* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document 8930377
Survey
Document related concepts
Photon polarization wikipedia , lookup
Woodward effect wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Nordström's theory of gravitation wikipedia , lookup
History of quantum field theory wikipedia , lookup
Superconductivity wikipedia , lookup
Time in physics wikipedia , lookup
Noether's theorem wikipedia , lookup
Speed of gravity wikipedia , lookup
Lorentz force wikipedia , lookup
Mathematical formulation of the Standard Model wikipedia , lookup
Maxwell's equations wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Electric charge wikipedia , lookup
Transcript
Electrostatics of dielectrics Polarization and bound charge density A dielectric material contains permanent charge distributions. We approximate the charges which are bound to the ions and are not free to move around, by tiny dipole moments. The potential created by those dipole moments is found as follows. The potential at a point at r , due to an electric dipole p at a point r ' is p ⋅ (r − r ' ) Φ (r ) = 3 | r − r '| The potential due to the superposition of all dipoles in the matter is P (r ' ) ⋅ (r − r ' ) Φ ( r ) = ∫ dr ' | r − r ' |3 where P (r ) is the mean dipole moment density, and is the polarization field. We now transform the integrand P ( r ' ) ∇'⋅P ( r ' ) P(r ' ) ⋅ (r − r ' ) 1 = P ( r ' ) ⋅ ∇' = ∇ '⋅ − | r − r ' |3 | r − r '| | r − r '| | r − r '| to obtain (using the Gauss theorem) ∇ '⋅P ( r ' ) σb ρ b (r ' ) 1 Φ (r ) = ∫ da ' nˆ ⋅ P (r ' ) − ∫ dr ' ≡ ∫ da ' + ∫ dr ' | r − r '| | r − r '| S | r − r '| | r − r '| S The bound surface charge density is σ b = nˆ ⋅ P and the bound charge density is ρ b ( r ) = −∇ ⋅ P ( r ) Exercise: What is the averaged electric field at the center of an empty sphere of radius a having bound surface charge density? 1 Solution: Let us assume that the polarization P is along the z-direction, and use spherical coordinates. From the above expression for the potential we find σ E (r ) = −∇ ∫ da ' b = ∫ da ' σ b | r − r '| S r − r ' r − r' = da ' nˆ ⋅ P 3 | r − r ' |3 ∫ | r − r '| ( ) We now put here r=0. Since nˆ ⋅ P = − P cosθ ' , we find that the integration over the surface leaves only the z-component of the field at the center to be non-zero. That component is π E z (0) = 2π ∫ dθ ' sin θ ' (cos θ ' ) 2 P = 0 4π P 3 The displacement field Inside the matter, the charge density consists of the free (to move) charges and the bound charges. The total charge density is ρ (r ) = ρ f (r ) + ρ b ( r ) = ∇ ⋅ E (r ) / 4π It follows that (when the bound charges are approximated by the dipoles) ρ f (r ) = ∇ ⋅ P (r ) + ∇ ⋅ E (r ) / 4π = ( ) 1 1 ∇ ⋅ E ( r ) + 4πP ( r ) ≡ ∇ ⋅ D(r ) 4π 4π and hence by Gauss's theorem ∇ ⋅ D = 4πρ f ˆ D ⋅ n da = 4 π d r ρ ( r ∫ ∫ f ) S The boundary conditions on a surface between two different media are: The tangential components of E are continuous, the normal component of D jumps by 4πσ f . In linear media there is a linear relation between Pi = ∑ χ ij E j , E and P , i, j = x, y , z j 2 where χ ij is the dielectric susceptibility tensor. In highly symmetric materials χ ij = δ ij χ . As a result, there is also a linear connection between the electric and the displacement fields, given by the dielectric tensor ε ij , Di = ∑ ε ij E j = ∑ (δ ij + 4πχ ij )E j j j Again, in highly symmetric materials ε ij = δ ij ε . Exercise: A dielectric sphere of radius a, made of a linear material of dielectric constant ε , is placed in a uniform electric field. Find the potential everywhere in space. Solution: Let us take the z-direction as the direction of the applied electric field, E0 = zˆE0 . Such a field, as we know, is created by a potential Φ 0 ( r ) = − zE 0 = − r cos θE0 . Since there are no free charges in the system we have that ∇ ⋅ D = 0. Because the medium is linear, this equation becomes ∇ ⋅ D = ∇ ⋅ εE = ε∇ ⋅ E = 0 , or simply, ∇ ⋅ E = 0. The electric field E is created by the potential that we need to find, Φ , namely, E = −∇Φ . In this problem, this potential satisfies the Laplace equation, ∇ 2 Φ = 0 . Obviously, we try to solve the equation using spherical coordinates. Since there is no dependence on the angle ϕ , it follows that the general solution is ∞ Φ in ( r , θ ) = ∑ An r n Pn (cos θ ) ∞ n =0 1 Φ out ( r , θ ) = ∑ Bn r n + C n n +1 Pn (cos θ ) r n =0 As r that → ∞ , the effect of the dielectric sphere tends to zero, and therefore we expect Φ ( r → ∞, θ ) → − r cos θE0 3 B1 = − E , Bn ≠1 = 0 Hence Let us now apply the boundary conditions on the surface of the sphere, tangential electric field there is continuous, namely − r = a . The C 1 ∂Φ in ( a, θ ) 1 ∂Φ out ( a, θ ) =− ⇒ An = Bn + 2 nn+1 a ∂θ a ∂θ a The perpendicular displacement field is continuous, namely −ε ∂Φ in ∂r =− r =a ∂Φ out C ⇒ εnAn = nBn − ( n + 1) 2 nn+1 ∂r a n ≠ 1 , for which Bn = 0 , this two equations have no non-trivial solution. Hence we find that n=1, and solve for the coefficients A1 and C1 in terms of E0 . The solution is It is easy to see that for 3 E r cos θ ε +2 0 ε − 1 a3 Φ out ( r , θ ) = − E0 r cos θ + E cos θ ε + 2 0 r 2 Φ in ( r , θ ) = − Inside the sphere, the potential is Φ in ( r , θ ) = − 3 E z ε +2 0 4 and consequently the electric field is along the z-direction, and its magnitude is smaller than that of the applied field (since ε > 1 ). Outside the sphere the potential can be ε − 1 a3 Φ out ( r , θ ) = − E0 z + E z ε + 2 0 r 3 written as We see that the field outside the sphere consists of the applied field (the first term) and a second term, ε − 1 3( z / r )rˆ − zˆ ε − 1 3 cos θrˆ − zˆ Eout = E0 + E0 a 3 = E0 + E0 a 3 3 ε +2 r ε +2 r3 The second term is the field due to a dipole of the following value ε −1 3 p= a E0 zˆ ε +2 This means that the applied field polarizes the sphere and makes it into a dipole. We can calculate the polarization field, which is the volume density of the dipole. In our case, the result is simple, ε −1 3 P= E zˆ ε + 2 4π 0 In this example, the polarization field is uniform. Therefore, there is no bound charge density (since ∇ ⋅ P = 0 ), but there is bound surface charge density, given by σ b = nˆ ⋅ P , which in our case becomes σb = ε −1 3 E cos θ ε + 2 4π 0 One can check that the total surface bound charge vanishes. Exercise: An electric field line is passing through a planar interface between two insulating media with dielectric constants ε1 and ε 2 . Find the angle it makes with the normal to the interface. Solution: By the boundary conditions, we have 5 E t1 = Et 2 ε 1 E n1 = ε 2 E n 2 and hence Et1 Et 2 tan θ1 tan θ 2 = ⇒ = ε 1 E n1 ε 2 E n 2 ε1 ε2 The Clausius-Mossotti formula—the dielectric susceptibility In this model the atoms or the molecules forming the medium are treated as little conducting spheres, of radius a. The radius of each sphere, a, is chosen such that the 4πa 3 / 3 is equal to the mean atomic volume, that is, to the inverse 3 density, 1 / n . In other words, n 4πa / 3 = N, the number of atoms in the material (or the corresponding volume, number of molecules). We know that when a sphere is placed in an electric field it acquires a dipole moment, and we expect that this dipole moment is proportional to the electric field. Let us denote the proportionality constant by α ( α is called the atomic polarizability). Namely, p = αE What is the electric field that acts on each sphere? That field consists of the applied field, E , and the field exerted by all other spheres in the system (which are polarized, and therefore produce a field). We now attempt to find the latter field. We focus on a single sphere, and draw a fictitious big sphere around it. That fictitious sphere is filled with other spheres, each one exerting the field of a dipole. When all these 6 dipole fields are added up, the result is zero. For example, if all dipoles are along the zdirection, their total field is E dip 3 z i2 − ri 2 2 zi2 − xi2 − yi2 = p∑ = p∑ =0 5 5 r r i i i i since on the average we cannot distinguish between x, y, and z. Hence the field due to the dipoles in the fictitious sphere averages to zero. The contribution of all other (far away) dipoles is approximated by some average polarization gives rise to the field 4π P . Hence we find that 3 P , which (as we have found above) 4π p = α (E + P) 3 The average polarization field P on the other hand, is just P = np Therefore, 4π P = nα ( E + P) 3 Solving this equation, P=E ε −1 nα ≡ χE ≡ E 1 − nα ( 4π / 3) 4π It follows that the atomic polarizability is given by α= 3 ε −1 1 4π ε + 2 n which is the Clausius-Mossotti formula. 7 Origin of atomic (or molecular) polarizability Polar molecules The energy of a polar molecule of dipole moment p in an external electric field E is U = − E ⋅ p . The average total dipole moment of a system of polar molecules is given by (take the electric field along the z-direction) π p = pz = p ∫e −U / kT sin θ cos θdθ 0 π ∫e −U / kT sin θdθ 1 = p coth a − , a = pE / kT a 0 (Langevin's equation). [ Note that π ∫ dθ sin θe a cos θ 1 = ∫ dµe aµ = 0 −1 ( ) 1 a 2 e − e − a = sinh a ] a a At low enough temperatures this gives just p; at high temperatures pz p2 = pa / 3 = E, 3kT at low temperatures pz = p . Induced dipole moments Let us represent each molecule by harmonically-bound charges. For each charge e mω 02 r = eE so that e2 p= E mω 02 8 The molecular polarizability is then (the sum is over all the charges composing the molecule) α =∑ j e 2j m jω 2j The polarization field is (n is the number density of the molecules) e 2j P = n∑ E 2 j m jω j The displacement field is e 2j D = E + 4πn ∑ E 2 j m jω j and the dielectric constant is given in terms of the plasma frequencies ε = 1 + 4πn ∑ j e 2j m j ω 2j ≡1+ ∑ j Ω 2j ω 2j Induced dipole moments by time-dependent fields Let the applied field be E (t ) = E0 e − iωt Neuton second law for a charge e is then d 2r dr 2 m( 2 + γ + ω 0 r ) = eE dt dt where we have added dissipation, characterized by phenomenological damping. The solution is 9 e 1 r= E m ω02 − ω 2 − iωγ and the dipole moment is e2 1 p= E m ω 02 − ω 2 − iωγ When the charge belongs to free electrons, we can relate γ to the conductivity, σ , ne 2 γ = mσ and we may assume that there is no restoring force (namely, ω 0 = 0 ). When the charge belongs to an ion, we can neglect the damping. It follows that for a material made of equal density, n, of ions and free electrons, the polarization field is e2 1 e2 1 + P=n E n E M ω 02 − ω 2 m − ω 2 − iωγ The (frequency-dependent) dielectric function is 2 Ω ion 4πne 2 / M 4πne 2 / m 4πσ − =1+ 2 +i ε (ω ) = 1 + 2 2 2 ω0 − ω iωγ (1 − iω / γ ) ω0 − ω 1 − iω / γ It is a complex function! The electrostatic energy in dielectric media We have found that in free space, the energy needed to establish a charge density W = Since in the matter ρ is 1 dr ρ ( r ) Φ ( r ) ∫ 2 ∇ ⋅ D = 4πρ , we may transform this expression 10 W = 1 8π 1 d r ∇ ⋅ D ( r ) Φ ( r ) = ∫ 8π ( ) 1 d r ∇ ⋅ D ( r ) Φ ( r ) + ∫ 8π ( d r D ( r ) ⋅ − ∇ Φ (r ) ∫ ) Applying Gauss’s theorem, the first term vanishes, while the second yields W = 1 8π d r ∫ D(r ) ⋅ E (r ) Let us now find the energy it costs to insert a dielectric body into a uniform field (assuming the dielectric body is linear). The fields before the body is inserted are E0 (r ), D0 (r ) = E0 (r ) , and therefore the electrostatic energy then is W0 = 1 8π d r ∫ D0 (r ) ⋅ E0 (r ) After the body has been inserted, the energy is as given above, and hence the energy cost is W − W0 = 1 8π 1 d r D ( r ) ⋅ E ( r ) − ∫ 8π d r ∫ D0 (r ) ⋅ E0 (r ) We may now write the integrand in the form D ⋅ E − D0 ⋅ E0 = ( E + E0 ) ⋅ ( D − D0 ) + E ⋅ D0 − E0 ⋅ D The volume integral of the first term here vanishes, since one may write E + E0 = −∇F , because the rotor of the fields vanishes, and therefore 1 8π 1 d r ( − ∇ ⋅ F ( r ))( D − D0 ) = ∫ 8π Because both the divergence of d r ∫ F (r )∇ ⋅ ( D − D0 ) = 0 D0 and of D are given by the same charge density that creates the uniform field. The volume integral of the second term, W − W0 = 1 8π 1 d r ( E ( r ) ⋅ D ( r ) − E ( r ) ⋅ D ( r )) = 0 0 ∫ 8π d r E ( r ) ⋅ ( E ( r ) − D (r )) ∫ 0 11 gives W − W0 = − 1 dr P ( r ) ⋅ E 0 ( r ) 2∫ and, in fact, the integration is over the dielectric body alone, since only there exists the polarization field. Example: Let us take our results above for the dielectric sphere, and calculate the energy. We have found Eout ε −1 Ein = E0 1 − ε + 2 ε − 1 a3 2 cos θrˆ + sin θθˆ = E0 + E0 3 ε +2 r ( Din = εEin ) Dout = Eout ε −1 3 P= E ε + 2 4π 0 and Let us now find the energy cost. W − W0 = − 1 1 3 2 ε −1 1 dr P ( r ) ⋅ E 0 ( r ) = − E0 = ∫ ∫ 2 2 sphere 4π ε + 2 8π ∫E sphere 2 0 ε −1 ( −3) ε +2 Exercise: Find the potential energy of a point charge in vacuum placed at a distance d from a semi-infinite dielectric medium of dielectric constant ε . Solution: We use cylindrical coordinates, with the z-axis passing through the point charge. Let us denote by σ b (ρ ) the bound surface charge density (we assume that the medium does not carry free charges). 12 Consider the normal components of the electric field and the displacement right at z=0. E n1 = − + At z = 0 we have while at z = 0− En 2 = − (ρ (ρ qd 2 + d2 qd 2 +d ) ) 2 3/ 2 3/ 2 + 2πσ d ( ρ ) − 2πσ d ( ρ ) Therefore, by continuity of the corresponding displacement vectors, we find σ d (ρ ) = 1−ε qd 2π (1 + ε ) ρ 2 + d 2 ( ) 3/ 2 The electric field the bound charge distribution is applying at the point charge is normal (by symmetry) and its magnitude is given by E = ∫ dϕρdρσ d ( ρ ) (ρ d 2 +d ) 2 3/ 2 = 1− ε ρ qd 2 ∫ dρ 1+ ε ρ2 + d2 ( ) 3 = 1− ε q 1 + ε 4d 2 The force on the point charge is just Eq. The work required to bring the point charge from infinity to a distance d from the medium is hence 1 − ε q dz 1 − ε q W = − ∫ dzF ( z ) = − = 2 ∫ 1 + ε 4 z 1 + ε 4d ∞ ∞ d d Exercise: Find the potential created by a dipole moment located at the center of a sphere of radius a and dielectric constant ε1 which is placed in a medium of dielectric constant ε2 . Solution: Let us take the dipole moment along the z-direction and use spherical coordinates. The general solution of the potential is 13 +1 r a Φ in (r , θ ) = ∑ A + B P (cos θ ) a r =0 ∞ +1 r a Φ out (r , θ ) = ∑ C + D P (cos θ ) a r =0 ∞ At the center of the sphere there is only the potential of the dipole. Therefore, as r p cos θ ε1 r 2 p B1 = 2 . In addition, C = 0 a ε1 → 0, Φ in (r , θ ) → and hence all B = 0 , except to ensure a well-behaved potential at infinity. The boundary condition on the tangential component of the electric field at r=a then gives that A = D = 0 except for D1 and A1 . The boundary condition on the normal component of the displacement field gives ε 1 (− A1 + 2 B1 ) = ε 2 2 D1 Applying again the boundary condition on the tangential component of the electric field gives A1 + B1 = D1 Solving these equations, Φ in (r , θ ) = p cos θ 1 2(ε − ε 2 ) 2 + 3 1 r ε1 r a (ε 1 + 2ε 2 ) Φ out (r , θ ) = 3 p cos θ 1 r2 ε 1 + 2ε 2 14