Problem Set #4: 4.1,4.7,4.9 (Due Monday, March 25th)

Transcription

1 Chapter 4 Multipoles, Dielectrics Problem Set #4: 4.,4.7,4.9 (Due Monday, March 5th 4. Multipole expansion Consider a localized distribution of charges described by ρ(x contained entirely in a sphere of radius R. Then the potential far away from the sphere can be approximated by the so-called multipole expansion in spherical harmonics, i.e. Φ(x l l0 m l 4π l q Y lm (θ, φ lm. (4. r l where l 0is the monopole term, l are the dipole terms, l are the quadrupole terms, l are the octupole terms etc. Our task is to find the so-called multipole moments q lm for a given ρ(x. Since, and, according to (.76, x x 4π Φ(x l l0 m l l ρ(x x x d x (4. r< l r> l Y lm(θ,φ Y lm (θ, φ, (4. then we get Φ(x l l0 m l 4π Y lm (θ, φ l r l Y lm (θ,φ r l ρ(x d x (4.4 44

2 CHAPTER 4. MULTIPOLES, DIELECTRICS 45 where q lm Y lm (θ,φ r l ρ(x d x. (4.5 It will be convenient to define q ρ(x d x (4.6 p x ρ(x d x (4.7 Q ij (x ix j r δ ij ρ(x d x (4.8 where Q is traceless (i.e. Q Q Q 0. Then the first few multipole moments in rectangular coordinates are given by q 00 ρ(x d x q 4π 4π q,± sin(θ e iφ r ρ(x d x (p x ip y q 0 cos(θ r ρ(x d x 4π 4π p z 5 5 q,± sin (θ (cos(φ i sin(φ r ρ(x d x (Q iq Q π π 5 5 q,± sin(θ cos(θ (cos(φ i sin(φ r ρ(x d x (Q iq 5 5 q 0 ( cos (θ r ρ(x d x 6π 6π Q. (4.9

3 CHAPTER 4. MULTIPOLES, DIELECTRICS 46 and the expansion (4.4 inrectangularcoordinatestakesthefollowingform ( ( Φ(x q 00 ɛ 0 r 4π ɛ 0 r cos θ q 0 ( ( sin ɛ 0 r θeiφ q sin ɛ 0 r θe iφ q, ( 5 ( cos θ ( q 5ɛ 0 r 0 5 sin θ cos 6π 5ɛ 0 r θeiφ q ( 5 sin θ cos 5ɛ 0 r θe iφ q, ( 5 5ɛ 0 r π sin θe iφ q ( 5 5ɛ 0 r π sin θe iφ q, q r p ( z cos θ (px ip y e iφ (p x ip y e iφ ] sin θ r Q ( cos θ ( (Q iq e iφ (Q iq e iφ ] sin θ cos θ 4r ( (Q iq Q e iφ (Q iq Q e iφ ] sin θ... 8r ] q r p x x i x j Q r ij.... (4.0 r 5 i,j It will also be useful to have a multipole expansion of the electric field at x due to charge configuration at x 0. From(. weobtain E Φ ˆr r Φ ˆθ r θ Φ ˆφ r sin θ φ Φ l q lm ɛ 0 l r l l0 m l ˆr(l ˆθ im ˆφ θ sin θ ] Y lm (θ, φ.(4.

4 CHAPTER 4. MULTIPOLES, DIELECTRICS 47 Then the monopole contribution is and the dipole contribution is E ɛ 0 q, ɛ 0 q,0 ɛ 0 q, E 0 q 00 ɛ 0 r Y 00ˆr q (4. ˆr ˆr ˆθ θ ˆφ i sin θ ˆr ˆθ ] θ 4π cos θ ˆr ˆθ θ ˆφ i sin θ r r r We can now rearrange the terms ] sin θe iφ E r ˆr p z cos θ p x sin θ cos φ p y sin θ sin φ] ] sin θeiφ (4. ˆθ p z sin θ p x cos θ cos φ p y cos θ sin φ]ˆφ p x sin φ p y cos φ] (ẑ cos θ ŷ sin θ sin φ ˆx sin θ cos φ p z cos θ p x sin θ cos φ p y sin θ sin φ] ( ẑ sin θ ŷ cos θ sin φ ˆx cos θ cos φp z sin θ p x cos θ cos φ p y cos θ sin φ] (ŷ cos φ ˆx sin φp x sin φ p y cos φ] ẑ p z ( cos θ ŷ p y sin φ ( sin θ ˆx p x cos φ ( sin θ ẑ cos θ(p x sin θ cos φ p y sin θ sin φŷ sin θ sin φ(p z cos θ p x sin θ cos φ ˆx sin θ cos φ(p z cos θ p y sin θ sin φ ˆxp x cos φ ŷp y sin φ (ẑ cos θ ŷ sin θ sin φ ˆx sin θ cos φ(p z cos θ p x sin θ cos φ p y sin θ sin φ p to find a compact expression E ˆr (p ˆr p r. (4.4 This is the dipole electric field away from the dipole, but the following equation is valid everywhere, ( x x p x x p x x E x x p δ(x x (4.5 x x ɛ 0 which was obtained by substituting x x r into (4.4.

5 CHAPTER 4. MULTIPOLES, DIELECTRICS 48 Note that the expansion in spherical coordinates and thus the values of multipol moments depend on the choice of origin. However, one can show that that the lowest non-vanishing multipole moment is independent of the origin (See Problem 4.4. In general the potential and electric field of the l-pole scale as 4. Dielectric media Φ r l (4.6 E. (4.7 rl So far we discussed the electrostatic problems in the vacua or in the perfect conductors. However, most of materials are not prefect conductors, but are the so-called ponderable media, and their response to external electric field must be take into account. To obtain a macroscopic description of a ponderable media one needs to average (or coarse-grain over macroscopically small, but microscopically large regions. In electrostatics the averaging is particularly simple since the curl equation is homogeneous, i.e. E micro 0 E macro 0. (4.8 and the solution methods using electric potential can still be used to solve electrostatic problems. The charges in the ponderable media can be considered fixed in space (i.e. do not move on macroscopic distances but they can can still be slightly displaced to gain non-zero multipole moments with dipole moment being the most dominant. The overall effect of the dipoles is an induced electric polarization defined macroscopically as a sum over averaged dipole moment density, P(x N i p i, (4.9 i where p i is the dipole moment and N i is the number density of the ith type of molecule. When an excess charge is added, the ponderable material gains anetcharge, ρ(x N i (x e i ρ excess (x (4.0 i but the majority of positive and negative charges are usually canceled since e i 0 and ρ(x ρ excess (x. (4.

6 CHAPTER 4. MULTIPOLES, DIELECTRICS 49 Consider a small volume V of the ponderable media. Then the electric potential at some distant point x can be expanded into multipole moments as Φ ] ρ(x x x V P(x (x x V (4. x x If we treat V as an infinitesimal d x then Φ(x ρ(x d x x x P(x (x x x x d x ρ(x x x P(x ] ( x x ] d x x x ρ(x P(x ]. (4. It is convenient to define the so-called displacement, ɛ 0 D(x, astheapplied electric field in the absence of dielectric material such that the resulting electric field E(x ɛ 0 D(x ɛ 0 P(x. (4.4 Then the free (or excess charge distribution ρ(x would give rise to, ɛ 0 D(x, i.e. D ρ (4.5 and the polarization charge distribution ρ pol (x would give rise to polarization, ɛ 0 P(x, i.e. P ρ pol. (4.6 and, of course, the total charge gives rise to the total electric field E(x, E ρ ρ pol ɛ 0. (4.7 The simplest type dielectric is a dielectric whose response to the applied field is linear and isotropic, i.e. P ɛ 0 χ e E, (4.8 where χ e is called the electric susceptibility. (However, there are other materials that could exhibit a spontaneous polarization such as paraelectric, ferroelectric, etc. Then the displacement can be written as D ɛe (4.9

7 CHAPTER 4. MULTIPOLES, DIELECTRICS 50 where ɛ ɛ 0 ( χ e (4.0 is the electric permittivity. If the dielectric is homogeneous (uniform then the Gauss s law in differential form reads as E ρ (4. ɛ and all of the solutions obtained in the previous sections are validwitha trivial replacement ɛ ɛ 0. If the medium, for example, consists of two regions with different dielectric properties, then we can derive boundary conditions that D and E must satisfy. An integral over a volume enclosing the boundary gives Dd x ρd x (4. ˆn (D D σ (4. where σ is the surface charge density of free (or excess charges, or Pd x ρ pol d x (4.4 ˆn (P P σ pol (4.5 where σ pol is the surface charge density of polarization charge. Moreover, an integral around a closed loops leads to ( E ˆn da 0 (4.6 E dl 0 ( Boundary-Value problems ˆn (E E 0. (4.8 Consider a point charge embedded in a semi-infinite dielectric ɛ at a distance d from the plane which separates another semi-infinite dielectric ɛ.it convenient to use the cylindrical coordinates with the plane of separation at z 0and the point charge placed at (z d. Theappropriateequationsare E ρ ɛ for z>0 (4.9 E 0 ɛ for z<0 (4.40 E 0. (4.4

8 CHAPTER 4. MULTIPOLES, DIELECTRICS 5 The last equation implies that solutions can be expressed in term of electric potential, i.e. E Φ. (4.4 To solve the problem we can place image charge q at (z d, then ( Φ q 4πɛ q for z>0 (4.4 ρ (d z ρ (d z and Then the boundary conditions ( Φ q 4πɛ ρ (d z for z<0. (4.44 leads to lim ɛ E z lim z 0 z 0 ɛ lim E ρ lim z 0 z 0 z Φ lim ɛ z 0 ρ Φ lim z 0 z Φ lim E z z 0 (4.45 ρ Φ lim ρ z 0 (4.46 whose solutions are q q q (4.47 (q q ɛ q ɛ (4.48 ( q ɛ ɛ ɛ ɛ ( q ɛ ɛ ɛ q (4.49 q. (4.50 Therefore Φ q 4π ɛ ρ (d z ( ɛ ɛ ɛ (ɛ ɛ ρ (d z (4.5 for z>0 and Φ q ( (4.5 4π ɛ ɛ ρ (d z for z<0.

9 CHAPTER 4. MULTIPOLES, DIELECTRICS 5 The polarization charge density inside either dielectric (and away from the point charge is zero P ɛ 0 χ e E 0. (4.5 but at the boundary surface from (4.5 and(4.8 weget σ pol ẑ (P P ẑ (ɛ 0 χ E ɛ 0 χ E ẑ ((ɛ ɛ 0 E (ɛ ɛ 0 E q ( 4π (ɛ ɛ 0 ɛ ɛ (ɛ ɛ 0 ( ɛ ɛ ɛ ɛ (ɛ ɛ (ρ d / q π ɛ0 (ɛ ɛ d. ɛ (ɛ ɛ (ρ d / (4.54 Consider another example of a uniform electric field E 0 E 0 ẑ (4.55 directed along the z axis corresponding to the electric potential Φ 0 E 0 z E 0 r cos θ E 0 rp (cos θ. (4.56 We are to calculate changes to (4.55and(4.56duetopresenceofadielectric sphere of radius a and permittivity ɛ placed in the origin. From the azimuthal symmetry of the problem and from the finiteness of the electric field at r 0and r the solutions inside and outside of the sphere are given respectively by d Φ in A l r l P l (cos θ (4.57 l0 and Φ out E 0 rp (cos θ B l r (l P l (cos θ. (4.58 By matching the boundary conditions at the surface of the sphere according to (4. and(4.8 weget lim ɛer r a in ɛ lim r a lim Eθ r a in lim r a l0 r Φ in ɛ 0 lim r a r θ Φ in lim r a r Φ out ɛ 0 lim Er r a out (4.59 r θ Φ in lim Eθ r a out. (4.60

10 CHAPTER 4. MULTIPOLES, DIELECTRICS 5 Then by equating coefficients of each of the Legendre polynomials we get from (4.59 and from (4.60 B ɛa ɛ 0 E 0 ɛ 0 (4.6 a B l ɛla l (l for l (4.6 a l A E 0 B a (4.6 A l B l a l for l. (4.64 Clearly the equations (4.6 and(4.64 cannot be satisfies unless simultaneously unless A l B l 0. (4.65 And the remaining two equations (4.6 and(4.6 are solved when A ɛ/ɛ 0 E 0 (4.66 B ɛ/ɛ 0 ɛ/ɛ 0 E 0, (4.67 and therefore Φ in ɛ/ɛ 0 E 0r cos θ, (4.68 Φ out E 0 r cos θ ɛ/ɛ 0 ɛ/ɛ 0 E a 0 cos θ. r (4.69 Note that the potential inside of the sphere corresponds to a uniform electric field directed along the z axis with a magnitude smaller than the unperturbed electric field, E in ɛ/ɛ 0 E 0 < E 0 (4.70 and the potential outside of the sphere perturbed by polarized dielectric. One can also solve an electrostatic problem of the effect of a cavity in dielectric media ɛ on the uniform electric field. The solutions are given by a trivial replacement ɛ ɛ 0 and ɛ 0 ɛ in the above discussion.

11 CHAPTER 4. MULTIPOLES, DIELECTRICS Electrostatic Energy It was previously derived the the potential energy of a distribution of charges is given by W ρ(xφ(x dx. (4.7 This formula was derived from the work required to assemble a configuration of charges by brining them from infinite. This might not hold in dielectric media since an additional work is required to polarize the dielectric media. Consider a small change (linear response in the potential energy due to change of the distribution of excess charges ρ, δw δρ(xφ(xd x (4.7 where ρ D (4.7 δρ δd. (4.74 For a localized δρ we can integrate by parts to obtain δw Φ δdd x (4.75 ( Φ δdd x E δdd x. However, for linearly polarizable media, and or E δd ɛ D δd ɛ ( δw δ W δ (D D δ (E D (4.76 E Dd x (4.77 E Dd x. (4.78 which is the same as (4.7 withe Φ and D ρ. Thismeansthat for linear media (and only for linear media the potential electrostatic energy defined using microscopic and macroscopic descriptions is the same.