Section 4: Electrostatics of DielectricsDielectrics and Polarizability There are two large classes of substances: conductors and insulators (or dielectrics). In contrast to metals where charges are free to move throughout the material, in dielectrics all the charges are attached to specific atoms and molecules. These charges are known as bound charges. These charges are able, however, to be displaced within an atom or a molecule. Such microscopic displacements are not as dramatic as the rearrangement of charge in a conductor, but their cumulative effects account for the characteristic behavior of dielectric materials. When an external electric field is applied to a dielectric material this material becomes polarized, which means that acquires a dipole moment. This property of dielectrics is known as polarizability. Basically, polarizability is a consequence of the fact that molecules, which are the building blocks of all substances, are composed of both positive charges (nuclei) and negative charges (electrons). When an electric field acts on a molecule, the positive charges are displaced along the field, while the negative charges are displaced in a direction opposite to that of the field. The effect is therefore to pull the opposite charges apart, i.e., to polarize the molecule. It is convenient to define the polarizability α of an atom in terms of the local electric field at the atom: p = α Eloc . where p is the dipole moment. For a non-spherical atom α will be a tensor. (4.1) There are different types of polarization processes, depending on the structure of the molecules which constitute the solid. If the molecule has a permanent moment, i.e., a moment is present even in the absence of an electric field, we speak of a dipolar molecule, and a dipolar substance. Fig.4.1 (a) The water molecule (b) CO2 molecule. An example of a dipolar molecule is the H2O molecule in Fig.4.1a. The dipole moments of the two OH bonds add vectorially to give a nonvanishing net dipole moment. Some molecules are nondipolar, possessing no permanent moments; a common example is the CO2 molecule in Fig.4.1b. The moments of the two CO bonds cancel each other because of the rectilinear shape of the molecule, resulting in a zero net dipole moment in the absence of electric field. Despite the fact that the individual molecules in a dipolar substance have permanent moments, the net polarization vanishes in the absence of an external field because the molecular moments are randomly oriented, resulting in a complete cancellation of the polarization. When a field is applied to the substance, however, the molecular dipoles tend to align with the field. The reason is that the energy of a dipole p in a local external field Eloc is U = −p ⋅ Eloc . It has a minimum when the dipole is parallel to the field. This results in a net non-vanishing dipole moment of the material. This mechanism for polarizability is called dipolar polarizability. 1 The third type of polarizability arises because the individual ions or atoms in a molecule are themselves polarized by the field. The field displaces Na+ and Cl. But in substances in which such bonds are missing . Thus the Na+ ion is polarized because the electrons in its various shells are displaced to the left relative to the nucleus.4.4. because the field tends to displace the positive ion Na+ to the right (see Fig. (2) which is the sum of the electronic. Fig. again because of the relative displacement of the orbital electrons. 4.3. each of the Na + and Cl . This excludes ferroelectricity from discussion.such as Si and Ge . the greatest contribution comes from the dipolar part. but otherwise is no real restriction provided the field 2 . We assume that the response of the system to an applied field is linear. the total polarizability is given by α = αe + αi + αd .2 Ionic polarization in NaCl. as in NaCl.ions are polarized. ionic substances the electronic part is often of the same order as the ionic. The electronic contribution is present in any type of substance.ions in opposite directions. as we shall see. In general.2). as shown in Fig. respectively. then the field tends to stretch the lengths of these bonds.3 Electronic polarization: (a) Unpolarized atom. Polarization If electric field is applied to a medium made up of large number of atoms or molecules. for instance. changing the bond length. The effect of this change in length is to produce a net dipole moment in the unit cell where previously there was none. This occurs in NaCl. Since the polarization here is due to the relative displacements of oppositely charged ions. Electronic polarizability arises even in the case of a neutral atom. we speak of ionic polarizability.4. The relative magnitudes of the various contributions in (2) are such that in nondipolar. We are clearly speaking here of electronic polarizability. the charges bound in each molecule will respond to applied field which will results in the redistribution of charges leading to a polarization of the medium. This is the case for water. In dipolar substances.ionic polarizability is absent. Fig. The polarization is a macroscopic quantity because it involves averaging of the dipole moments over a volume which contains many dipoles. but the presence of the other two terms depends on the material under consideration. and the negative ion Cl. In the case of NaCl. resulting in a stretching in the length of the bond. and dipolar polarizabilities. or dipolar. (b) Atom polarized as a result of the field. ionic. Ionic polarizability exists whenever the substance is either ionic. however. because in each of these classes there are ionic bonds present. The electric polarization P(r') is defined as the dipole moment per unit volume.to the left.If the molecule contains ionic bonds. therefore. as in H2O. r') caused by the configuration of moments in δV is given without approximation by δΦ (r . 4πε 0 V ⎩ ⎣ ⎦⎭ 3 (4. (4.5) To simplify this equation we use the identity r − r′ r − r′ 3 = −∇ 1 1 .6) where ∇′ implies differentiation with respect to r'.7) We can take this integral by parts. We now treat δV as (macroscopically) infinitesimal. r′) = P (r′) ⋅ ( r − r′ ) ⎤ 1 ⎡ ρ (r′) δV + δV ⎥ . ⎟⎟ + ′ r − r ⎠ (4.1). The first term is the contribution from free charges and the second term is due to a volume distribution of dipoles. ⎠ ⎦⎥ (4. the contribution to the potential δΦ(r. An important point to note that the electric field which enters eq. The macroscopic field is the average over volume with a size large compared to an atomic size. which results in 3 . This allows us to rewrite Eq. ⎢ 3 4πε 0 ⎢⎣ r − r′ r − r′ ⎥⎦ (4.3) P = ε 0 χeE .5) as follows: Φ (r ) = ⎡ ρ (r′) ⎛ 1 + P(r′) ⋅ ∇′ ⎜ ⎢ ∫ ⎜ 4πε 0 V ⎣⎢ r − r′ ⎝ r − r′ 1 ⎞⎤ 3 ⎟⎟ ⎥ d r ′ .(4.9) We can now use the divergence theorem to transform the third term in eq. ⎢ 3 4πε 0 V∫ ⎣⎢ r − r′ r − r′ ⎦⎥ 1 Φ (r ) = (4. Then the induced polarization P is parallel to E with a coefficient of proportionality that is independent of direction: (4. As a further simplification we suppose that the medium is isotropic. put it equal to d3r'.strengths do not become extremely large.9) to the integral over surface of the dielectric. (4. = ∇′ r − r′ r − r′ (4. The constant χe is the electric susceptibility of the medium. The free charge contained in volume δV is ρ(r') δV and the dipole moment of δV is P(r')δV. (4. If there are no higher macroscopic multipole moment densities. Now we look at the medium from a macroscopic point of view assuming that the medium contains free charges characterized by the charge density ρ and bound charges characterized by polarization P.3) is the a macroscopic electric field which is different from a local electric field entering eq. We can build up the potential and the field by linear superposition of the contributions from each macroscopically small volume element δV at the variable point r'. Taking into account ⎧⎪ P (r′) ∇′ ⋅ ⎨ ⎪⎩ r − r′ ⎫⎪ ⎛ 1 ⎬ = P (r′) ⋅ ∇′ ⎜⎜ ⎪⎭ ⎝ r − r′ ⎞ 1 ∇′ ⋅ P (r′) .4) provided r is outside δV. and integrate over the volume of the dielectric to obtain the potential ⎡ ρ (r′) P (r′) ⋅ ( r − r′ ) ⎤ 3 + ⎥ d r′ .8) an integration by parts transforms the potential into Φ (r ) = 1 ⎧⎪ ρ (r′) ⎡ P(r′) ⎤ ⎪⎫ 1 ∫ ⎨⎪ r − r′ − r − r′ ∇′ ⋅ P(r′) + ∇′ ⋅ ⎢⎢ r − r′ ⎥⎥ ⎬⎪ d r ′ . the polarization of the medium produces an effective charge which can be interpreted as a macroscopic bound charge or polarization charge. Consider a small pill box enclosing a small section of the volume and surface of the polarized material. In this case the average polarization charge inside the dielectric is zero.12) is explicitly included in −∇ ⋅ P (r ) due to the abrupt change of the polarization at the surface.4. In this case the surface integral (the third term in this equation) vanishes due to the assumption of the finite volume of the dielectric. (a) bulk charge density due to the divergence of polarization. the surface polarization charge (4. The expression for the potential then becomes 1 d 3r ′ Φ(r ) = (4. it will contain equal amount of positive and negative charges and the net charge will be zero. the polarization charge appears on the surface on the dielectric.5 Assume that polarization P(r) has discontinuity at the surface as is shown in Fig.(4. The bulk charge density is given by ρ P (r ) = −∇ ⋅ P (r ) . as is seen in Fig.4b.4a. Fig.5.4 Origin of polarization-charge density. a b Fig. (4. This can be seen from the following consideration. because if we take a macroscopic volume. Therefore.11) The presence of the divergence of P in the effective charge density can be understood qualitatively. (b) surface charge density due to uncompensated charges of the surface.4. where the tails of the dipoles are concentrated.12) This contribution is present even for the uniform polarization within a finite volume. The surface charge density is σ P ( r ) = P (r ) ⋅ n . (4. there is an excess of negative charge. there is a net positive (negative) charge on the surface which is not compensated by charges inside the dielectric. On the other hand if we consider a volume including a boundary perpendicular to the direction of polarization. There are two contributions to the bound charge – bulk and surface.13) ∫ r − r′ [ ρ (r′) − ∇′ ⋅ P(r′)] . In deriving Eq. From the divergence theorem we have 4 . If the polarization is nonuniform there can be a net increase or decrease of charge within any small volume. at the center of this region.Φ (r ) = 1 ∫ 4πε 0 V d 3r ′ 1 ρ (r′) − ∇′ ⋅ P (r′)] + [ r − r′ 4πε 0 v∫ S P (r′) ⋅ n da .10) As follows from this expression. r − r′ (4. 4. 4πε 0 all space In this case.10) we can integrate over all the space. For example in Fig.4. S V This is particularly useful way to represent Gauss’s law because it makes reference only on free charges.17) Thus. (4. D = ε 0E + P . Using the electric displacement the Gauss’s law takes the form ∇⋅D = ρ . 5 . This is the consequence of the charge conservation – by inducing an electric polarization in a material we do not change the total charge. Connecting D and E is necessary before a solution for the electrostatic potential or fields can be obtained.23) v∫ D ⋅ nda = ∫ ρ (r )d r .21) Because this field is generated is generated by free charges only. For a linear response of the system (4. (4. We can. the polarization charge can always be represented by ρ P = −∇ ⋅ P .18) It is important to note that the total polarization charge is always equal to zero.19) We can therefore.14) S If the region is small enough this results in ( Pout ⋅ n − Pin ⋅ n ) da = ∇ ⋅ Pd 3r .∫ ∇ ⋅ P(r)d r = ∫ P(r) ⋅ nda . Mathematically this fact can be easily seen from eq. (4.3) the displacement D is proportional to E. make a general statement that the presence of the polarization produces an additional polarization charge so that the total charge density becomes ρtotal = ρ free + ρ P = ρ − ∇ ⋅ P .24) ε = ε 0 (1 + χ e ) .20) It is convenient to define the electric displacement D.15) Taking into account that Pout = 0 and Pin = P we have P ⋅ nda = −∇ ⋅ Pd 3r . (4.25) is the electric permittivity. (4. (4. (4. (4. 3 V (4.18) – the integration of the polarization charge over any closed surface which enclosed the volume of a polarized material gives zero according to the divergence theorem.(4. rewrite the expression for the divergence of E as follows: ∇⋅E = 1 ε0 [ ρ − ∇ ⋅ P] . εr = ε/ε0 = 1 + χe is called the dielectric constant or relative electric permittivity. where D = ε 0E + P = ε 0E + ε 0 χ eE = ε E . therefore. (4. (4. and we still have the surface polarization charge so that on the surface σ P da = ρ P d 3r .16) Thus −∇ ⋅ P must have a delta-function at the surface.22) In the integral form it reads as follows: 3 (4. A sphere with a radial polarization distribution 6 (4. then ε is independent of position. ++++ E P −−−− Fig. In the absence of free charges the electric displacement is zero and therefore D = ε 0 E + P = 0 everywhere. The ratio of the electric field between the plates before and after the dielectric was introduced is Edielectic Edielectic Edielectic 1 = = = . within the slab the electric field is E = − P / ε 0 . D responds to only free charges. Assume that a battery voltage is applied to the plates of the capacitor so that the plates acquire a surface charge σ. The reduction can be understood in terms of a polarization of the atoms that produce fields in opposition to that of the given charge.7b). Since E = ( D − P ) / ε 0 . except that the electric fields produced by given charges are reduced by a factor ε/ε0. we see that E decreases in magnitude.7a).28) . In the absence of dielectric material the polarization is zero in all the space and therefore ⎧0. ε (4.4. The Gauss’s law (4. Capacitor with a dielectric material inside Another simple example is the parallel plate capacitor (Fig. 1. but also uniform. inside Fig. so it changes.27) b Now we remove the battery so that the surface charge is fixed and slide a dielectric slab between the plates (Fig.4.26) In this case all problems in that medium are reduced to those with no electric polarization. One immediate consequence is that the capacitance of a capacitor is increased by a factor of ε/ε0 if the empty space between the electrodes is filled with a dielectric with dielectric constant ε/ε0.4. thus it is unchanged by the introduction of the dielectric slab. Therefore. Now we consider a couple of simple examples.6). A slab with a uniform polarization pointing perpendicular to the surface Assume that we have a slab of a dielectric which has a uniform polarization pointing in the z direction (Fig.7 a (4. whereas outside the slab E = 0 . Evacuum D / ε 0 ε Edielectic / ε 0 ε r where εr is the dielectric constant. giving rise to surface polarization charges. D = ε 0E = ⎨ ⎩ σ .4. and P is parallel to E.6 2.If the dielectric is not only isotropic. E responds to all charges. 3. The dielectric material obtains a uniform polarization. However.4. This result can be interpreted in terms of surface polarization charges σ P = +P on the top surface and σ P = −P on the bottom surface resulting in the electric field E = − P / ε 0 in the slab and E = 0 outside the slab. outside .22) can then be written ∇⋅E = ρ . ( 2 r ∂r r (4. inside (4. the polarization is P = D − ε 0E = εq q q rˆ − 0 2 = 2 4π r 4πε r 4π q ⎛ ε − ε 0 ⎞ rˆ ⎛ ε 0 ⎞ rˆ ⎜1 − ⎟ 2 = ⎜ ⎟ . which gives a correct electric field (4.8 This polarization of obviously produces a surface polarization charge on the surface of the sphere which is equal to σ P = P ⋅ n = P .4. Since the system has a spherical symmetry. (4. Consider a positive point charge q placed at the origin of an infinite dielectric medium of electric permittivity ε. 4π r 2 (4.31) the total surface polarization charge is 4π R 2 P .33) Consequently the electric field is E= D ε = q 4πε r 2 rˆ .21). Thus. This leads to D= q rˆ . the total bulk polarization charge is ⎛ 2P ⎞ 2 2 ⎟ 4π r dr = −4π R P . outside P=⎨ ⎩ Prˆ .34) if we use this ε εr charge in Coulomb’s law. (4.36) We have therefore a polarization at the origin which has an opposite sign to the free charge q.(4.22) to find the electric displacement D: v∫ S D ⋅ nda = q .32) where the integration is performed over a sphere of radius r centered at the origin.29) Fig. (4. We need to find polarization and polarization charges. the polarization is ⎧ 0. The magnitude of the polarization is constant and only the direction charges. ε ⎠r 4π ⎝ ε ⎠ r 2 ⎝ (4. The sum of ε q the free and polarization charge is 0 q = .Another simple example is a sphere with the radial distribution of polarization (Fig. We note that the total polarization charge is not zero because the system 7 .35) The polarization charge is ρ P = −∇ ⋅ P = − q ⎛ ε − ε 0 ⎞ ⎛ rˆ ⎞ ⎛ ε − ε0 ⎞ 3 ⎜ ⎟ ∇ ⋅ ⎜ 2 ⎟ = −q ⎜ ⎟δ (r ) . 4.18) which leads to ρ P = −∇ ⋅ P = − P∇ ⋅ rˆ = − P 1 ∂ 2 2P r )=− . 4π ⎝ ε ⎠ ⎝ r ⎠ ⎝ ε ⎠ (4.34) According to eq. Polarization of a point charge. r ⎠ 0⎝ R 3 ∫ ρPd r = ∫ ⎜ − (4. Indeed. we can use a Gauss’s law (4.8). The bulk polarization charge can be found using eq.4. The total polarization charge is zero.30) We see that the bulk polarization charge is distributed over all the sphere and diverges at the center of the sphere. According to the Gauss’s law v∫ S D ⋅ nda = σ A . Now we consider a rectangular contour C such that it is partly in one medium and partly in the other and is oriented with its plane perpendicular to the surface.9. Boundary conditions For solving electrostatics problems one needs to know boundary conditions for the electric field. a.41) . (4. one above the surface. Consider a boundary between different media. and the other below the surface. Consider a small pillbox. if we consider two points. (4. 4. Eq.38) tells us that there is a discontinuity of the D ⊥ at the interface which is determined by the surface charge. Since the curl of electric field is zero we have v∫ E ⋅ dl = 0 . This implies that E2& = E1& . then b Φ (b) − Φ (a) = − ∫ E ⋅ d l .38) where D ⊥ is the component of the electrical displacement perpendicular to the surface. b.considered is infinite and hence the positive polarization charge are located at infinity. half in one medium and half in the other.39) (E & 2 − E1& ) l . as is shown in Fig. where E & is the component of electric field parallel to the surface.9 Schematic diagram of the boundary surface between different media. (4. The contribution from the top and bottom surfaces to the integral gives A ( D2 − D1 ) ⋅ n . A E2 D2 σ E1 D1 l Fig.4.e. For the rectangular contour C of infinitesimal height this integral is equal to (4. The boundary region is assumed to carry idealized surface charge σ. The electrostatic potential is continuous across the boundary. resulting in D2⊥ − D1⊥ = σ .37) where the integral is taken over the surface of the pillbox and A is the are of the pillbox lid. (4. a 8 (4. the tangential component of electric field is always continuous.40) i. with the normal it to its top pointing from medium 1 into medium 2. Indeed. In the limit of zero thickness the sides of the pillbox contribute nothing to the flux. E can be obtained from the potential E = −∇Φ . the integral does too.As the path length shrinks to zero. ε2 (4. We must find the appropriate solution to the equations: and ∇⋅E = ρ . Fig. all the methods of solving boundary value problems in electrostatics discussed in preceding sections can readily be extended to dielectric materials. So far the procedure is completely analogous to the problem with a conducting material in place of the dielectric ε2 for z < 0. z<0.4. (4.4. as shown in Fig.4.46) (4.47) where we took in to account that σ = 0.42) Therefore.45) subject to the boundary conditions (4. But we now must specify the 9 . z >0. Boundary value problems with dielectrics Since the presence of dielectrics equations for the electric field are ∇ × E = 0 and ∇ ⋅ E = ρ / ε .4.48) where R1 = s 2 + (d − z ) 2 and R2 = s 2 + (d + z ) 2 .38) and (4. z ) will be Φ= 1 ⎛ q q′ ⎞ ⎜ + ⎟. (4. a distance d away from a plane interface that separates the first medium from another semi-infinite dielectric ε2.φ . We attempt to use the image method by locating an image charge q' at the symmetrical position A' shown in Fig.10a.10b.40) at z = 0: D2⊥ = ε 2 E2 z = D1⊥ = ε1E1z E2 x = E1x E2 y = E1 y (4.43) ∇⋅E = ρ .44) ∇×E = 0 . The surface may be taken as the plane z = 0.10b Example 1: To illustrate the method of images for dielectrics we consider a point charge q embedded in a semi-infinite dielectric. z > 0 . the electrostatic potential defined by E = −∇Φ obeys the Poisson equation ∇ 2Φ = − ρ / ε . Below we consider a few examples. with dielectric constant ε1. Then for z > 0 the potential at a point P described by cylindrical coordinates ( s. ε1 (4. 4πε1 ⎝ R1 R2 ⎠ (4.10a Fig. Since ∇ × E = 0 everywhere. 4πε 2 R1 1 (4.56) . ε1 1 ∂Φ . z < 0.11. z<0.55) (4.54) 2ε 2 q′′ = q ( ε1 + ε 2 ) We can now obtain the potential which is given by Φ1 ( z ) = (ε − ε ) 1 ⎛ q q ⎜ + 1 2 4πε1 ⎜⎝ s 2 + (d − z ) 2 ( ε1 + ε 2 ) s 2 + (d + z ) 2 Φ2 ( z) = ⎞ ⎟.potential for z < 0. ⎟ ⎠ 2ε 2 q . 10 (4. z > 0 .50) ∂⎛1⎞ ∂⎛ 1 ⎞ −s . which is given by Es = − results in ( q + q′ ) = q′′ .46) and (4.53) ε2 These equations can be solved to yield the image charges q' and q": q′ = ( ε1 − ε 2 ) q ( ε1 + ε 2 ) (4.52) and that the condition of continuous tangential component of E (4.51) and Using these equations we find that the condition for the continuous normal component of D (4. (4. Since there are no charges in the region z < 0. ⎜ ⎟ =− ⎜ ⎟ = 2 ∂z ⎝ R1 ⎠ z =0 ∂z ⎝ R2 ⎠ z =0 ( s + d 2 )3/ 2 (4. ⎜ ⎟ = ⎜ ⎟ = 2 ∂s ⎝ R1 ⎠ z =0 ∂s ⎝ R2 ⎠ z =0 ( s + d 2 )3/ 2 (4. 2 4πε 2 ( ε1 + ε 2 ) s + (d − z ) 2 1 For the two cases ε 2 > ε 1 and ε 2 < ε 1 the field lines of D are shown qualitatively in Fig. which is determined by the z-component of electric field E z = −∂Φ / ∂z .4.49) Now we try to pick the image charges in such a way that the boundary conditions (4.47).47) are satisfied. These conditions involve the following derivatives: d ∂ ⎛1⎞ ∂⎛ 1 ⎞ . leads to q − q′ = q′′ . it must be a solution of the Laplace equation without singularities in that region. Clearly the simplest assumption is that for z < 0 the potential is equivalent to that of a charge q" at the position A of the actual charge q: Φ= q′′ .46). s ∂s (4. 57) where n = − zˆ is the unit normal pointing from dielectric 1 to dielectric 2. however. where. The latter can be calculated by generalizing eq.(4. 2 2π ε1 ( ε1 + ε 2 ) ⎡ s + d 2 ⎤ 3/ 2 ⎣ ⎦ 11 (4. as z passes through z = 0.(4. (4. This implies that there is a polarization-surface-charge density on the plane z = 0. the electric susceptibility χ e is related to the dielectric constant as follows χ e = ε r − 1 . Since Pi = ( ε i − ε 0 ) Ei = − ( ε i − ε 0 ) ∇ ⋅ Φ i z =0 . Lines of electric displacement for a point charge embedded in a dielectric 1 near a semi-infinite slab of dielectric 2.58) we find from eqs.4. χ e takes a discontinuous jump. σ P = − P1z + P2 z = q ε 0 ( ε1 − ε 2 ) d .Fig.59) ⎦ = − (ε 2 − ε 0 ) z =0 2ε 2 (d − z ) 4πε 2 ( ε1 + ε 2 ) ⎡ s 2 + (d − z ) 2 ⎤ 3/ 2 ⎣ ⎦ q =− z =0 q (ε 2 − ε 0 ) d . Therefore.25).55) and (4. At the interface between the two dielectrics.61) . P = ε 0 χ e E . (4. −∇ ⋅ P = −ε 0∇ ⋅ ( χ eE ) . (4. except at the point charge q.12) to the case of two dielectrics: σ P = ( P1 − P2 ) ⋅ n .60) 2π ( ε1 + ε 2 ) ⎡ s 2 + d 2 ⎤ 3/ 2 ⎣ ⎦ Therefore. (4.2) is the polarization in the dielectric i at z = 0. The polarization-charge density is given by −∇ ⋅ P . according to eq. Inside either dielectric. and Pi (i =1.56) that P1z = − ( ε1 − ε 0 ) d Φ1 dz = − ( ε1 − ε 0 ) z =0 q ε 2 ( ε1 − ε 0 ) d − 2 2π ε1 ( ε1 + ε 2 ) ⎡ s + d 2 ⎤ 3/ 2 ⎣ P2 z = − ( ε 2 − ε 0 ) d Φ2 dz ⎧ ⎫ ( ε1 − ε 2 ) q ⎪ (d − z ) (d + z ) ⎪ − ⎨ ⎬ = 4πε1 ⎪ ⎡ s 2 + (d − z ) 2 ⎤ 3/ 2 ( ε1 + ε 2 ) ⎡ s 2 + (d + z ) 2 ⎤ 3/ 2 ⎪ ⎦ ⎣ ⎦ ⎭ z =0 ⎩⎣ (4. In the regions where χ e and hence ε is constant it can be taken out of differentiation which leads to −∇ ⋅ P = −ε 0 χ e∇ ⋅ E = 0 .11. such that Δχ e = (ε 1 − ε 2 ) / ε 0 . Consequently the problem is one of solving the Laplace equation with the proper boundary conditions at r = a. The tangential component of the electric field should be continuous on the surface and hence − 1 ∂Φ in a ∂θ =− r =a 1 ∂Φ out a ∂θ .4. The second illustration of electrostatic problems involving dielectrics is that of a dielectric sphere of radius a with dielectric constant ε r = ε / ε 0 placed in a uniform electric field. Bl and Cl are to be determined from the boundary conditions. as indicated in Fig.64) l =0 The coefficients Al .62) Example 2.61) approaches the value appropriate to a conducting surface. For the first boundary condition (4.66) r =a Since there are no free surface charge the normal component of the electric displacement should also be continuous on the surface and hence −ε ∂Φ in ∂r = −ε 0 r =a ∂Φ out ∂r .63) and (4. r a (4. σP = − q d .63) this leads 12 .64) in these boundary conditions.67) r =a Now we substitute the series (4. From the axial symmetry of the geometry we can take the solution to be of the form a Legendre polynomial expansion. Fig. 2 2π ⎡ s + d 2 ⎤ 3/ 2 ⎣ ⎦ (4. From the boundary condition at infinity suggesting that Φ → − E0 r cos θ .In the limit ε 1 = ε 0 and ε 2 → ∞ the electric field inside dielectric 2 becomes very small and hence it behaves much like a conductor. r < a (4.12 Both inside and outside the sphere there are no free charges. which at large distances from the sphere is directed along the z axis and has magnitude E0.4. (4. The other coefficients are determined from the boundary conditions at r = a.63) l =0 ∞ Φ out (r .65) we find that the only nonvanishing Bl is B1 = − E0 . θ ) = ∑ ( Bl r l + Cl r − l −1 ) Pl (cos θ ) .θ ) = ∑ Al r l Pl (cos θ ) . Then the surface-charge density (4.12. r > a (4. The solution is different for the region inside and outside the sphere and take the form: ∞ Φ in ( r . (4. r < a εr + 2 ε r − 1 a3 Φ out (r .72) The potential is therefore Φ in (r . the coefficient of each Legendre function must be equal and therefore ε 2C A1 = − E0 − 31 a ε0 C ε lAl = −(l + 1) 2ll+1 . r > a εr + 2 r2 (4.69) and (4.71) The second equations in (4.70) Since this equation must be valid for all θ. The dipole moment can be interpreted as the volume integral of the polarization P. l ≠ 1 a ε0 (4. the coefficient of each Legendre function derivative must be equal and therefore C A1 = − E0 + 31 a (4. l ≠ 1 a The second boundary condition (4.73) (4. (4.∞ ∑ Al al l =0 dPl (cos θ ) ∞ ⎛ 1 ⎞ dP (cos θ ) = ∑ ⎜ − E0 a lδ1l + Cl l +1 ⎟ l .64) leads to ∞ ∞ ⎛ l =0 ⎝ ε ∑ lAl a l −1Pl (cos θ ) = ε 0 ∑ ⎜ − E0δ1l − (l + 1)Cl l =0 1 ⎞ ⎟Pl (cos θ ) . a l +2 ⎠ (4.68) Since this equation must be valid for all θ.69) Cl Al = 2l +1 .71) can be satisfied simultaneously only with Al = Cl = 0 for all l ≠ 1 .θ ) = − E0 r cos θ + E0 cos θ .76) εr + 2 oriented in the direction of the applied field. The remaining coefficients are given in terms of the applied electric field E0 A1 = − C1 = 3 E0 εr + 2 εr −1 3 a E0 εr + 2 (4.θ ) = − 3 E0 r cos θ .74) It is easy to see that the potential inside the sphere describes a constant electric field parallel to the applied field with magnitude 3 Ein = E0 .75) εr + 2 Outside the sphere the potential is equivalent to the applied field E0 plus the field of an electric dipole at the origin with dipole moment: ε −1 3 p = 4πε 0 r a E0 . The polarization is 13 . (4. dθ a ⎠ dθ l =0 ⎝ (4. Following this reasoning we found that U= 1 ρ (r )Φ(r )d 3 r .67) shows that the results for the cavity can be obtained from those of the sphere by the replacement ε / ε 0 to ε 0 / ε .75).78) This can be thought of as producing an internal field directed oppositely to the applied field.P = ( ε − ε 0 ) E = 3ε 0 (ε r − 1) E εr + 2 0 .76). as sketched in Fig.4. (4. 2ε r + 1 (4. It is constant throughout the volume of the sphere and has a volume integral given by (4. bringing infinitesimal pieces of charge in from infinity.66) and (4. opposing.13. 4. showing the polarization on the left and the polarization charge with its associated. (4. Spherical cavity in a dielectric with a uniform field applied. Electrostatic energy in dielectrics In free space we derived the energy of a distribution of charge ρ (r ) by assembling the distribution little by little.14.13 Dielectric sphere in a uniform field E0.79) Similarly. In fact. parallel to E0. The polarization-surface-charge density is σ P = P ⋅ n = 3ε 0 (ε r − 1) E εr + 2 0 cos θ . for example. electric field on the right. the field outside is the applied field plus that of a dipole at the origin oriented oppositely to the applied field and with dipole moment ε −1 3 p = −4πε 0 r a E0 .4. (4. the field inside the cavity is uniform. so reducing the field inside the sphere to its value (4.14. inspection of boundary conditions (4. ∫ 2 14 (4. can be handled in exactly the same way as the dielectric sphere. as shown in Fig. 4. Thus. The problem of a spherical cavity of radius a in a dielectric medium with dielectric constant ε r = ε / ε 0 and with an applied electric field E0 parallel to the z axis. and of magnitude: Ein = 3ε r E0 .77) Fig.81) .80) 2ε r + 1 Fig. This corresponds to integrating from zero field up to the final field (a functional integration).84) Now and hence integrating by parts we obtain δ U = ∫ ∇ ⋅ (δ DΦ ) d 3r + ∫ δ D ⋅ Ed 3r . D U= 1 3 d r ∫ δ ( D ⋅ E) . When dielectrics are present we will use a somewhat different argument (which still corresponds to the same procedure). (4.82) where the integration is performed over all space. where δ D is the resulting change in D. The point is that this is the interaction energy of δρ (r ) with the sources already present (and which produce Φ(r ) ). which vanishes for a localized charge distribution if the integration is performed over all space: ∫ ∇ ⋅ (δ DΦ ) d r = v∫ (δ DΦ ) ⋅ nda = 0 . To first order in δρ (r ) . Imagine that some infinitesimal change in the charge density. (4.87) So far it implies to any material. so we can write the change in the energy as δ U = ∫ ( ∇ ⋅ δ D ) Φd 3 r . (4. The change in D which arises as a consequence of the change δρ (r ) in the charge density is related to the latter by the equation ∇ ⋅ ( D + δ D ) = ρ + δρ and therefore δρ = ∇ ⋅ δ D . δρ (r ) .85) The divergence theorem turns the first term into a surface integral. 3 V (4. 2 2 (4. (4.89) Now we build the free charge up from zero to the final configuration. Now. is made. 2∫ 0 15 (4.83) ∇ ⋅ (δ DΦ ) = ( ∇ ⋅ δ D ) Φ + δ D ⋅ ( ∇Φ ) = ( ∇ ⋅ δ D ) Φ − δ D ⋅ E . as we will see. then D = ε E . if the material is a linear dielectric. the work done is equal to δ U = ∫ δ D ⋅ Ed 3 r .88) Thus the change in the electrostatic energy is δU = 1 δ ( D ⋅ E ) d 3r . 2∫ (4. and it is not clear if this work is included in the equation above.90) . (4.This is in general not true in the presence of dielectrics (however. Suppose that there is initially some macroscopic charge density ρ (r ) . the change in energy of the system is δ U = ∫ δρ (r )Φ(r )d 3r .86) S Therefore. In the presence of dielectrics work must also be done to induce polarization in the dielectric. it may be true in some cases). and fields E(r ) and D(r ) . the interaction energy of δρ (r ) with itself is second-order in small (infinitesimal) quantities. so infinitesimal increments 1 1 δ ( D ⋅ E ) = δ ( ε E 2 ) = εδ E ⋅ E = δ D ⋅ E . potential Φ(r ) . we obtain U =− 1 1 1 D ⋅∇Φd 3r = − ∫ ∇ ⋅ ( DΦ ) d 3r + ∫ Φ∇ ⋅ Dd 3r .93) . ∫ 2 2 2 (4.92) Through the divergence theorem. ∫ 2 2 Thus for a linear dielectric. The second term becomes U= 1 1 Φ∇ ⋅ Dd 3r = ∫ ρ (r )Φ(r )d 3r . the first term yields a surface term which vanishes at infinity. 2∫ (4. Indeed. the original formula is valid.91) This result is valid only for linear media and leads to eq. 16 (4. using E = −∇Φ and integrating (4.91) by parts.81). (4.This leads to U= 1 D ⋅ Ed 3r .