier discussion:quite generally, provided you are far enough away, the potential decreases as r−2, i.e. faster thanthe potential of a single charge. The reason is that the positive and negative charge almost canceleach other, but not quite (they exactly cancel as d → 0, and that this cancelation improves whenthe observation point moves further away.

Exercises

1. (advanced only) Starting from Eq. (22), show using (4) that, in vector notation, the electricfield can be written as

E =1

4πε0

3(p · r)r− r2pr5

(23)

Hint: This is done most easily by writing the second factor in Eq. (22) as (p · r)/r3 =(pxx + pyy + pzz)/r3.

2. Using Eq. (23) show that at point (r, θ),

Er =1

4πε0

2p cos θ

r3(24)

Eθ =1

4πε0

p sin θ

r3(25)

and then show that the magnitude of the electric field at (r, θ) is given by

E =1

4πε0

p

r3

√3 cos2 θ + 1 (26)

3. Show the results above for Er and Eθ reduce to the correct values for Ex and Ey forθ = 0, π/2.

2.3 Torque on a dipole

When an electric dipole is in a uniform electric field the charges experience forces of equalmagnitude, but in opposite directions, as illustrated in Fig. 4. If the dipole moment vector p is

7

Figure 4: Schematic of a dipole in a uniform electric field. The two charges are subject to equal,opposite forces, leading to a torque.

not parallel to E, the dipole experiences a torque; if the angle between p and E is θ the torqueis given by (taking the centre of the dipole as origin)

τ = qE × d cos θ/2 + qE × d cos θ/2 = qdE cos θ. (27)

In terms of vectors the torque can be written as

τ = p×E. (28)

Note that the torque is a maximum when p ⊥ E, and is equal to zero when p ‖ E. If a dipole isfree to rotate it would oscillate about the θ = 0 position; if there is friction it settles down to anequilibrium orientation where p ‖ E.

2.4 Potential energy of a dipole

Because of the torque which tends to align the dipole with the electric field, a dipole has potentialenergy associated with its orientation in an electric field. Starting with θ = 0, the work requiredto rotate the dipole to angle θ is given by

U = 2qE

(d

2− d

2cos θ

)= pE − pE cos θ (29)

This expression assume that the potential energy is zero at θ = 0. It is customary, however, totake the potential energy to be zero at θ = π/2, in which case U = −pE cos θ, or in terms ofvectors

U = −p ·E (30)

2.5 Calculating E from V for an electric dipole

Given the form of equation 21. it is natural to use polar coordinates. As incremental displace-ments in the r and θ coordinates are δr and rδθ respectively, the radial and azimuthal componentsof the electric field are given by

Er = −∂V

∂r=

14πε0

2p cos θ

r3(31)

Eθ = −1r

∂V

∂θ=

14πε0

p sin θ

r3(32)

These results are consistent with the directly calculated cartesian components for θ = 0, π/2which were derived in Section 2.1.

8

Case 1: θ = 0

Ex = Er =1

4πε0

2p

r3(33)

Ey = Eθ = 0 (34)

Case 2: θ = π/2

Ex = −Eθ = − 14πε0

p

r3(35)

Ey = Er = 0 (36)

3 Charge multipoles–(advanced only, except the first paragraph)

You may ask why we even considered dipoles in the previous section. After all, it may not be clearwhere we can find equal and opposite charges that are close together. Now as it happens, and thisis an idea that we develop over the coming sections, the way we model atoms in electromagnetismis like tiny dipoles–in essence what happens is that the applied electric field pulls the nucleus andthe electrons in opposite directions, therefore creating a dipole. Therefore, any treatment of howelectric fields affect matter, and matter affects electric fields, needs to rely on the properties ofdipoles. This idea is further developed in Sections 4 and 5. The remainder of this section dealswith an associated problem that can be skipped by Regular students.Let us now consider a charge distribution ρ(r), that is confined to some limited region of spaceclose to r′, but is otherwise arbitrary, and let us calculate the associated electric potential at theorigin, which is far away from the charges. The exact result is

V =1

4πε0

∫ρ(r)r

dr. (37)

Now write r = r′ + ∆r, and so

V =1

4πε0

∫d∆r

ρ(r′ + ∆r)|r′ + ∆r| , (38)

where r′ is fixed and ∆r = (∆x, ∆y, ∆z is the vector pointing from r′ to r. We now make useof the knowledge that the charge distribution is close to r′ and that the observation point is faraway. We do so by expanding the denominator in a Taylor series where δr is the small parameter.The first term of this series is 1/r′ and it corresponds to assuming that all charges are locatedat r′. However, is often in physics that it is the second term that we are interested in. Let ustherefore push this to the next order:

1r′ + ∆r

=1r′

+(

∆x∂(1/r)

∂x+ ∆y

∂(1/r)∂y

+ ∆z∂(1/r)

∂z

) ∣∣∣r′

+ · · · (39)

where the vertical bar at the end indicates that the function is to be evaluated at r′. Now notethat

∂x

1r

= − 1r2

∂r

∂x= − x

r3, (40)

9

where the last equality can be checked by writing r =√

x2 + y2 + z2, doing the differentiation.We now substitute this into (39) and then find for V

V =1

4πε0

∫d∆r

ρ(r′ + ∆r)r′

− 14πε0

∫d∆r

ρ(r′ + ∆r)(r′)3

∆r · r′ + · · · , (41)

where ∆r · r′ = x′∆x + y′∆y + z′∆z.This may look somewhat messy, but recall that r′ is a constant vector that can be taken out ofthe integral. The first integral in (41) then simply gives the total charge q in the distribution.The interpretation of the second integral is slight more complicated. Let us rewrite (41) first as

V =1

4πε0

Q

r′− 1

4πε0

1(r′)3

r′ ·∫

d∆r ρ(r′ + ∆r)∆r. (42)

Let us have a closer look at the second integral. As an example, let us take ρ(r′+∆r) = Q[δ(r+)−δ(r−)], where δ is the Dirac delta-function. This gives Q(r+−r−), i.e., the dipole moment of thedistribution. Even if the distribution is more complicated, consisting of a continuous distributionof charge, the integral picks out the net dipole moment. For example, even if the total chargeQ = 0, it may be that the positive charge occurs predominantly at large x, and the negativecharge predominantly at small x. In this case the distribution has a net dipole moment pointingin the x direction. With this, then we write V as

V =1

4πε0

Q

r′+

14πε0

p · r(r)3

+ · · · , (43)

where the sign of the second term has flipped since we take the vector r now as pointing fromthe charges to the observer.As discussed, the first term corresponds to the effect of the net charge in the distribution, andif it is nonzero then at large distances it is the dominant contribution to the potential. For adistribution for which the net charge vanishes, the next term, involving the dipole moment ofthe distribution, is the dominant one. For a distribution for which p vanishes as well, the nexthigher order term is a quadrupole term, which we briefly discuss below. At sufficiently largedistances consecutive terms in the series become progressively smaller. A final property, whichis not proven here, is the following: in general the value of the dipole moment p is depends onthe position of the origin, except when the net charge Q = 0.As mentioned, an electric dipole is part of a multipole hierarchy, of which the first four are shownin Figure 5:

• monopole: a single charge, for which the electric field varies as 1/r2, and the electricpotential varies as 1/r;

• dipole: two separated charges of equal magnitude and opposite sign, for which at largedistances, the electric field varies as 1/r3, and the electric potential varies as 1/r2;

• quadrupole: two separated equal and opposite dipoles, for which at large distances, theelectric field varies as 1/r4, and the electric potential varies as 1/r3. We do not prove thespatial dependence of the field and potential here, but the argument is the same as thatbefore: a quadrupole can be considered to be two dipole close together and with oppositeorientation. Seen from a large distance the dipole almost cancel, but not quite, and thiscancelation improves when the distance to the observer increases.

10

+ +-

+-

+ -

+-

+ -

+

-

+-

monopole dipole quadrupole octopole

Figure 5: Illustration of the hierarchy of electric multipoles.

• octopole: two separated equal and opposite quadrupoles, for which at large distances (byinference), the electric field varies as 1/r5, and the electric potential varies as 1/r4.

The multipole concept provides a way of approximating the fields of an arbitrary charge distri-bution at large distances by a sum of successively higher order (and progressively smaller) terms.Thus the electric field magnitude and electric potential can be written as

E =1

4πε0

(A

r2+

B

r3+

C

r4+ . . .

)(44)

V =1

4πε0

(A′

r+

B′

r2+

C ′

r3+ . . .

)(45)

The first term is the field due to a point charge of magnitude equal to the net charge of thedistribution: Σqi for point charges and

∫ρdr′ for a continuous charge distribution.

The conclusion from this section is that when we have an arbitrary charge distribution then at alarge distance the dominant contribution to the potential or the field comes from the net charge.It leads to a 1/r dependence in the potential and 1/r2 in the field. If the charge distribution issuch that there is no net charge, then the dominant contribution comes from the dipole momentof the charge distribution, leading to a 1/r2 dependence in the potential and 1/r3 in the field.If in turn the dipole moment also vanishes, then the dominant term comes from the quadrupoledistribution of the charges, etc.

4 One dimensional treatment of dielectrics

In this section we apply the knowledge from Section 3 to the description of the response of a largenumber of atoms to an applied field. Here we consider dielectric media, media which do not havefree charge carriers. It therefore applies to insulating materials such as glass or various crystallinematerials, but not to semiconductors or metals. The latter will be discussed in subsequent partsof this Unit.Electric dipoles are interesting in their own right but are also essential for the understanding ofdielectric media: we think about dielectric media as consisting of atoms. An applied field pulls theelectrons and the nuclei in opposite directions (recall that F = qE) setting up a dipole. Note thatthe argument developed in Section 3 applies here. The net charge in a medium would be expectedto vanish, and thus the next highest term is the dipole term even if the charge distribution iscontinuous. The dipole moment might not be very large since the effective separation betweenthe positive and negative charges would be a fraction of the size of an atom, but that does notmatter here. Since there are lots of atoms the total effect might be substantial. To see this effect

11

Figure 6: One-dimensional geometry that we are considering: a medium is modeled as set ofdipoles, all pointing in the +x direction between a and b, with the observation point at x.

in an elementary way we consider a one-dimensional geometry so that full-blown vector calculusis not needed.Before commencing with the one-dimensional treatment, a brief comment regarding the assump-tions we are making here: we assume that the constituent molecules are non-polar, which meansthat in the absence of an applied electric field they have no dipole moment; in other words, theydo not have a permanent dipole moment. Rather, the atoms or molecules are assumed to have nodipole moment in the absence of an electric field–the dipole moments are induced by the appliedfield by pulling the positive and negative charges apart. The dielectric properties of materialsconsisting of polar molecules is discussed in Sec. 12.Let us consider a dipole p at position x′ pointing in the positive direction, and an observer locatedat x, as illustrated in Fig. 6. The potential dV at x due to this dipole is given by

dV =1

4πε0

p

(x− x′)2=

14πε0

pd

dx′

(1

x− x′

), (46)

where, since the derivative is taken with respect to the source point x′ the usual − sign does notshow.Now let us take a collection of dipoles on the x-axis located between a and b. The density ofdipoles, i.e., the dipole moment per unit length is P (x′). Note that P is generally not a uniformfunction of position since the density of dipoles (atoms) or the dipole moment per dipole (atom)may vary. The total potential V at position x due to all dipoles is then

V (x) =1

4πε0

∫ b

a

P (x′)d

dx′1

x− x′dx′. (47)

We integrate by parts to find

V (x) =1

4πε0

[P (x′)

1x− x′

]b

a

− 14πε0

∫ b

a

1x− x′

dP (x′)dx′

dx′. (48)

Now this is an interesting result: in both terms the distance between the dipole and the observerenters the denominator linearly, reminiscent of the way the potential of a point source is calcu-lated. Let us therefore look at the two terms more closely. If P was a uniform function thenthe second term would vanish and the first term would be the only one left. Its interpretation isquite easy. If P is uniform then in inside the “medium” the positive and negative charges canceleach other out. It is only at the edges where they do not: at x = a there is some bare negativecharge, whereas at x = b there is some bare positive charge. The bare charges each lead to acontribution to the potential at the observation point, described by the first term in (48).As mentioned, if P is uniform then inside the contribution from the end points are the onlycontribution to the potential at x. In contrast if P is not uniform then the second term in (48)

12

leads to a contribution from the bulk of the “medium.” The precise form of this may be not beobvious but it is clear that it must depend on the derivative of P .Having made this interpretation we note that the contribution to the potential at x by theone-dimensional medium consisting of tiny dipoles can be written as

V (x) =1

4πε0

σb(b)x− b

+1

4πε0

σb(a)x− a

+1

4πε0

∫ b

a

ρb(x′)x− x′

dx′. (49)

Here ρb = −dP/dx and σb = P sign(n), where sign(n) is the direction of the normal pointing outof the medium. The subscript b is short for “bound,” as discussed in the next paragraph.Now in a more general situation where we have free charges and also medium consisting ofdipoles, the total potential is simply given by the sum of the contributions from the free chargesand the bound charges associated with dipoles, and therefore associated with the response of themedium. The charges are “bound” in that are bound to atoms. This in contrast to free chargesin conductors that can roam freely. In the presence of a medium the ρ in Gauss’ Law needs toinclude both types of charges. We return to this in Section 7.We note that σb and ρb essentially describe the same phenomenon, namely that a nonuniformityin the distribution of P leads to net charges. The expression for ρb picks out the nonuniformitywithin the medium, whereas σb is the obvious and large nonuniformity that occur at the edgesof the medium! However, they describe the same physics.In drawing our previous conclusions we have been somewhat cavalier in that we only consideredthe effect of the bound charges at positions outside the one-dimensional medium. For the collec-tion of dipoles to be truly equivalent to the effect of free charges we must also consider internalpoints. However, this is tricky, since the local potential would vary enormously. For example,close to an atom the magnitude of the electric potential would be expected to be very large,whereas further away it might be expected to be smaller. However, it can be shown that theconclusion we reached earlier is universally valid.

5 Three dimensional treatment of dielectrics

The argument in the previous section was somewhat contrived in that we considered a one-dimensional geometry. The full three dimensional calculation proceeds similarly, except thatthe mathematics, which relies on vector calculus, is somewhat more complicated. However,the results are fully consistent with those in Section 4. In the three-dimensional case we dealwith a small volume, rather than with an interval. First of all we now define the macroscopicpolarization P to be dipole moment per unit volume, i.e.,

P = Np (50)

where N is the density of dipoles. Having defined P, we then find that the results from fullthree-dimensional calculations are

ρb = −~∇ ·P; σb = P · n, (51)

both of which are immediate generalizations of the one-dimensional results. In the first of thesethe del operator (defined in Eq. (5)) describes three-dimensional variations of the polarization.In the second expression n is the normal pointing out of the volume which we are considering.A characteristic of both these expressions is that variations of Px in the y-direction, say, do notcontribute to the bound charge. For the case of the surface density of bound charge it is obvious

13

why: the dipoles are oriented parallel to the surface which does not lead to an accumulation ofnet charge at the boundary.We can take this argument one step further. Let us suppose that, for whatever reason, P (x)depends on time. Then we may expect that the effect is similar to the effect of a current, in thiscase a “bound” current, since the associated charges remain bound to atoms, or a polarizationcurrent. It can be shown that the polarization current Jb is given by

Jb =∂P∂t

, (52)

which, since P has units of C/m2, has the expected units of A/m2, and which needs to be addedto possible currents associated with free charges. The polarization current can be somewhatunderstood as follows: consider a medium with an applied field that is turned on. As a con-sequence the atoms or molecules form small dipoles where there were none before. In thinkingabout this argument it is easiest to assume that the electrons are fixed by the crystal structureand that the nuclei move to setup the dipole moments. The movement of the positive chargesclearly generates a current, the polarization current.

6 Response of atoms and molecules

In the previous section we started we the assumption that the dipole moment per unit volumeP had some value, from which we then determined the bound charge density. In this sectionwe deal with the first question: given some electric field E, what is P? This is not entirelystraightforward since the field that is setup by one of the dipoles would be expected to affect theother dipoles as well. In other words, the externally applied field E may differ from the localfield Eloc the dipoles see. The local field can be thought of as a combination of the applied fieldand an internally generated field due to the atomic dipoles. To lowest order one might expectthat

p = αEloc, (53)

where α the polarizability, a property of the individual atoms or molecules, describes the easewith which the positive and negative charges can be pulled away from each other. The key pointhere is that p and Eloc are assumed to be proportional.

d

E

.+

electron cloud

Figure 7: Induced electric dipole in an atom in the presence of the local electric field.

We can estimate the polarizability α for hydrogen as follows. Assume that the electron chargeis uniformly spread with density ρ over a spherical volume of radius a. An external electric field

14

Eloc causes the centres of the positive and negative charges to be displaced by an amount d.If we assume that the electrons are fixed by the lattice in which the atom is located, then thenucleus moves by d. The distance d is determined by the condition that the external field Eloc

balances the field by the electrons. The latter can be calculated using Gauss’ law. At a positiond from the centre of a uniform charge distribution the field due to the electrons has a strength

4πd2E =4πρd3

3ε0, (54)

and so, since this field at equilibrium equals the external field Eloc

Eloc =ρd

3ε0=

Qd

4πε0a3=

p

4πε0a3, (55)

where Q is the total negative charge of the distribution. Therefore we find for the polarizability

α =p

Eloc= 4πε0a

3, (56)

which is the result we were after. Taking a ≈ 10−10 m, we find α/4πε0 ≈ 10−30 m3. Thiscompares with the measured value of 0.66× 10−30 m3 for hydrogen.

7 The electric displacement field D

From our studies we found that an distribution of dipoles leads to the presence of bound chargesat the edges and at positions where the distribution is inhomogeneous. Now we also know thatthe electric field is associated with the presence of charges; the key point here is that the natureof the charges, whether free or bound, does not matter for the electric field. Thus in Gauss’ law,which is a direct consequence of Coulomb’s law, the total charges ρf + ρb needs to be used, i.e.,

∮E · dA =

1ε0

∫(ρf + ρb)dr, (57)

where ρf and ρb are understood to include possible surface charges, or, in differential form,

~∇ ·E =ρf + ρb

ε0. (58)

Now since from (51) ρb = −~∇ ·P, we can rewrite this as

~∇ ·E =1ε0

ρf − 1ε0

~∇ ·P. (59)

Now this is interesting: we have seen that E arises from free and bound charges, whereas Parises from bound charges only. The relation allows us to define a third field, a field that arisesfrom free charges only. This field, the electric displacement D, is defined as

D = ε0E + P (60)

so that~∇ ·D = ρf (61)

15

Now we have gone through this argument using the differential form of Gauss’ law. The integralform, with which we are more familiar reads

∮D · dA =

∫ρf dr (62)

So, in summary, we have identified three fields, the electric field E, which arises from all charges,the polarization P which arises from the bound charges, and the electric displacement D, whicharises from the free charges only. The three fields are related by Eq. (60).

8 Response of the medium

We now have collected a lot of puzzle pieces. We have identified three fields and their sources,and their relation (60). We also understand how the atoms and molecules that make up themedium respond to an external field. However, one of the key relations is still missing. Insome sense it is easy to see where: Eq. (60) is a single relation between three fields and canbe considered to define D, and even though we understand the microscopic response of atoms,we do not yet know the macroscopic response of a medium. Such relation, which define theelectric properties of a medium, and which thus distinguish vacuum, from metal and water, sayare called constitutive relations, since they refer to the constitution (the make-up) of a medium.The constitutive relation is the missing puzzle piece and we now set out to discuss it.Earlier we studied the microscopic response of single atoms and found that, to lowest order,p = αEloc according to Eq. (53). The most straightforward way to obtain a macroscopic responseis to combine Eqs. (50) and (53), thereby ignoring the difference between the applied and thelocal field, to find

P = NαE. (63)

As we discussed, the difference between Eloc and E is the effect of the response of the dipoles tothe external field. The difference between these two fields can be ignored for very dilute systems,particularly gases.However, for dense system this is simply not good enough. Now it can be shown that under someconditions Eloc ≈ E + P/(3ε0) (essentially by the method described in Appendix A). For morecomplicated situations a numerical calculation is required. However, in general we can say that

Eloc = E + KP/ε0 (64)

where K is some positive constant.We now combine Eqs (53) and (64), and remember that P = Np to find the constitutive relation

P =Nα

1− (NαK/ε0)E ≡ ε0χE (65)

where χ is the susceptibility of the medium, the macroscopic version of the polarizability. Thisexpression is consistent with our earlier arguments: the total, macroscopic polarization of volumeis the sum of all the dipoles moments present in that volume. In a gas or other dilute medium,each dipole has the strength αE (assuming the dipoles are identical) and the total polarizationis thus NαE. In a dense medium the dipoles interact with each other so that the direct propor-tionality no longer holds. This effect is described by the denominator of Eq. (65). It is commonto use the constitutive relation between D and E, which, by combining (60) and (65) can bewritten as

D = ε0εrE, where εr = 1 + χ (66)

16

material εr

vacuum 1.0air (1 atm) 1.00059teflon 2.1perspex 3.4glass 5-10neoprene 6.7quartz 4.3ethanol 28methanol 33water 80

Table 1: Dielectric constant values for a number of different media.

which defines the dielectric constant εr,1 and which is the constitutive relation we are after.The key aspect of this relation is that D and E are directly proportional to each other. Theproportionality constant, which, apart from a factor ε0, is the dielectric constant, can either becalculated using (65), or can simply be measured. Some values of the dielectric constant aregiven in Table 1.It should be stressed here that constitutive relation (66) (or, equivalently (65)), have a differentstatus than the other equations that we have encountered. The others describe the universalbehaviour of electric fields and how they interact with matter. In contrast, Eqs. (66) and (65)describe the response of a large but finite class of materials. Both are needed to do meaningfulwork. Later on in this part of the course we briefly discuss materials that have more complicatedconstitutive relation. Perhaps the analogy with mechanics is the most obvious, in that theequations in this section are the equivalent of Newton’s equation, and they have thus universalvalidity. The constitutive relations are somewhat like the relation F = −kx, which applies toideal springs, but not to other objects.

8.1 Polarizability and dielectric constant

We found that for dense media Eq. (65) holds, with K a constant where K = 1/3 for theparticular case of spherical molecules or atoms. By inverting this relation and by using (66) wefind that

α =ε0

Nεr − 1

Kεr + (1−K)=

3ε0

Nεr − 1εr + 2

(67)

where the second equality applies to the case of K = 1/3, which is applicable to sphericalcavities. Equation (67) is the famous Clausius-Mossotti equation which relates the microscopicatomic/molecular polarizability to the macroscopic counterpart. If εr ≈ 1, then εr + 2 ≈ 3 itreduces to the dilute medium form, Eq. (63). In spite of its somewhat crude origins the Clausius-Mossotti equation is quite widely valid when applied to gases and (non-polar) liquids. The simpletheory however fails for crystalline solids where the dipole interaction is more complicated becauseof orientational effects.

1In some text books the symbol for the dielectric constant is κ or K. However, in the scientific literature thesymbol εr is invariably used, and we use this convention here.

17

Examples

1. For nitrogen gas at one atmosphere and 0 C, N = 2.52× 1025 m−3, εr = 1.00058. Sincethe gas is dilute, both formulas give α/4πε0 = 1.83 × 10−30 m3. For comparison, theexperimental value is 1.74× 10−30 m3.

2. Suppose that in a sample of diamond, P = 8.0 × 10−6 C.m−2. For diamond the atomicweight is 12, the atomic number is 6, and its density is 3.51 × 103 kg.m−3. The numberdensity is given by

N =3.51× 103

12× 1.66× 10−27= 1.76× 1029 m−3 (68)

Therefore,

p =P

N =8.0× 10−6

1.76× 1029= 4.5× 10−35 Cm (69)

If p = qδ, where q = 6 × 1.6 × 10−19 C, δ = 4.7 × 10−17 m. Given that for diamond,εr = 5.5, the value for α given by the dilute formula, as expected, does agree with resultobtained using the Clausius-Mossotti equation.

α/4πε0 = 0.81× 10−30 m3 (Clausius−Mossotti equation) (70)α/4πε0 = 2.03× 10−30 m3 (dilute formula) (71)

9 Gauss law in the presence dielectrics

Section 1.1 had a brief reminder of the use of Gauss’ law in vacuum. Let us now see how itis applied in the presence of dielectric media. To do this we need to remind ourselves that wehave control over the free charges, but not over the bound charges, since they merely appear inresponse to the applied field. We therefore need to deal in the first instance with the electricdisplacement field D since its component normal to the surface does not depend on bound chargesat all. So rather than Eq. (57) for E, we use Eq. (62) for D.As an example, let us consider a dielectric sphere of radius R and dielectric constant εr, with a(free) charge Q at its centre. Our challenge is to find the electric field E everywhere. To do this,we use Gauss’ law, which should be possible since the geometry has a high level of symmetry.Recall from the argument in the previous paragraph that we do not know the bound charges,only the free charges and so it is best to apply Eq. (62) for D. This is straightforward and weimmediately find that

D =Q

4πr2r, (72)

where r is the unit vector pointing in the r direction. Note that D is orthogonal to the surfaceand so here this field does not depend on the bound charges at all. Now we have found D it iseasy to find E using Eq. (66). Thus, for r < R we have

E =Q

4πε0εrr2r, (73)

while for r > R

E =Q

4πε0r2r. (74)

Though we have considered only a single example here, it shows quite generally how to approachproblems involving dielectrics.

18

10 Energy density of the electromagnetic field

10.1 Energy density in vacuum

In this section we study the energy density of the electromagnetic field and how it affected bythe presence of a dielectric medium. In a sense it is a generalization of the type of problemsthat we discussed at the end of the previous section. However, we first consider free chargesonly. We start with the following thought experiment: we have a set of M discrete charges Qj

that initially are all at infinity. Because of this they are independent of each other and in thisconfiguration we take the potential energy to be zero.

M = 2: Let us first consider the case in which M = 2. We let the charges, which initially wereat infinity, to have a finite mutual distance r12. The way to envisage this is to assumethat charge 1 is fixed2 and then to bring charge 2 slowly in from infinity. In this case thepotential energy is

W2 =Q1Q2

4πε0r12. (75)

M = 3: Let us now bring in the third charge Q3. Let charges Q1 and Q2 be fixed. Then theadditional potential energy that is acquired is

W ′3 =

Q1Q3

4πε0r13+

Q2Q3

4πε0r23. (76)

and so the total potential energy is

W3 =Q1Q2

4πε0r12+

Q1Q3

4πε0r13+

Q2Q3

4πε0r23. (77)

Arbitrary M : From the previous simple cases we see that

WM =1

4πε0ΣM

i,j>i

QiQj

rij=

12

14πε0

ΣMi,j 6=i

QiQj

rij, (78)

where the conditions on the sum require j to be larger than i, and unequal to i, respectively,and where the factor 1/2 expresses the fact the each ij term has acquired an identical jicounterpart. Let us now carve out of this double sum the terms for which i = 1:

Wi=1 =Q1

2

(Q2

4πε0r12+

Q3

4πε0r13+ · · ·

). (79)

Here the expression in brackets represents the potential at Q1 due to the all other charges.Denoting this potential by V1, this can be written as

Wi=1 =12Q1V1. (80)

Had we not taken i = 1 but any other, arbitrary value for i = l, say, then we would havefound QlVl/2, where, again, Vl is the potential at charge l due to al other charges.

We can now therefore write the total potential energy of the system as

WM =12ΣM

i=1QiVi, (81)

which is the result that we were after.2The force that is needed to keep charge 1 fixed does not do any work and it therefore does not contribute to

the potential energy.

19

We now found an expression for the potential energy of a set of discrete charges. However, thisbegs the following question: “how did those discrete charges come together in the first place?”,and “do we not count the potential energy associated with these charges?”. In fact, this latterpotential energy is automatically included if we write the continuous analogue of Eq. (81)

W =12

∫V ρdr, (82)

where now the charges are taken to be continuously distributed and so all charge accumulationis included. In fact, it can be shown that while Eq. (81) can be positive or negative (which iseasy to see when M = 2), below we see that Eq. (82) can never be negative. The differenceis the large potential energy that is needed to bring microscopic charges together to form themacroscopic charges.To rewrite Eq. (82) in a more useful form we use the one-dimensional approach as we did earlier.To do this we write the equation as

W =12

∫V (x)ρ(x)dx. (83)

Now we know from Section 1.3 that in one dimension ρ = ε0dE/dx, and we also know thatE = −dV/dx and so we can write

W = −ε0

2

∫V

d2V

dx2dx. (84)

We now integrate by parts and find

W = −ε0

2

[V

dV

dx

]+

ε0

2

∫ (dV

dx

)2

dx. (85)

The first term is evaluated at the edges of the interval that we are considering, provided that itcontains all the charges. However, nothing stops us from making this interval larger and larger.All this time the magnitude of the integrand decreases as |x|−3, and so the contribution fromthe first term is negligible. In the second term, which, by the way, is never negative, we replacedV/dx by −E to get the final result. Had we considered a full three-dimensional geometry thenthe analysis would have been essentially the same and we would have found

W =ε0

2

∫E2dr, (86)

and the electric energy density per unit volume is thus ε0E2/2.Let us finish with a brief comment on the approach taken in this section. We started by calcu-lating the mechanical energy associated with bringing charges together from infinity. We theninterpreted this mechanical energy as having been used to set up the associated electric field,and so the energy is thought of as residing in the field. This type of approach is similar tothat in earlier E&M courses which usually start out with the force between two charges. Theinterpretation of this mechanical result is then to introduce an electric field through the relationF = qE, and so the mechanical result is seen as the interaction between a charge and a field.

10.2 Energy density in dielectrics

Now we have evaluated the energy density in vacuum let us now see how it is affected by thepresence of a dielectric. Fortunately we have done most of the hard work so this is relatively

20

easy. In fact Eq. (82) is still valid except that we replace ρ, which previously was the only chargeavailable, is replaced by ρf , the free charge. We think about this as again bringing the chargesin from infinity, but the potential is now affected by the dielectric. Thus

W =12

∫V (x)ρf (x)dr. (87)

By the same argument as before, this can be rewritten in terms of the field, and we find

W =12

∫D ·E dr, (88)

which reverts to result (86) in the absence of the dielectric. The energy density per unit volumeis thus

D ·E2

=ε0εr E ·E

2=

D ·D2ε0εr

, (89)

where the last two equalities are only valid for the dielectric media that we have been dealingwith until now.

11 Capacitors

We have developed a powerful machinery to deal with dielectric media. Let us now use thisin the simple case of a parallel plate capacitor. Although it is our intention is to look at theeffect of dielectric material between the capacitor plates, we begin by recalling the properties ofa vacuum capacitor. For a parallel plate capacitor consisting of two plates of area A, separatedby a distance d, one with a charge of +Q and the other −Q, the electric field between the platesis given by

E =σ

ε0=

Q

ε0A(90)

The potential difference is given by

V = Ed =Qd

ε0A, (91)

from which we get the capacitance of the parallel plate capacitor

C =Q

V=

ε0A

d. (92)

11.1 Dielectric-filled capacitor

Let us consider a capacitor that is filled with a dielectric with dielectric constant εr. There is anumber of ways to deal with this problem, but, as in Section 9, it is probably easiest to realizethat displacement field D is unaffected by the dielectric as its sources are free charges only. FromEq. (61), and the same mathematical argument as applied earlier to E, we note that D = σf .Therefore, since D = ε0εrE, this can be written as E = σf/(ε0εr), and thus

V =σfd

ε0εr, (93)

since the potential is by definition the integral over the electric field and not over the displacementfield. We next introduce the total charge Q = σfA and so

V =Qd

ε0εrA, (94)

21

and so the capacitance is

C =Q

V=

ε0εrA

d, (95)

and thus the dielectric increase the capacitance by a factor εr. The physical reason for this canbe seen in Fig. 8: in the capacitor the dielectric becomes polarized with positive bound surfacecharges appearing at the negatively charged capacitor plate, and vice versa. These bound chargesset up a field that partly cancels the field due to the free charges, hence, for the same amount offree charge, the potential goes, down, and thus the capacitance goes up see (Fig. 9).

+++++++++

---------

+

+

+

+

+

-

-

-

-

-

-Q +Q

A

d

Figure 8: Schematic of a dielectric-filled capacitor

E+++ + + + + + + +

----------

P

A

δ

Figure 9: Polarisation of the dielectric between the capacitor plates

As a final calculation we determine the potential energy U of the capacitor. Since D = σf = Q/Awe find from the last of (89) that

U =Q2d

2Aε0εr. (96)

Then using Eq. (95) we find that

U =12CV 2, (97)

which in fact is valid for any type of capacitor, not just for parallel plate capacitors. Too seethis, briefly, we note that U =

∫QdV = C

∫V dV which gives the answer above.

22

Examples

1. Determine the capacitance of a parallel plate capacitor, with plates of area A and spacingd, if a slab of dielectric with dielectric constant εr and thickness x (< d) is adjacent to oneof the plates.

Using D makes this problem quite straightforward. Suppose the magnitude of the chargeon each plate is Q; this is free charge and determines D which is given by

D = σf =Q

A(98)

It follows that the electric fields in the dielectric-free and dielectric-filled regions are re-spectively Q/ε0A and Q/εrε0A. The potential difference between the plates is therefore

V =Q

ε0A(d− x) +

Q

εrε0Ax = Q

(d− x

ε0A+

x

εrε0A

)(99)

With a little algebra we find

C =Q

V=

(d− x

ε0A+

x

εrε0A

)−1

=κε0A

x + εr(d− x)(100)

Note that this result has the correct limits as x → 0 (no dielectric) and x → d (completelyfilled with dielectric).

2. Suppose a dielectric slab (εr = 2.50) is inserted between the plates of an electrically isolatedcharged capacitor, which has a vacuum capacitance of 100 pF capacitor and is charged to100 V. Compare the energy stored before and after, and explain any difference.

When the dielectric is inserted the capacitance increases by a factor εr to 250 pF. As thecapacitor is electrically isolated, the charge on the plates remains unchanged, with a valueof

Q = CV = 100× 10−12 × 100 = 1.00× 10−8 C (101)

As Q remains constant we use U = Q2/2C to calculate the stored energy. Before insertionof the dielectric slab,

U =Q2

2C=

10−16

2× 100× 10−12= 5.00× 10−7 J (102)

When the dielectric is inserted C increases by a factor εr = 2.5, while as noted Q remainsconstant. The energy stored is therefore less by a factor of 2.5, i.e 2.00 × 10−7 J. Whathappened to the energy? It has been consumed doing work to draw in the dielectric slab.In the absence of friction the dielectric slab would accelerate into the capacitor, overshootand oscillate back and forth. With friction it is drawn in and the excess energy dissipatedas heat.

3. Repeat the above problem, this time with the voltage kept constant.

The capacitance again increases by a factor εr, so the charge Q = CV increases from itsinitial value (1.00× 10−8 C) to Q = 2.50× 10−8 C. As V remains constant it is convenientto calculate the energy stored using U = 1/2CV 2, which gives

U =12CV 2 =

250× 10−12 × 104

2= 1.25× 10−6 J (103)

23

Thus the energy has increased from the initial value of 5.00× 10−7 J to 1.25× 10−6 J, i.e.by an amount of 7.50×10−7 J. Where does this energy come from? In this case, to keep Vconstant the capacitor must remain connected to a power supply, from which charge flowsto bring the charge up to its new higher value. The energy transferred from the powersupply is

U = ∆Q V = 1.50× 10−8 × 100 = 1.50× 10−6 J (104)

This is in fact twice the amount required by the capacitor, so again the dielectric slab isdrawn into the capacitor.

12 Dielectric materials containing polar molecules

Molecules which have an intrinsic (i.e. permanent) dipole moment (such as water) are calledpolar molecules. In the absence of an electric field, such dipoles in a gas or liquid have randomorientations due to thermal motion, and hence P = 0. If an electric field is applied the dipolestend to align with the field, but thermal motion prevent complete alignment. In fact it is clearthat the degree of alignment depends upon the relative values of the energy associated with thedipole in the electric field (∼ pE), where p is the intrinsic dipole moment, and the thermal energy(∼ kT ), i.e. upon the ratio pE/kT .The polarisation of the medium at temperature T is given by the Langevin equation (see Ap-pendix C), shown in figure 10

P = Np

(coth

(pE

kT

)− 1

pE/kT

)(105)

If pE/kT ¿ 1, then

P → Np

(pE

3kT

)(106)

while for pE/kT À 1,P → Np (107)

i.e., the dipoles are completely aligned with the electric field.

12.1 Frequency dependence of polarisation

As induced dipole moments require displacement of electrons only, the dielectric constant of amaterial consisting of non-polar molecules remains constant to very high frequencies. Dielectricconsisting of polar molecules can have large dielectric constants due to alignment of the intrinsicdipole moments. However as the frequency of the applied field is increased the dipoles cannot re-orient quickly enough to follow the field. Thus with increasing frequency the dielectric constantfalls to a value corresponding to induced polarisation due to displacement of electrons only. Aswater molecules have an intrinsic dipole moment, water provides a good example: at dc εr = 80,while at 600 THz (500 nm) εr = 1.77. The latter value characterises the response of water tovisible light frequencies. We will see later when we discuss electromagnetic waves that refractiveindex and dielectric constant are related simply

n =√

εr (108)

Thus at optical frequencies, the refractive index of water is√

1.77 = 1.33.

24

0 5 10 150

0.2

0.4

0.6

0.8

1

pE/kT

Pol

aris

atio

n, P

/np

Figure 10: The Langevin equation. The dashed lines show the limiting behaviour for pE/kT ¿ 1and pE/kT À 1.

13 Other dielectric behaviour

13.1 Anisotropic dielectric media

In an isotropic material, for which all directions are equivalent, the vectors E, P and D areparallel, and ε, εr and χe are constants. We consider non-linear materials, for which the latteris not necessarily true, later. In an anisotropic material P and D is not necessarily parallel toE, nor to each other. In this case the constants ε, εr and χe become tensors. For example,instead of P = ε0χeE, we have (neglecting for notational simplicity the subscript e for electricsusceptibility)

Px

Py

Pz

= ε0

χxx χxy χxz

χyx χyy χyz

χzx χzy χzz

Ex

Ey

Ez

(109)

Thus the constant χ is replaced by a 3× 3 matrix X, called in this physical context a 2nd ranktensor. Equation 109 can be written as

P = ε0XE (110)

Similarly, D = ε0(1 + χe)E becomes

Dx

Dy

Dz

= ε0

1 + χxx χxy χxz

χyx 1 + χyy χyz

χzx χzy 1 + χzz

Ex

Ey

Ez

(111)

For particular materials some of the tensor elements may be zero. However, P and D and Eare parallel to each other only is χij = 0 for i 6= j and the χii are all equal, i.e. the material isisotropic.

25

13.1.1 Nonlinear dielectric media

For a nonlinear material, P is a non-linear function of E:

P = ε0(χ(1)E + χ(2)EE + χ(3)EEE + . . .) (112)

where we wrote EE since the square of a vector is not defined (of course the square of itsmodulus, a scalar, is well defined). For typical field χ1E À χ2E

2 À χ3E3 À · · · . Amongst

many others things, transparent nonlinear dielectric materials are used to generate harmonics ofintense light from lasers. To see this consider the χ(2) term in the expansion above. If the electricfield varies as cos(ωt), then the nonlinear polarization associated with this field is proportionalto cos2(ωt) = [1 + cos(2ωt)]/2. Let us forger about the constant part here. Rather, the keyobservation is that even though the field has a frequency ω, the response of the medium P hasa component that varies at 2ω.3 If an experiment is well designed, this can lead to efficient“frequency doubling.” For example, infrared light at 1064 nm from a pulsed Nd-YAG laser canbe efficiently frequency doubled to green light at 532 nm with tens of % of efficiency using asuitable nonlinear crystal.

13.2 Piezoelectricity

Piezoelectric materials have the property that mechanical stress produces polarisation withinthe medium, resulting in an electric field in the medium and therefore a potential differencebetween opposite faces. Stress causes a small distortion of the crystal structure, which in certainmaterials can lead to a dipole moment. The inverse behaviour also occurs: the application ofa potential difference produces stress and hence strain (fractional change in dimensions) of thesample. Thus piezoelectric materials can be used as electromechanical transducers to transforma force to a potential difference and conversely a potential difference to a dimension change.Examples of common piezoelectric materials include quartz (SiO2), barium titanate (BaTiO3)and lead zirconate titanate (PZT: ceramics with varying proportions of Pb, Zr and Ti).Applications of piezoelectric materials include:

• Stress → potential difference: gas lighters, microphones

• Potential difference → strain: nano-position controllers, speakers, alarms, scanningtunneling microscopes

A particularly important application of quartz is as a circuit component for stabilisation of elec-tronic oscillators (quartz oscillators). Because of its piezoelectricity the electrical oscillations canbe locked to a mechanical resonance of the crystal. Examples include applications which requirehigh stability electric oscillations, such as clocks and watches and the generation of clock frequen-cies in digital circuitry such as microprocessors, In critical applications the crystal mechanicalresonance frequency is further stabilised within high tolerances by keeping its temperature con-stant.In general the direction of the induced polarisation (and hence induced electric field E) is notparallel to the stress T. The relation between these two vector quantities involves a tensor ( a3× 3 matrix). Thus

E = eT (113)3This, by the way, is the hallmark of nonlinear effects. Recall that linear effects cannot change frequencies.

Linear effects can affect different frequencies in different ways, but cannot generate frequencies that were notthere before–only nonlinear effects can do so.

26

where e is the piezoelectric tensor. In expanded notation,

(Ex, Ey, Ez) =

exx exy exz

eyx eyy eyz

ezx ezy ezz

Tx

Ty

Tz

(114)

This, in general, one component of E depends on all components of T, for example,

Ex = exxTx + exyTy + exzTz (115)

or in condensed notation Ei = eijTj where i = x, y, z and repeated subscripts imply summation.

13.3 Ferroelectric and antiferroelectric materials

A ferroelectric crystal has a permanent polarisation, Thus even in the absence of an appliedelectric field the center of positive charge is displaced from the centre of negative charge. Allferroelectric materials are piezoelectric (e.g. BaTiO3 is ferroelectric), but not all piezoelectricmaterials are ferromagnetic (e.g. quartz). The behaviour of ferromagnetic materials parallelsthat of ferromagnetic materials (whence the name, despite the absence of iron), which will becovered in more detail later in this module.Antiferroelectric materials have neighbouring domains with opposite polarisation so that thenett polarisation is zero. When stress is applied the cancelation is destroyed resulting in a nettpolarisation.

13.4 Electrets

These are materials which contain molecules with permanent dipole moments, and sufficientlylow melting points (e.g. waxes) so that the dipoles can be aligned in the molten state, and thealignment frozen in when the material solidifies.

Appendices

A Electric field at the centre of a spherical cavity in adielectric (advanced)

If there is a spherical cavity in a polarised dielectric there is bound surface charge on the surfaceof the cavity given by P · n. Recall that n is the outward pointing normal, i.e. the normal thatpoints from the medium towards the air. In this particular case, therefore, n points into thesphere and the surface charge on the right-hand side of the sphere is thus positive, and that onthe negative side is negative. The electric field at the centre can be calculated by first determiningthe field due to the bound surface charge on the ring surface element shown in figure 11. Theelectric field on the axis of a uniform ring of charge q, of radius r, on the axis at a distance dfrom the centre of the ring is given by

E =q cos θ

4πε0r2where cos θ =

d

r(116)

27

θ

r

P

+

+

+

+

+

+

+

-

-

-

-

-

-

-

Figure 11: Calculation of the electric field at the centre of a spherical cavity in a dielectric

For the ring element shown we replace q by the bound surface charge density P cos θ multipliedby the surface area of the ring element, 2πr sin θ × rdθ. Thus the field at the centre due to thering of bound charge is given by

dE =2πr sin θ × rdθ P cos θ cos θ

4πε0r2=

P

2ε0cos2 θ sin θdθ (117)

The resultant field at the centre of the sphere is obtained by integrating over θ = 0 → π:

E =∫ π

0

P

2ε0cos2 θ sin θ =

P

2ε0

[− cos3 θ

3

0

=P

3ε0(118)

B Derivation of the Clausius-Mossotti equation for a ran-dom medium (advanced)

The derivation of the Clausius-Mossotti equation is subtle and is based on an argument originallydue to Lorentz. The aim of the exercise is determine the difference between the applied fieldand the field experienced by a particular dipole inside the medium. The two are likely to bedifferent since the the each dipole is surrounded by other dipoles which are very close by, whichhave strong localized fields. Because of thermal effects, these strong field are likely to vary withtime. So the first part in the argument is to average over position and over time so that thestrong fields average out.The next step is to determine the average contribution of the surrounding dipoles. Here we useLorentz’ argument: let us distinguish the dipoles which are far away from those which are closeby. We will treat the former in some average sense, while the nearby dipoles need to be treatedone-by one. In order to determine which dipoles are far away and which are close by we drawan imaginary sphere around the dipole of interest. The dipoles falling within the sphere areconsidered close by, while the other are considered far away. Of course the radius of the spherehas to be chosen appropriately, but that can be done.We now consider the dipoles which are close-by. According to one of the exercises in Section 2

28

the electric field of a dipole is given by

E(r) =1

4πε0

3(p · r)r− |r|2pr5

(119)

Let us now assume that we are dealing with a random medium such as a glass. The atoms canthen be considered to be located at arbitrary positions. To get the field at our particular dipole,therefore, we need to average over all positions (the radius of the sphere needs to be chosensufficiently large that it contains a sufficient number of atoms to make the averaging sensible.Let us start with determining the average x-component of the electric field from (119)

〈Ex〉 =1

4πε0

⟨3(pxx + pyy + pzz)x− |r|2px

r5

⟩(120)

where the 〈·〉 indicates a spatial average. Let us look at the numerator

3(px〈xx〉+ py〈yx〉+ pz〈zx〉)− 〈r2〉px. (121)

Now the x and y positions of the atoms are uncorrelated, and thus 〈xy〉 = 〈x〉〈y〉, and since thepositions are random 〈x〉 = 〈y〉 = 0. The xz term is dealt with in a similar way, but the xx termis not. This is easy to see: x2 is never negative, and so its average cannot vanish (of course it itnot always zero either). So the numerator can be written as

3px〈xx〉 − 〈r2〉px. (122)

But〈r2〉 = 〈x2 + y2 + z2〉 = 〈x2〉+ 〈y2〉+ 〈z2〉 = 3〈x2〉, (123)

since x, y and z are equivalent directions. Therefore 〈Ex〉 = 0, and the same is true for the otherfield components. We have thus found that for a medium where the atoms are located in randompositions, the average contribution of the near by dipole averages to zero. It should be notedthat the same conclusion is reached for many other situations, for example when the dipoles arelocated on a cubic grid. So the only contribution to the field comes from the dipoles which arelocated far away. However, we already now their contribution from Eq. (118): E = P/(3ε0), andthis is the result we are after.This argument leaves a substantial number of untied knots. To name but one: if only the dipoleslocated far away contribute to the electric field, then surely the boundaries of the medium areimportant too. We disregard all these issue and simply make the point that the result we havefound here is qualitatively correct. We should not take the factor 1/3 too seriously, but thegeneral form of the result is correct.

C Derivation of the Langevin equation (Advanced)

The dielectric consists of permanent dipoles which have a range of orientations determined bythe Boltzmann distribution. Consider a dipole making an angle θ to the electric field, with anenergy −pE cos θ. The proportion of dipoles with this energy is proportional to the solid angledΩ = 2π sin θdθ. As the component of the dipole in the direction of the electric field is p cos θ,the polarisation per unit volume is given by

P = A 2πp

∫ π

0

exp(−pE cos θ

kT

)cos θ sin θdθ (124)

29

where A is a constant obtained by noting that

N = A 2π

∫ π

0

exp(−pE cos θ

kT

)sin θdθ (125)

where N is the density of dipoles. The first integral can be done by parts; the second is straight-forward. The results are (where a = pE/kT )

P = A 2πp

(1a2

)(a(ea + e−a)− (ea − e−a)) (126)

N = A 2π

(1a

)(ea − e−a) (127)

from which we find

P = Np

(coth

(pE

kT

)− 1

pE/kT

)(128)

30