Chapter 3 Special Techniques). Laplace 3. Laplace s Equation) (3.. Introduction) Lapla

Transcription

1 2 25

2 Chapter 3 Special Techniques). Laplace 3. Laplace s Equation) (3.. Introduction) Laplace 3..2 Laplace s Equation in One Dimension) Laplace 3..3 Laplace s Equation in Two Dimensions) Laplace (3..4 Laplace s Equation in Three Dimensions) Boundary Conditions and Uniqueness Theorems (3..6 Conductors and the Second Uniqueness Theorem) Neumann (3.2 The Method of Images) (3.. The Classic Image Problem) (Sec Induced Surface Charge) (Sec Force and Energy) (3.2.4 Other Image Problems) Separation of Variables Cartesian Coordinates Spherical Coordinates Multipole Expansion Approximate Potential at Large Distance The Monopole and Dipole Terms The Electric Field of a Dipole Chapter 4 Electric Fields in Matter Polarization Dielectrics Induced Dipoles Alignment of Polar Molecules Polarization The Field of a Polarized Object Bound Charges Physical Interpretation of Bound Charges The Electric Displacement Gauss 4.3. Gauss s Law in the Presence of Dielectrics A Deceptive Parallel Boundary Conditions Linear Dielectrics Susceptibility, Permittivity, Dielectric Constant Boundary Value Problems with Linear Dielectrics Energy in Dielectric Systems i

3 Forces on Dielectrics Chapter 6 Magnetic Fields in Matter Magnetization Multipole Expansion of the Vector Potential Diamagnets, Paramagnets, Ferromagnets Torques and Forces on Magnetic Dipoles Magnetization The Field of a Magnetized Object Bound Currents Physical Interpretation of Bound Currents H 6.3 The Auxiliary Field H Ampère Linear and Nonlinear Media Magnetic Susceptibility and Permeability A Deceptive Parallel Boundary Conditions Chapter 7 Electrodynamics Maxwell Maxwell s Equations Maxwell Maxwell s Equations in Matter Boundary Conditions ii

4 Chapter 3 Special Techniques) Sec.. Poisson Laplace Laplace Earnshaw Sec..2 Sec..3 Laplace Sec..4. Laplace 3. Laplace s Equation).. (3.. Introduction) ρ(r) Coulomb E(r) E(r) = r r 4πɛ 0 r r 3 ρ(r )dτ (dτ = dx dy dz ) (.) (.) (.) Gauss E da = Q enc ɛ 0 S V E = V E V (r) = 4πɛ 0 r r ρ(r )dτ (.2) (.2) (.) ρ Gauss Gauss V E = ρ ɛ 0 (.3) 2 V = ɛ 0 ρ (.4) Poisson Poisson (.2) (ρ = 0)

5 ρ =0 ρ =0 V =0 ) Poisson Laplace x, y, z 2 V x V y 2 2 V =0 (.5) + 2 V =0 (.6) z2 Laplace Laplace Laplace Laplace Laplace..2 Laplace 3..2 Laplace s Equation in One Dimension) V x Laplace d 2 V dx 2 =0 V (x) =mx + b (.7) m, b Laplace. x,x 2 V V (x )=V, V(x 2 )=V 2 m =(V V 2 )/(x x 2 ), b =(V 2 x V x )/(x x 2 ) (.8) 2. x V dv/dx(= E) V (x )=V, dv/dx x=x = E m = V + E x (.9). V (x) a V (x + a) V (x a) V (x) = [V (x + a) V (x a)] 2 2. Laplace () x = x 0 V (x) V (x 0 ) x 0 V (x) V (x 0 ) V (x 0 ) < 0 Laplace V (x) =0, V (x) =x 4 x =0 (.7) Laplace 2

6 ..3 Laplace 3..3 Laplace s Equation in Two Dimensions) V x, y Laplace 2 V x V =0 (.0) y2 Laplace V = x 2 y 2, V = xy, V = e kx cos ky, V = e ky cos kx (.). (x, y) V (x, y) R V V (x, y) = Vdl (.2) 2πR 2. V () (.) (.2) circle..4 Laplace (3..4 Laplace s Equation in Three Dimensions) Laplace () r V r R V V = 4πR 2 Vda (.3) (2) () V sphere V r 0 V (r 0 )=V max r 0 V V (r) <V max r 0 V 4πR 2 Vda< sphere 4πR 2 V max da = V max sphere 4πR 2 da = V max (.4) sphere r = r 0 V (r 0 )=V max () () q R q z (0, 0,z) r =(x,y,z ) r V = q (.5) 4πɛ 0 x 2 + y 2 +(z z) 2 x = R cos φ sin θ, y = R sin φ sin θ, z = R cos θ (.6) 3

7 .: x 2 + y 2 +(z z) 2 = z 2 + R 2 2zR cos θ (.7) V = q 4πɛ 0 z2 + R 2 2zR cos θ V da = R 2 sin θdθdφ V ave = q R 2 4πR 2 sin θdθdφ 4πɛ 0 z2 + R 2 2zR cos θ = [ q π 4πR 2 2πR 2 z2 + R 4πɛ 0 zr 2 2zR cos θ] 0 = q q [(z + R) (z R)] = 4πɛ 0 2zR 4πɛ 0 z (.8) (.9) q V V Gauss Gauss E da = Q enc =0 (.20) E da > 0 (.20) Earnshaw Earnshaw 4

8 ..5 境界条件と一意性の定理 3..5 Boundary Conditions and Uniqueness Theorems V を一意に決めるためには Laplace 方程式に適当な境界条件を課す必要がある ここで問題となるのは 解を決めるために必要かつ十分な境界条件とはどのようなものであるか ということである 一次元の場合 は一般解が具体的に分っているので 簡単であったが 二次元 三次元の Laplace 方程式は偏微分方程式で あるため 境界条件の与え方は自明ではない Laplace 方程式の解を一意に決めるための条件は 一意性の 定理によって表される 図.2:! 第一の一意性の定理 (First uniqueness theorem) " ある領域 V における Lapalce 方程式の解は 境界面 S におけるポテンシャル V を定めれば一意に決まる # $ 第一の一意性定理の証明 図..5 のように領域 V と境界面 S が与えられているものとする もしも Laplace 方程式が 二つの解 V, V2 を持つとしよう 2 V = 0, 2 V2 = 0 (.2) これら二つの解はどちらも境界 S において同じポテンシャルの値を持つものとする Laplace 方程式の解が 一意に定まることを示すには 必ず V = V2 であることを示せばよい そのために 二つの差 V3 V V2 (.22) を考える これもまた Laplace 方程式に従うことは容易にわかる 2 V 3 = 2 V 2 V 2 = 0 (.23) 境界面 S 上では V = V2 であるから ここでは V3 = 0 である しかし Laplace 方程式は極大も極小も許さ なず 最大値も最小値も境界でのみ取ることができる したがって V3 の最大値も最小値も両方とも 0 でな ければならない よって V の全ての領域において V3 = 0 でなければならない これより V = V2 (.24) であることが示された このように 境界面においてポテンシャルの値を定める境界条件の与え方をディリクレ Dirichlet の境 界条件という 5

9 V 0 V = V 0 ( Laplace Sec Poisson ρ V Poisson 2 V = ρ (.25) ɛ 0 V,V 2 2 V = ρ, 2 V 2 = ρ (.26) ɛ 0 ɛ 0 V 3 V V 2 Laplace ( 2 V 3 = V V 2 = ρ ) ρ =0 (.27) ɛ 0 ɛ 0 S V = V 2 V 3 =0 V 3 =0 V V (a) ρ (b) V..6 (3..6 Conductors and the Second Uniqueness Theorem) V Dirichlet V Q Q 2 (Seoncd uniqueness theorem) V ρ : Gauss E = ɛ 0 ρ, E 2 = ɛ 0 ρ, (.28) Gauss Gauss E da = Q i, E 2 da = Q i (.29) S i ɛ 0 S i ɛ 0 6

10 .3: S i i Gauss E da = Q tot, E 2 da = Q tot (.30) ɛ 0 ɛ 0 S S S E 3 = E E 2 (.3) E 3 =0 (.32) E 3 da =0 (.33) i Q i V 3 = V V 2 V V 2 V 3 =0 (V 3 E 3 )=V 3 ( E 3 )+E 3 ( V 3 )= (E 3 ) 2 (.34) E 3 =0 E 3 = V 3 (V 3 E 3 )dτ = V 3 E 3 da = (E 3 ) 2 dτ (.35) V S V V 3 V 3 =0 V 3 E 3 da = V 3 E 3 da (.36) V 3 V 3 (.33) 0 (E 3 ) 2 dτ =0 (.37) V (.37) 0 0 E 3 =0 E = E 2 7

11 .4:.4 ±Q Laplace.5: Green [T 2 U +( T ) ( U)]dτ = (T U) da (.38) V S T = U = V 3 Prob Neumann Laplace ρ E = V n Neumann Prob.3.4 8

12 .2 (3.2 The Method of Images) method of images) Sec (3.. The Classic Image Problem).6:.6 d q q V = q 4πɛ 0 r r q Coulomb (.) (.2) Poisson (0, 0,d) q z>0 ρ(r) =qδ(x)δ(y)δ(z d). z =0 V = V 0 Poisson Poisson Poisson Poisson.7 +q (0, 0,d) q (0, 0, d) [ ] V (x, y, z) = q 4πɛ 0 x2 + y 2 +(z d) q 2 x2 + y 2 +(z + d) 2 (.39) z >0 (0, 0,d) +q q Poisson (.39),2.7 9

13 .7: (.6) z 0 z <0 z 0 (.39) (a) (b) +q +q q.8: (a) (b).8 Poisson.2.2 (Sec Induced Surface Charge) σ = ɛ 0 V n (.40) V/ n V z σ = ɛ 0 V z (.4) z=0 0

14 (.39) V z = 4πɛ 0 { } q(z d) [x 2 + y 2 +(z d) 2 ] + q(z + d) 3/2 [x 2 + y 2 +(z + d) 2 ] 3/2 (.42) σ(x, y) = qd 2π(x 2 + y 2 + d 2 ) 3/2 (.43) q >0.9 x = y =0 0! / (q / d 2 ) r / d (r=(x 2 +y 2 ) /2 ).9: Q = σ(x, y)dxdy (.44) x = r cos φ,y = r sin φ dxdy = rdrdφ Q = 2π dφ 0 0 [ ] qd rdr 2π(r 2 + d 2 ) = qd 3/2 r2 + d 2 0 = q (.45) q.2.3 (Sec Force and Energy) q q E = V q q Coulomb F = 4πɛ 0 q 2 (2d) 2 ẑ (.46).6).7) 2d +q, q W = 4πɛ 0 q 2 2d (.47)

15 (a) (b) +q +q F q.0: (a) (b) W = 4πɛ 0 q 2 4d (.48) W = ɛ 0 2 E 2 dτ (.49) z <0 z>0 z <0 E =0 q F (.46) 2 W = d F dl = d 4πɛ 0 q 2 4z 2 dz = 4πɛ 0 ] d [ q2 = q 2 4z 4πɛ 0 4d (.50) q 0 0 q q q.2.4 (3.2.4 Other Image Problems) Ex.3.2.(a) R a q q.(b) q = R a q (.5) Griffiths (.49) (.47) 2 2

16 P r r 2 O a (a) (b).: b = R2 a P V = ( ) q + q 4πɛ 0 r r 2 r q r 2 q (.52) (.53) r 2 = r 2 + a 2 2ar cos θ, r 2 2 = r 2 + b 2 2br cos θ (.54) [ V = 4πɛ 0 = q 4πɛ 0 [ q r2 + a 2 2ar cos θ R a ] q r2 +(R 2 /a) 2 2(R 2 /a)r cos θ ] r2 + a 2 2ar cos θ a2 r 2 /R 2 + R 2 2ar cos θ (.55) r = R θ V =0 0 (.55) (.52) b<r Poisson q q q Coulomb F = 4πɛ 0 qq (a b) 2 = 4πɛ 0 qr a(a R 2 /a) 2 = 4πɛ 0 qra (a 2 R 2 ) 2 (.56) 3

17 Separation of Variables Laplace Schrödinger V σ Laplace V (x, y, z) V (x, y, z) =X(x)Y (y)z(z) (r, θ, φ) V (r, θ, φ) =R(r)Y (θ, φ) Sec..3. Sec Cartesian Coordinates.2 xz y =0 y = a x =0 z V 0 (y).2: : z V z x, y Laplace 2 V x V =0 (.57) y2 Laplace (i) V =0 when y =0 (ii) V =0 when y = a (iii) V = V 0 (y) when x =0 (iv) V 0 as x (.58) (iv) 0 Laplace V (x, y) =X(x)Y (y) (.59) 4

18 (.58) (.59) (.59) (.59) Lapalce (.57) V = XY Y d2 X dx 2 d 2 X X dx 2 + Y + X d2 Y =0 (.60) dy2 d 2 Y =0 (.6) dy2 x y f(x)+g(y) =0 (.62) f g (.6) d 2 X X dx 2 = C, Y d 2 Y dy 2 = C 2, C + C 2 =0 (.63) C,C 2 C > 0,C 2 < 0 C = k 2 = C 2 k>0 d 2 X dx 2 = k2 X, d 2 Y dy 2 = k2 Y (.64) X(x) =Ae kx + Be kx, Y(x) =C sin ky + D cos ky (.65) V V (x, y) =(Ae kx + Be kx )(C sin ky + D cos ky) (.66) (iv) x X 0 A =0 (i) Y (0) = 0 D =0 (ii) Y (a) =C sin ka =0 sin ka =0 k k = nπ a, (n =, 2, 3 ) (.67) n =0 (.68) k =0 y V =0 k n BC C V (x, y) =Ce kx sin ky (.68) k (.67) C > 0,C 2 < 0 C < 0,C 2 > 0 Laplace (.66) x y V (x, y) =(Asin kx + B cos kx)(ce ky + De kx ) (.69) (iii) x V =0 (i) (ii) V (x, 0) = V (x, a) =0 C > 0,C 2 < 0 5

19 Laplace (.68) k (.67) (.68) V n (x, y) =e nπx/a sin(nπy/a), (n =, 2, 3, ) (.70) (iii) V 0 (y) sin(nπy/a) n V 0 (y) (.68) Laplace V,V 2 2 (α V + α 2 V 2 )=α 2 V + α 2 2 V 2 (.7) α, α 2 Laplace Laplace V,V 2,V 3, V = α V + α 2 V 2 + α 3 V 3 + Laplace α, α 2, 2 V = 2 (α V + α 2 V 2 + α 3 V 3 + )=α 2 V + α 2 2 V 2 + α 3 2 V 3 + =0 (.72) (.68) V (x, y) = C n e nπx/a sin(nπy/a) (.73) n= (i),(ii),(iv) C n (iii) V (0,y)= C n sin(nπy/a) =V 0 (y) (.74) n= (.74) V 0 (y) Fourier 2 Fourier (.73) C n V 0 (y) C n Griffith Fourier s trick ) (.74) sin(n πy/a) n 0 a a a C n sin(nπy/a)sin(n πy/a)dy = V 0 (y)sin(n πy/a)dy (.75) n= 0 (.75) a 0, (n n) sin(nπy/a)sin(n πy/a)dy = (.76) 0 a 2, (n = n) n n = n (.75) =(a/2)c n C n = 2 a a 0 0 V 0 (y)sin(nπy/a)dy (.77) V 0 (y) Laplace V 0 y x =0 (.77) C n = 2V 0 a a 0 sin(nπy/a)dy = 2V 0 ( cos nπ) = nπ 6 0, (n ) 4V 0 nπ (n ) (.78)

20 (.73) V (x, y) = 4V 0 π n=,3,5, n e nπx/a sin(nπy/a) (.79).3.4 (.79) x =0 Fourier n = n = V/V x/a y/a.3: V (x, y).2 V / V y / a.4: V (0,y) Fourier (.79 sin x =Im ( e ix) (.79) ( ) { } 4V 0 V (x, y) =Im π n e nπx/a e inπy/a 4V 0 =Im [e π( x+iy)/a] n π n n=,3,5, ln( + z) =z 2 z2 + 3 z2 + = n=,3,5, ( ) n zn n n= (.80) (.8) 7

21 n=,3,5, n zn = +z ln 2 z [ ] 2V0 +eπ( x+iy)/a V (x, y) =Im ln π e π( x+iy)/a u, v Im [ln(u + iv)] = tan (v/u) (.82) (.83) +e π( x+iy)/a e = [ + eπ( x+iy)/a ][ e π( x iy)/a ] π( x+iy)/a [ e π( x+iy)/a ][ e π( x iy)/a ] = +2ie πx/a sin(πy/a) e 2πx/a 2e πx/a cos(πy/a)+e 2πx/a = sinh(πx/a)+i sin(πx/a) cosh(πx/a) cos(πy/a) Im [ln ] eπ( x+iy)/a +e π( x+iy)/a V (x, y) = 2V 0 π tan [ ] sin(πx/a) = tan sinh(πx/a) [ ] sin(πy/a) sinh(πx/a) (.84) (.85) (.86).3 {f n } f n(x)f m (x)dx =0 (n m) (.87) {f n } {f n } ψ(x) {f n } f n(x)f m (x)dx = δ nm (.88) ψ(x) = n c n f n (x) (.89) {f n } {f n } c n (.89) f m (.88) fm(x)ψ(x)dx = c n fm(x)f n (x)dx = c m (.90) n (.89) [ ] ψ(x) = fn(x )f n (x) ψ(x )dx (.9) n ψ(x) fn(x )f n (x) =δ(x x ) (.92) n (.92) 8

22 Spherical Coordinates.5: Laplace 2 V = ( r 2 r 2 V ) + r r x = r cos φ sin θ, y = r sin φ sin θ, z = r cos θ (.93) r 2 sin θ ( sin θ V ) + θ θ r 2 sin 2 θ 2 V =0 (.94) φ2 (.94) z V φ Laplace r 2 ( r 2 V ) + r r r 2 sin θ (.96) (.95) V ( d r 2 dr ) + R dr dr Θ sin θ d R dr ( sin θ V ) =0 (.95) θ θ V (r, θ) =R(r)Θ(θ) (.96) d dθ ( r 2 dr ) = dr Θ sin θ ( sin θ dθ ) =0 (.97) dθ ( d sin θ dθ ) dθ dθ (.98) r θ l(l + ) ( d r 2 dr ) = l(l + ), R dr r ( d sin θ dθ ) = l(l + ) (.99) Θ sin θ dθ dθ R d R dr ( r 2 dr ) = l(l + ) (.00) dr 9

23 R = Ar l + B r l+ (.0) A, B A, B (.0) (.00) r 2 dr dr = r2 [lar r (l + ) B r l+2 ]=larl+ (l + ) B r l (.02) ( d r 2 dr ) = d [lar l+ (l + ) Br ] dr dr dr = l(l + )Ar l + l(l + ) B = l(l + )R rl+ (.03) ( d sin θ dθ ) = l(l + ) (.04) Θ sin θ dθ dθ l 0 θ 2π Legendre Θ(θ) =P l (cos θ) (.05) Legendre φ 2 Legendre Legendre Rodorigues Legendre P l (x) = ( ) l d 2 l (x 2 ) l (.06) l! dx P 0 (x) = P (x) =x P 2 (x) =(3x 2 )/2 P 3 (x) =(5x 3 3x)/2 P 4 (x) = (35x 4 30x 2 + 3)/8 P 5 (x) = (63x 5 70x 3 + 5x)/8 (.07) P l (x) x l l l (.06) (/2 l l!) P l () = Legendre P l (x) x P l (x)p l (x)dx = π Laplace ( V (r, θ) = Ar l + 0 P l (cos θ)p l (cos θ)sinθdθ = 2 2l + δ ll (.08) B r l+ ) P l (cos θ) (.09) (.09) V (r, θ) = l=0 ( A l r l + B ) l r l+ P l (cos θ) (.0) A l Legendre (.08) 20

24 Example 3.6 () R V 0 (θ) r =0 (.0) l B l =0 V (r, θ) = V (R, θ) = A l r l P l (cos θ) (.) l=0 A l R l P l (cos θ) =V 0 (θ) (.2) l=0 Legendre A l V 0 (θ) A l (.2) P l (cos θ)sinθ θ π A l R l dθ sin θp l (cos θ)p l (cos θ) = l=0 Legendre (.08) 0 A l R l 2Rl 2l + = A l = 2l + 2R l π π (.) 0 0 π 0 dθ sin θp l (cos θ)v 0 (θ) (.3) V 0 (θ)p l (cos θ)sinθdθ (.4) V 0 (θ)p l (cos θ)sinθdθ (.5) V 0 (θ) =k sin 2 (θ/2) (.6) sin 2 (θ/2) = ( cos θ)/2 P 0 (x) =,P (x) =x (.6) V 0 (θ) = k 2 ( cos θ) =k 2 [P 0(cos θ) P (cos θ)] (.7) A l (.5) Legendre (.08) A 0 = k 2, A = k 2R, A l =0(l 0, ) (.8) (.) V (r, θ) = k [ P 0 (cos θ) ] 2 R P (cos θ) = k ( R ) 2 cos θ (.9) Example 3.7 (2) R V 0 (θ) 0 r V =0 l A l =0 B l V (r, θ) = r l+ P l(cos θ) (.20) l=0 2

25 V (R, θ) = (.2) P l (cos θ)sinθ θ B l = 2l + 2 l=0 (.20) B l R l+ P l(cos θ) =V 0 (θ) (.2) π R l+ V 0 (θ)p l (cos θ)sinθdθ (.22) 0 Example 3.8 E = E 0 ẑ R.6 z z.6: V (R, θ) =V 0 E = E 0 ẑ V E 0 z + C (.23) C xy V (z = 0) 0 C =0 3 (i) V = V 0 when r = R (ii) V E 0 r cos θ for r R (iii) 0 (.24) (iii) (i)-(iii) (.0) A l,b l 3 Griffiths V 0 =0 xy 0 xy C = V 0 =0 V 0 22

26 (ii) r B l A l r l P l (cos θ) = E 0 r cos θ (.25) l=0 P (x) =x (.25) A l r l P l (cos θ) = E 0 rp (cos θ) (.26) l=0 P l (cos θ)sinθ θ l = l (i) V (R, θ) = A = E 0, A l =0(l ) (.27) l=0 Legendre B 0 ( A l R l + B ) l R l+ P l (cos θ) =V 0 (.28) R = V 0, A l R l + B l =0(l 0) (.29) Rl+ (.27) A 0 =0 (.27) B 0 = RV 0, B = A R 3 = E 0 R 3, B l =0(l 0, ) (.30) (.27) (.30) (.0) V (r, θ) = RV 0 r ( ) E 0 r R3 r 2 cos θ (.3) (iii) V 0 ) V σ(θ) = ɛ 0 RV 0 r = ɛ 0 r=r r 2 + ɛ 0 E 0 (+2 R3 r 3 cos θ V 0 = ɛ 0 r=r R +3ɛ 0E 0 cos θ (.32) da =2πR 2 π sin θdθ (.32) 0 Q = daσ =4πRɛ 0 V 0 (.33) Q =0 V 0 =0 ( ) V (r, θ) = E 0 r R3 r 2 cos θ (.34) σ(θ) ==3ɛ 0 E 0 cos θ (.35) (0 θ π/2) (π/2 θ π) (.34) V = V ex + V sphere V ex = E 0 r cos θ, V sphere = E 0 R 3 cos θ (.36) r2 V ex V sphere 23

27 Q (.3) V (r, θ) = Q ) 4πɛ 0 r E 0 (r R3 r 2 cos θ (.37) Q Laplacian (.93) Laplacian 2 V = r 2 ( r 2 V ) + r r r 2 sin θ gradient ( sin θ V ) + θ θ 2 V r 2 sin 2 θ φ 2 (.38) V = V r ˆr + V r θ ˆθ + V r sin θ φ ˆφ (.39) ˆr, ˆθ, ˆφ r, θ, φ ˆr = ˆθ = r r r r r θ r θ = cos φ sin θˆx +sinφsin θŷ + cos θẑ (.40) = cos φ cos θˆx +sinφcos θŷ sin θẑ (.4) ˆφ = r φ r φ (.39) V/ r = sin φˆx + cos φŷ (.42) V r = V x x r + V y y r + V z r = V z r r (.43) (.40) V r r/ r = = V ˆr r r = V ˆr (.44) V θ = V x x θ + V y y θ + V z r = V z θ θ (.45) (.4) r/ θ = r V θ = V ˆθ r θ = r V ˆθ (.46) V φ = V x x φ + V y y φ + V z z r = V φ φ (.47) (.42) r/ φ = r sin θ V = V ˆφ r φ φ = r sin θ V ˆφ (.48) ˆr, ˆθ, ˆφ V =( V ˆr)ˆr +( V ˆφ)ˆφ +( V ˆθ)ˆθ (.49) 24

28 (.44),(.46),(.48) (.39) Laplace = ˆr r + ˆθ r θ + ˆφ r sin θ φ (.50) 2 V = V (.5) (.50) 2 V = ( ˆr r + ˆθ r = 2 V r 2 + r V θ θ + ˆφ r sin θ ) ( V φ 2 V φ 2 + V r ) + 2 V r 2 θ 2 + r 2 sin 2 θ ( r ˆθ ˆθ θ + r sin θ ˆφ ˆθ φ r ˆr + V r θ ˆθ + ) V r sin θ φ ˆφ ( r ˆθ ˆr θ + r sin θ ˆφ ˆr ) φ + V r sin θ φ ( r ˆθ ˆφ θ + r sin θ ˆφ ˆφ φ ) (.52) (ˆr, ˆθ, ˆφ) r 0 (.40)-(.42) (.52) = 2 V r 2 = r 2 r (.38) ˆθ ˆr θ =, ˆθ ˆθ θ =0, ˆθ ˆφ θ =0, ˆr ˆφ =sinθ (.53) φ ˆθ ˆφ = cos θ (.54) φ ˆφ ˆφ φ + 2 V r 2 θ V r 2 sin 2 θ φ V r r + ( r 2 V ) ( + r r 2 sin θ V ) + sin θ θ θ =0 (.55) cos θ V r 2 sin θ θ 2 V r 2 sin 2 θ φ 2 (.56) Ledengre V V (r, θ, φ) =Y (θ, φ)r(r) (.57) Y (θ, φ) [ ( sin θ ) + 2 ] V sin θ θ θ sin 2 θ φ 2 Y (θ, φ) = l(l + )Y (θ, φ) (.58) Y (θ, φ) =Θ(θ)Φ(φ) (.59) d 2 Φ Φ dφ 2 = Θ sin θ d ( sin θ dθ ) + l(l + ) sin 2 θ (.60) dθ dθ 25

29 φ θ m 2 Φ Θ Φ sin θ ( d sin θ dθ dθ dθ Φ φ m d 2 Φ dφ 2 = m2 Φ(φ) (.6) ) ] + [l(l + ) m2 Θ =0 (.62) sin 2 θ Φ(φ) =e ±imφ (.63) Φ(φ +2pi) =Φ(φ) e im(φ+2π) = e im(φ) (.64) Θ θ z = cos θ m =0, ±, ±2, (.65) Θ(θ) =P (cos θ) =P (z) (.66) dz = sin θdθ P [ d ( z 2 ) dp ] ] + [l(l + ) m2 dz dz z 2 P =0 (.67) Legendre z m =0 P (z) [ d ( z 2 ) dp ] + l(l + )P =0 (.68) dz dz Legendre Legendre (.68) P (z) =z α j=0 a j z j (.69) { (α + j)(α + j )aj z α+j 2 [(α + j)(α + j + ) l(l + )] a j z α+j} =0 (.70) j=0 z z z 0 ( ) z α 2,z α 0 α(α )a 0 =0 (.7) (α + )αa =0 (.72) a 0 0 α =0 α = (.73) a 0 α =0 α = (.74) 26

30 a j+2 = [ ] (α + j)(α + j + ) l(l + ) a j (.75) (α + j + )(α + j + 2) (.73) (.74) a 0 a 0 a 0 0,a =0 α α =0 α = (.76) (.75) α =0 α = α =0, α = (i)l z 2 < (ii) z = ± l =0, α = a 0 = P (z) =z + 3 z3 + 5 z5 + (.77) Q 0 (z) = 2 ln +z z (.78) z = ± l z = ± z (.75) j = j max 0 (α + j max )(α + j max + ) = l(l + ) (.79) j j max a j =0 α j max 0 l 0 (.79) α =0 j max = l α = j max = l P (z) z l Legendre P l (z) Legendre P 0 (z) = P (z) =z P 2 (z) =(3z 2 )/2 P 3 (z) =(5z 3 3z)/2 P 4 (z) = (35z 4 30z 2 + 3)/8 P 5 (z) = (63z 5 70z 3 + 5z)/8 (.80) P l (z) z l l l P l () = Legendre Rodorigues (.8) (/2 l l!) P l () = Legendre P l (z) z P l (z)p l (z)dx = P l (z) = ( ) l d 2 l (z 2 ) l (.8) l! dz π 0 P l (cos θ)p l (cos θ)sinθdθ = 2 2l + δ ll (.82) 27

31 m 0 Legendre (.67) Legendre (m = 0) n l, m ()l 0 (2)m l, l +,, 0,,l,l Legendre Pl m (z) m >0 Pl m (z) =( ) m ( z 2 m/2 dm ) dz m P l(z) = ( )m 2 l l! ( z 2 m/2 dl+m ) dz l+m (z2 ) l (.83) (.67) m 2 m Pl m (z) = P m l (z) m Legendre Pl m (z)pl m (z)dz = 2 2l + (l + m)! (l m)! δ ll (.84) Y lm (θ, φ) =( ) (m+ m )/2 (2l + )(l m )!4π(l + m )!P m l (cos θ)e imφ (.85) l =0,, 2,, m = l, l +,,l,l (.86) Y ml π 0 sin θdθ 2π l =0,, 2 l =0 : Y 00 = 4π 0 dφy l m (θ, φ)y lm(θ, φ) =δ ll δ mm (.87) 3 3 l = : Y ± = 8π sin θe±iφ, Y 0 = 4π cos θ 5 l =2 : Y 2±2 = 32π sin2 θe ±2iφ, Y 2± = 5 Y 20 = 6π (3 cos2 θ ) 5 sin θ cos θe±iφ 8π (.88) 28

32 Multipole Expansion Approximate Potential at Large Distance V Q/4πɛ 0 r Q 0 Example ±q d (electric dipole) +q d/2 θ r + r d/2 q r.7: +q r +, q r V (r) = ( q q ) 4πɛ 0 r + r ( r ± = r 2 +(d/2) 2 rd cos θ = r 2 d ) d2 cos θ + r 4r 2 r d (d/r) 2 ( d ) /2 r ± r r cos θ ( ± d2r ) r cos θ (.89) (.90) (.9) d cos θ (.92) r + r r2 V (r) qd cos θ (.93) 4πɛ 0 r2 29

33 .8: /r 2 /r monopole quadrupole /r 3 octopole /r 4.8 z z P P r r ẑ r r r q r θ r q r (a) (b).9: z q. 4.9(a) z r q r P r >r r θ V (r, θ) = q 4πɛ 0 n=0 r n r n+ P n(cos θ) (.94) z φ Laplace φ 4 Griffiths 30

34 (.0) 0 A n =0 B n V (r, θ) = r n+ P n(cos θ) (.95) n=0 B n P z θ =0 Legendre P n () = r r = r (.99) (.96) V (r, 0) = V (r, 0) = V (r, 0) = n=0 B n r n+ (.96) q 4πɛ 0 r r (.97) ) ( r = r r q 4πɛ 0 r ( ) r n (.98) n=0 r ( ) r n (.99) B n = qr n 4πɛ 0 (.200) (.200) (.96) θ (.94) n=0 r P q r θ rr r.20: q z (.94).20 P r q r θ rr P V (r) = q 4πɛ 0 V (r) = n=0 r n r n+ P n(cos θ rr ) (.20) q 4πɛ 0 r r ( r r ) r>r 5 (.202) r r = r n r n+ P n(cos θ rr ) (.203) n=0 5 (.203) Legendre r r =(r 2 + r 2 2rr cos θ rr ) /2 r /r (.203) P (cos θ rr ) r r Legendre 3

35 P r r dτ r r θ rr.2: ρ(r) ρ(r).2 r V (r) = 4πɛ 0 (.203) V (r) = 4πɛ 0 n=0 r n+ r r ρ(r )dτ (.204) r n P n (cos θ rr )ρ(r )dτ (.205) [ V (r) = ρ(r )dτ + 4πɛ 0 r r 2 + r 3 r 2 ( 3 2 cos2 θ rr 2 r cos θ rr ρ(r )dτ ) ] ρ(r )dτ + (.206) (.205) /r multipole expansion (n = 0) (monopole) ( /r) (n = ) (dipole) ( /r 2 ) (n = 2) (quadrupole) ( /r 3 ) (n = 3) (octopole) ( /r 4 ) The Monopole and Dipole Terms (.205) V mon (r) = Q 4πɛ 0 r (.207) Q = ρdτ (.207) V dip (r) = 4πɛ 0 r 2 r cos θ rr ρ(r )dτ (.208) 32

36 r cos θ rr = ˆr r V dip (r) = 4πɛ 0 r ˆr 2 r ρ(r )dτ (.209) p r ρ(r )dτ (.20) (.209) V dip (r) = 4πɛ 0 p ˆr r 2 = 4πɛ 0 p r r 3 (.2).22: N ρ(r) = N i= q iδ(r r i ) r i i (.20) p = N q i r i (.22).22 ±q i= p = qr + qr = q(r + r )=qd (.23) d = r + r (.2) V dip (r) = 4πɛ 0 qd ˆr r 2 = 4πɛ 0 qd cos θ r 2 (.24) θ d r (.93) (.93) (.2) pure dipole p = qd d The Electric Field of a Dipole (.2) E dip = V dip = ( ) p r 4πɛ 0 r 3 = [ ( )] (p r) 4πɛ 0 r 3 + p r r 3 (.25) 33

37 (p r) = (p x x + p y y + p z z)=p (.26) ( ) r 3 = [(x 2 + y 2 + z 2 ) 3/2 ]= 3r r 5 (.27) E dip = [3(p ˆr)ˆr p] (.28) 4πɛ 0 r3 p z V dip = E dip (r, θ) = ˆr p 4πɛ 0 r 2 = p cos θ 4πɛ 0 r 2 (.29) E r = V 2p cos θ = r 4πɛ 0 r 3 (.220) E θ = V r θ = p sin θ 4πɛ 0 r 3 (.22) E φ = V =0 r sin θ φ (.222) p 4πɛ 0 r 3 (2 cos θˆr +sinθˆθ) (.223) /r 3 /r 2 /r 4 /r 5 /r /r 2 /r 3 /r 4.23:.23(a) p ẑ.23(b) r d 34

38 2 Chapter 4 Electric Fields in Matter Polarization Dielectrics conductors insulators) dielectrics 2..2 Induced Dipoles E atomic polarization E p p = αe (2.) α atomic polarizability 2. H He Li Be C Ne Na Ar K Cs α/4πɛ 0 [0 30 m 3 ] Handbook of Chemistry and Physics, 79th ed. (Boca Raton: CRC Press, Inc., 997) Example 4. 2.(a) +q a q 35

39 (a) (b) 2.: 2.(b) d Prob.2.2 q q E e = 4πɛ 0 qd a 3 (2.2) E qe e = qe E = 4πɛ 0 qd a 3 (2.3) p = qd =(4πɛ 0 a 3 )E (2.4) α =4πɛ 0 a 3 (2.5) Bohr a 0 = m a a 0 α/4πɛ 0 = m 3 (2.6) CO 2 α = C 2 m/n α = C 2 m/n p = α E + α E (2.7) E E p E p x = α xx E x + α xy E y + α xz E z p y = α yx E x + α yy E y + α yz E z (2.8) p z = α zx E x + α zy E y + α zz E z 36

40 α ij α ij principal axis α xy, α yz, α xx, α yy, α zz 2.2: CO Alignment of Polar Molecules 2.3: H 2 O p 2.3 p = C m polar moleculde 2.4 ±q F + = qe(r + ) F = qe(r ) F = q[e(r + ) E(r )] (2.9) r ± ±q r ± = ±d/2 E F + = qe F = qe F =0, N =(r + F + )+(r F )=[(d/2) (qe)] + [( d/2) ( qe)] = qd E (2.0) 37

41 2.4: E p = qd N = p E (2.) N p E F + F (2.9) F = q[e(r + ) E(r )] = q(e + E ) (2.2) E ± = E(r ± )=E(±d/2) d E + E (d )E (2.3) F =(p )E (2.4). E ρ(r) F = ρ(r)e(r)dτ (2.5) ρ 0 E E(r) E(0) + (r )E (2.6) F = E(0) [( ρ(r)dτ + ) ] rρ(r)dτ E = QE(0) + (p )E (2.7) F =(p )E ( ) N = r ρ(r )Edτ = r ρ(r )dτ E = p E (2.8) (2.4) F = (p E) r (p )p 38

42 E(r) ρ(r) U = ρ(r)v (r)dτ (2.9) V V (r) V (0) + r V (2.20) U V (0) ρ(r)dτ + V rρ(r)dτ = QV (0) p E(0) (2.2) U = p E (2.22) p 2 r p 2.5: p p 2 r p p 2 (.28) (2.22) - E = 4πɛ 0 r 3 [3(p ˆr)ˆr p ] (2.23) U = 4πɛ 0 r 3 [p p 2 3(p ˆr)(p 2 ˆr)] (2.24) Polarization

43 E E =0 2.6: E E =0 2.7: polarization P (2.25) dielectric polarization N p i (i =, 2, N) V P N i= P = p i (2.26) V (freeze in) Sec.2.2 Sec.2.3 P P 40

44 The Field of a Polarized Object Bound Charges P r r r O r p V 2.8: P = r p r V (r) = 4πɛ 0 (r r ) p r r 3 (2.27) 2.8) dτ p = Pdτ (2.28) V (r) = 4πɛ 0 ( ) r r = r r r r 3, ( ) P r r V (r) = 4πɛ 0 ( A)dτ = A da V (r) = 4πɛ 0 r r P da 4πɛ 0 S V V (r r ) P r r 3 dτ (2.28) ( ˆx x + ŷ y + ẑ ) z (2.29) ( ) P r r dτ (2.30) ( ) ( ) = P r r + r r P (2.3) V r r ( P)dτ (2.32) σ b = P ˆn (2.33) 4

45 ρ b = P (2.34) V (r) = 4πɛ 0 S σ b r r da + 4πɛ 0 V ρ b r r dτ (2.35) P O r r r ρ b r dτ V r da S r r σ b 2.9: (2.35) σ b ρ b (2.33),(2.34) bound charges polarization charge (2.28) ρ b, σ b Gauss Example 4.2 R 2.0: 42

46 2.0 z P ρ b σ b = P ˆn = P cos θ (2.36) θ.3.2 Griffiths Ex.3.9 V (r, θ) = P 3ɛ 0 r cos θ, P R 3 cos θ, 3ɛ 0 r2 for P P r, 3ɛ 0 V (r, θ) = P ˆr R 3 3ɛ 0 r 2, for r R r R for r R for r R (2.37) (2.38) E = V = 3ɛ 0 P, for r<r (2.39) p 2. V = p ˆr, for r R (2.40) 4πɛ 0 r2 p = 4 3 πr3 P (2.4) 2.: 43

47 2.2.2 Physical Interpretation of Bound Charges +q q ±q 2.2: (a) (b) 2.3: P σ b 2.3 P 2.3(a) d, A p = P (Ad) ±q d p = qd q = PA (2.42) σ b = q A = P (2.43) 2.3(b) A end A = A end cos θ σ b = q A end = P cos θ = P ˆn (2.44) (2.33) 44

48 p p p 2 q +q q +q q 2 +q 2 q 2 +q 2 q = q q 2 p 2 2.4: 2.5: P 2.4 P ρ b A d P P 2 q =(P P 2 )A (2.45) Ad ρ = q Ad = P P 2 d (2.46) d 0 x ρ = P x x (2.47) 2.6 (2.47) ρ = P (2.48) ρ b (2.34) 45

49 2.6: S P V 2.7: ρ b σ b σ b da + ρ b dτ =0 (2.49) S V σ b = P ˆn σ b da = P ˆnda = P da (2.50) S S S (2.50) σ b da = ( P)dτ (2.5) S V ρ b = P (2.49) 46

50 Example 4.3 R Example 4.2 q q 2.8 z z d σ b 2.8: ρ r 0 E(r) = ρ 3ɛ 0 (r r 0 ) (2.52) r + r q q q E(r) = 4πR 3 (r r + ) /3 3ɛ 0 4πR 3 (r r )= /3 3ɛ 0 4πR 3 (r r + )= qd /3 3ɛ 0 4πɛ 0 R 3 (2.53) p = qd =( 4 3 πr3 )P E = 3ɛ 0 P (2.54) ±q V = 4πɛ 0 p ˆr r 2 (2.55) 47

51 The Electric Displacement 2.3. Gauss 4.3. Gauss s Law in the Presence of Dielectrics ρ b = P σ b = P ˆn free charge true chage) ρ = ρ b + ρ f (2.56) E Gauss ɛ 0 E = ρ = ρ b + ρ f = P + ρ f (2.57) (2.57) P divergence electric displacement Gauss (ɛ 0 E 0 + P) =ρ f (2.58) D ɛ 0 E + P (2.59) D = ρ f (2.60) Gauss D da = Q fenc (2.6) S Q fenc = V ρ f dτ Gauss ρ f ρ b (2.60) σ b ρ b Gauss (2.6) Example 4.4 : 2.9 λ a D 2.9 Gauss s L Gauss (2.6) D(2πsL) =λl (2.62) D = λ 2πsŝ (2.63) P =0 2.9: E = D = λ for s>a (2.64) ɛ 0 2πɛ 0 sŝ, P 48

52 A Deceptive Parallel (2.60) Gauss ρ ρ f ɛ 0 E D D E D Coulumb D(r) r r 4π r r 3 ρ f (r )dτ (2.65) 2.20: D E A A A E =0 Gauss E = ρ/ɛ 0 D D = ɛ 0 ( E)+( P) = P (2.66) P P = P P P dl 0 Stokes P 0 P D 0 D Gauss (2.6) D P =0 D D Boundary Conditions D above σ f σ f Dabove D below D below 2.2: 2.22: E Griffiths Sec E above E below = ɛ 0 σ (2.67) E above E below =0 (2.68) D above = ɛe above + P above (2.69) 49

53 D below = ɛe below + P below, (2.70) (2.67) (2.68) D Dabove Dbelow = ɛ(eabove Ebelow)+P above Pbelow (2.7) (2.67) D above D below = σ + P above P below (2.72) P above = P above ˆn above = σ b,above (2.73) Pbelow = P above ˆn below = σ b,below (2.74) ˆn above, ˆn below σ b,above, σ b,below σ b = σ b,above + σ b,below Pabove Pbelow = (σ b,above + σ b,below )= σ b (2.75) D above D below = σ σ b (2.76) σ = σ b + σ f Dabove Dbelow = σ f (2.77) D above D below = ɛ 0(E above E below )+P above P below (2.78) (2.68) D above D below = P above P below (2.79) Gauss 2.2 Gauss Gauss (2.6) D da = Q fenc = σ f A (2.80) S ɛ 0 (2.77) 2.22 D E dl D dl = ɛ 0 E dl + ɛ 0 (2.79) P dl = P dl (2.8) Linear Dielectrics Susceptibility, Permittivity, Dielectric Constant D Guass Gauss 50

54 P P = ɛ 0 χ e E (2.82) χ e electric susceptibility χ e (2.82) linear dielectrics (2.82) E E 0 P (2.82) E 0 P E ρ f D D D = ɛ 0 E + P = ɛ 0 E + ɛ 0 χ e E = ɛ 0 ( + χ e )E (2.83) D E ɛ D = ɛe (2.84) ɛ = ɛ 0 ( + χ e ) (2.85) (permittivity) χ e =0 ɛ = ɛ 0 ɛ 0 (2.85) ɛ 0 ɛ r =+χ e = ɛ ɛ 0 (2.86) (relative permittivity) (dielectric constant) Material Dielectric Constant Material Dielectric Constant Vacuum Benzene 2.28 Helium Diamond 5.7 Neon.0003 Salt 5.9 Hydrogen Silicon.8 Argon Methanol 33.0 Air (dry) Water 80. Nitrogen Ice ( 30 C) 99 Water vapor (00 C) KTaNbO 3 (0 C) 34, C Handbook of Chemistry and Physics, 79th ed. (Boca Raton: CRC Press, Inc., 997) 5

55 Example 4.5 : 2.23 a Q b ɛ 2.23: Gauss (2.6) D E = D = E = P = D =0 (2.84) E Q 4πɛr ˆr, 2 for a<r<b Q ˆr (r >a) (2.87) 4πr2 Q 4πɛ 0 r 2 ˆr, for r>b (2.88) V = = Q 4π 0 E dl = b ( ɛ 0 b + ɛa ɛb ( Q 4πɛ 0 r ) 2 ) dr a b ( ) Q 4πɛr 2 dr 0 a (0)dr (2.89) E P = ɛ 0 χ e E = ɛ 0χ e Q ˆr (2.90) 4πɛr2 ρ b = P =0 (2.9) ɛ 0 χ e Q 4πɛb 2 = χ e Q 4π( + χ e )b 2 σ b = P ˆn = ɛ 0χ e Q 4πɛ 0 a 2 = χ e Q 4π( + χ e )a 2 (2.92) /4πɛ 0 (Q/r 2 ) /4πɛ(Q/r 2 ) a<r<b a<r<b 52

56 2.24: Q b = χ e +χ e Q (2.93) Q b Q a Q Q b =/( + χ e )Q =(ɛ 0 /ɛ)q b Q b 2.24 r >b Q a <r<b Q Q b =(ɛ 0 /ɛ)q (2.88) E D D E D =0 D = ɛ(r)e = ɛ(r)( E)+ ɛ(r) E = ɛ(r) E (2.94) 2.20 D 0 ɛ Guass D = ɛ( E) =0 (2.95) D = ɛ E = ρ f (2.96) ρ f E vac E vac ɛ 0 ɛ 2.25: E = ɛ 0 ɛ E vac = ɛ r E vac (2.97) 53

57 q E = q ˆr (2.98) 4πɛ r Example 4.6 ɛ r 2.26: /ɛ r V /ɛ r C = Q/V C = ɛ r C vac (2.99) (2.82) P x = ɛ 0 (χ exx E x + χ exy E y + χ exz E z ) P y = ɛ 0 (χ eyx E x + χ eyy E y + χ eyz E z ) P z = ɛ 0 (χ ezx E x + χ ezy E y + χ ezz E z ) (2.00) χ exx, χ exy, susceptibility tensor (2.82) isotropic 54

58 Boundary Value Problems with Linear Dielectrics ρ b ( χ ) e ρ b = P = ɛ 0 ɛ D = ( χe +χ e ) ρ f (2.0) ρ f 2 ρ b =0 ρ = ρ f + ρ b =0 Laplace Laplace (2.77) ɛ above E above ɛ below E below = σ f (2.02) ɛ above V above n ɛ below V below n = σ f (2.03) V above = V below (2.04) Example R E 0 Chapter 3 Example 3.8 E 0 0 E 0 V in (r, θ) (r R) V out (r, θ)(r R) Laplace (i)v in = V out, at r = R (ii)ɛ V in V out = ɛ 0 r r, at r = R (iii)v out E 0 r cos θ, for r R (2.05) 2.27: (ii) (2.03) σ f =0 Laplace V in (r, θ) = A l r l P l (cos θ) (2.06) l=0 2 σ b χ e 55

59 (iii) V out (r, θ) = E 0 r cos θ + (i) A l B l 3 l=0 l=0 B l r l+ P l(cos θ) (2.07) A l R l B l P l (cos θ) = E 0 R cos θ + R l+ P l(cos θ) (2.08) l=0 A l R l = B l R l+, A = E 0 R + B R 2 for l (ii) ɛ r la l R l = (l + )B l R l+2, for l ɛ r A = E 0 2B R 3 A l = B l =0, for l, A = 3 ɛ r +2 E 0, B = ɛ r ɛ r +2 R3 E 0 (2.09) (2.0) (2.) V in (r, θ) = 3E 0 ɛ r +2 r cos θ = 3E 0 ɛ r +2 z (2.2) E = 3 ɛ r +2 E 0 (2.3) Example xy z <0 ɛ = ɛ 0 ( + χ e ) d (0, 0,d) q xy E σ b σ b (x, y) =P ˆn = P z = ɛ 0 χ e E z (x, y, 0 ) (2.4) E z (x, y, 0 ) z =0 z 0 q 2.28: 3 P (cos θ) =cosθ l Legendre V in V out 56

60 Coulomb q (x, y, 0) z z E (x, y) = qd (2.5) 4πɛ 0 (x 2 + y 2 + d 2 ) 3/2 E (x, y) = σ b(x, y) 2ɛ 0 (2.6) E(x, y, 0 )=E (x, y)+e (x, y) = qd 4πɛ 0 (x 2 + y 2 + d 2 ) σ b(x, y) (2.7) 3/2 2ɛ 0 (2.4) [ σ b = ɛ 0 χ e qd 4πɛ 0 (x 2 + y 2 + d 2 ) σ ] b 3/2 2ɛ 0 (2.8) σ b σ b (x, y) = 2π ( χe ) χ e +2 qd (x 2 + y 2 + d 2 ) 3/2 (2.9) q σ b Coulumb σ b z>0 (0, 0, d) q b = (χ e /χ e + 2)q 2.29(b) z <0 (0, 0,d) q b 2.29(c) (a) (b) (c) 2.29: (a) (b)z >0 (0, 0, d) q b (c)z <0 (0, 0,d) q b [ ] V above (x, y, z) = q 4πɛ 0 x2 + y 2 +(z d) + q b 2 x2 + y 2 +(z + d) 2 (z >0) (2.20) 57

61 [ ] V below (x, y, z) = q + q b 4πɛ 0 x2 + y 2 +(z d) 2 (z <0) (2.2) z =0 (i) V above (x, y, 0) = V below (x, y, 0) V above (ii) ɛ 0 z = ɛ V below (2.22) z=0 z z=0 (i) (ii) V above z = z=0 { } q(z d) 4πɛ 0 [x 2 + y 2 +(z d) 2 ] + q b (z + d) = 3/2 [x 2 + y 2 +(z + d) 2 ] z=0 d (q q b ) 3/2 4πɛ 0 (x 2 + y 2 + d 2 ) 3/2 (2.23) = { } (q + q b )(z d) = z=0 4πɛ 0 [x 2 + y 2 +(z d) 2 ] z=0 d (q + q b ) (2.24) 3/2 4πɛ 0 (x 2 + y 2 + d 2 ) 3/2 V below z ( ɛ(q + q b )=ɛ χ e ɛ 0 (q q b )=ɛ 0 ( + χ e χ e +2 (ii) χ e +2 ) ( ) 2 q = ɛ χ e +2 (2.25) ) q =2ɛ 0 χ e + χ e +2 q = ɛ 2 χ e +2 q (2.26) ( Vabove σ b = ɛ 0 z V ) below z=0 z = d [q q b (q + q b )] z=0 4πɛ 0 (x 2 + y 2 + d 2 ) = q bd (2.27) 3/2 2πɛ 0 (x 2 + y 2 + d 2 ) 3/2 dxdyσ(x, y) = q b (2.28) χ e χ e /(χ e + 2) 58

62 Energy in Dielectric Systems W = ɛ 0 2 E 2 dτ (2.29) 2.30 C Q W = 2 CV 2 (2.30) +Q ɛ V V = Q/C W = Q 0 ( q ) dq = Q 2 C 2 C = 2 CV 2 (2.3) Q 2.30: ɛ C = ɛ r C vac W = 2 ɛ rc vac V 2 (2.32) V W = ɛ r W vac (2.33) ɛ r W = ɛ 0 2 ɛ r E 2 dτ = 2 D Edτ (2.34) (2.34) q ɛ 2.3: 59

63 2.3 ρ f ρ f + ρ f W = ( ρ f )Vdτ (2.35) D = ρ f, ρ f = ( D) D D W = [ ( D)]Vdτ (2.36) [( D)V ]=[ ( D)]V +[( D)] V (2.37) W = { [( D)V ] [( D)] V }dτ = { [( D)V ]+[( D)] E}dτ (2.38) { [( D)V ]dτ = ( DV ) da (2.39) D = ɛe W = ( D) Edτ (2.40) 2 (D E) = (ɛe E) =ɛ( E) E =( D) E (2.4) 2 ( W = 2 ) D Edτ (2.42) W = D Edτ (2.43) 2 (2.29) (2.29) (2.43) (2.29) (2.43) (2.43) (2.40) (2.43) (2.4) 60

64 Forces on Dielectrics 2.32 w l d ±Q 2.32: 2.33 (fringing region) 2.33 E (fringing field) W dx dw = Fdx (2.44) F x F = dw dx (2.45) 6

65 Q F Q F 2.33: 2.34: W = 2 CV 2 = Q 2 2 C Q Q 2 F = dw dx = dc 2 C 2 dx = 2 V 2 dc dx (2.46) (2.47) x l x 2.35 C = ɛ 0wx d, C ɛw(l x) 2 = d C = C + C 2 = ɛ 0wx d + ɛ 0ɛ r w(l x) d (2.48) = ɛ 0w d [x + ɛ r(l x)] = ɛ 0w d [ɛ rl +( ɛ r )x] = ɛ 0w d (ɛ rl χ e x) (2.49) dc dx = ɛ 0χ e w (2.50) d F = ɛ 0χ e w 2d V 2 (2.5) x dw/dx V F = 2 V 2 dc dx (2.52) (2.47) (2.44) dw = Fdx+ VdQ (2.53) VdQ F = dw dx + V dq dx = 2 V 2 dc dx + V 2 dc dx = 2 V 2 dc dx (2.54) 62

66 x l x d ɛ : (2.47) Q F =(p )E F = [P(r) ]E(r)dτ (2.55) x F = F ˆx F = (P(r) )E x (r)dτ = (P ) V dτ (2.56) x V x = ( V )= x x E (2.57) F = P ( ) E dτ (2.58) x P = ɛ 0 χ e E P =0 x x = X (2.58) X+l F = ɛ 0 χ e dx X d 0 dy X x X + l, 0 y d, 0 w z (2.59) w x F = ɛ 0χ e 2 0 d 0 dze dy ( ) E = ɛ 0χ e x 2 w 0 X+l X dx d 0 dy w 0 dz x (E2 ) (2.60) dz[e 2 (X + l, y, z) E 2 (X, y, z)] (2.6) 63

67 E = V/d x = X x = X + l E(X, y, z) V/d, E(X + l, y, z) 0 (2.62) F = ɛ 0χ e wd 2 ( ) 2 V = ɛ 0χ e w d 2d V 2 (2.63) (2.58) (2.59) χ e r χ e (r) χ e (r) =χ e Θ(x X)Θ(X + l x)θ(y)θ(d y)θ(z)θ(w z) (2.64) Θ(x) ( ) E F = ɛ 0 χ e E dτ = ɛ 0 χ e x 2 x (E2 )dτ = ɛ 0 χe 2 x E2 dτ (2.65) dθ(x)/dx = δ(x) (2.65) x χ e(x, y, z) =χ e [δ(x X) δ(x X l)]θ(y)θ(d y)θ(z)θ(w z) (2.66) F = ɛ 0χ e 2 (2.6) d 0 dy w 0 dz[e 2 (X, y, z)] E 2 (X + l, y, z)] (2.67) 64

68 3 Chapter 6 Magnetic Fields in Matter Magnetization Multipole Expansion of the Vector Potential θ rr r r 3.: 3. A(r) = µ 0I 4π r r dl (3.) r r = ( ) r n P n (cos θ rr ) (3.2) r r A(r) = µ 0I 4π n=0 n=0 r n+ (r ) n P n (cos θ rr )dl (3.3) A(r) = µ [ 0I dl + 4π r r 2 r cos θ rr dl + ( 3 r 3 (r ) 2 2 cos2 θ rr ) ] dl + 2 (3.4) 65

69 dl =0 (3.5) monopole term Maxwell B =0 A dip (r) = µ 0I 4πr 2 r cos θ rr dl = µ 0I 4πr 2 (ˆr r )dl (3.6) (c r)dl = c da c a (3.7) A dip (r) = µ 0 m ˆr 4π r 2 (3.8) m magnetic dipole moment m I da = Ia (3.9) a R m = IπR 2 B dip (r) = A dip = µ 0 [3(m ˆr)ˆr m] (3.0) 4π r3 3.2(a) (3.0) (3.0) 3.2(b) (3.0) a Ia = 3.2: Diamagnets, Paramagnets, Ferromagnets magnetism U 66

70 magnetic dipoles magnetization E B paramagnets B diamagnets ferromagnets Torques and Forces on Magnetic Dipoles 3.3: C D B A 3.4:

71 3.4 z BC DA x AB CD N = af sin θˆx (3.) F F = IbB (3.2) N = IabB sin θˆx = mb sin θˆx (3.3) N = m B (3.4) m = Iab (3.4) Prob.6.2 (3.4) N = p E paramagnetism (a) (b) 3.5: ( ) F = I (dl B = I dl B =0 (3.5) 3.5(a) R I B 3.5(b) F = 2πIRBcos θ (3.6) 68

72 B m F = (m B) (3.7) Prob.6.4 (3.7) F = (p E) : 3.6 v R T =2πR/v I = e T = ev 2πR (3.8) m = I(πR 2 )ẑ = evrẑ (3.9) 2 e 2 4πɛ 0 R 2 = m v 2 e R (3.20) e(v B) B 3.7 v e 2 4πɛ 0 R 2 + e vb = m v 2 e R m e (3.2) 69

73 v = v v e 2 4πɛ 0 R 2 + evb + e vb = m v 2 e R + m 2v v e R (3.20) v B 2 (3.22) v = erb 2m e (3.23) 3.7 m = 2 e( v)rẑ = e2 R 2 4m e B (3.24) m B diamagnetism 3.7: Magnetization () (2) M (3.25) M magnetization vector P M 70

74 The Field of a Magnetized Object Bound Currents P r r r O r m dτ V 3.8: M(r) r m r A(r) = µ 0 m (r r ) 4π r r 3 (3.26) r dτ M(r )dτ 3.8 A(r) = µ 0 4π M(r ) (r r ) r r 3 dτ (3.27) A(r) = µ 0 4π r r = r r r r 3 ( ) M(r ) r r dτ (3.28) [ M(r ] ( ) ) r r = r r [ M(r )] + r r M(r ) (3.29) A(r) = µ 0 4π r r [ M(r )]dτ µ 0 4π ( v)dτ = A(r) = µ 0 4π V r r [ M(r )]dτ + µ 0 4π S [ M(r ] ) r r dτ (3.30) v da (3.3) r r [M(r ) da ] (3.32) J K A(r) = µ 0 J(r ) 4π r r dτ, A(r) = µ 0 K(r ) 4π r r da (3.33) J b = M (3.34) 7

75 K b = M ˆn (3.35) ˆn A(r) = µ 0 J b (r ) 4π r r dτ + µ 0 4π V S K b (r ) r r da (3.36) (3.36) J b = M K b = M ˆn J b K b bound currents magnetization currents) (3.27) 3.9 P O r r r r r r dτ V r J b S K b 3.9: Physical Interpretation of Bound Currents J b K b 3.0(a) 3.0(b) I (a) (b) 3.0: 3.: I M 3. a t m = Mat I m = Ia 72

76 I = Mt K b = I/t = M ˆn 3.0(b) K b K b = M ˆn (3.37) bound) 3.2: 3.2(a) y x I x =[M z (y + dy) M z (y)]dz = M z dydz (3.38) y (J b ) x = M z y (3.39) 3.2(b) z M y 3.2(a) x I x = [M y (z + dz) M y (z)]dy = M y dydz (3.40) z (J b ) x = M y z (3.4) (J b ) x = M z y M y z (3.42) J b = M (3.43) ρ ρ + J =0 (3.44) t 73

77 ρ/ t =0 J =0 A ( A) =0 J b = ( M) =0 (3.45) (3.29) (3.29) product rule [ (fa) = y (fa z) ] z (fa y) (fa) = f A + f( A) (3.46) [ ˆx + z (fa x) ] [ x (fa z) ŷ + x (fa y) ] y (fa x) ẑ (3.47) y (fa z) z (fa y)= f y A z + f A z y f z A y f A y z =( f A) x + f( A) x (3.48) z (fa x) x (fa z)= f z A x + f A x z f x A z f A z x =( f A) y + f( A) y (3.49) x (fa y) y (fa x)= f x A y + f A y x f y A x f A x y =( f A) z + f( A) z (3.50) (3.46) (3.3) (3.32) (3.3) V vdτ = S v da v v ˆx = v zŷ v y ẑ (v ˆx) = v z y v y z =( v) x (3.5) (v ˆx) da = v z da y v y da z = (v da) x (3.52) V V ( v) x dτ = (v da) x (3.53) S ˆx ŷ,ẑ ( v) y dτ = (v da) y (3.54) V S ( v) z dτ = (v da) z (3.55) V S ( v)dτ = (v da) (3.56) S 74

78 3.3 H 6.3 The Auxiliary Field H 3.3. Ampère J b = M K b = M ˆn [ free current J f ] Ampère J = J f + J b (3.57) µ 0 ( B) =J = J f + J b = J f + M (3.58) ( ) B M = J f (3.59) µ 0 H µ 0 B M (3.60) Ampère H = J f (3.6) H dl = I fenc (3.62) I fenc = J f da Ampèrian loop Ampère (3.6) (3.62) Ampère H Linear and Nonlinear Media Magnetic Susceptibility and Permeability B H (3.60) M M = χ m H (3.63) χ m χ m > 0 χ m < 0 (3.63) linear media B H (3.60) B = µ 0 (H + M) =µ 0 ( + χ m )H (3.64) B H B = µh (3.65) 75

79 µ µ µ 0 ( + χ m ) (3.66) permeability χ m =0 µ = µ 0 µ 0 permeability of free space Example (a) n χ m I Ampèrian loop 3.3(b) Ampère H dl = I fenc B H Griffiths Example 5.9 H = niẑ (3.67) B = µ 0 ( + χ m )niẑ (3.68) χ m > 0 B χ m < 0 B (a) (b) 3.3: K b = M ˆn = χ m (H ˆn) =χ m ni ˆφ (3.69) K b I : A Deceptive Parallel Ampère Ampère J J f B µ 0 H µ 0 H J f B B = J µ 0 (3.70) B =0 (3.7) 76

80 B Ampère (3.70) (3.7) Ampère B (3.7) H H = J f (3.72) H ( ) H = B M = M (3.73) µ 0 M B µ 0 H M Griffiths Prob.6.9 Prob.6.4 H =0 B =0 B = µ 0 M H =0 H =0 M M =0 M M B H Ampère H M =0 H H H = Boundary Conditions B above, H above l B above,h above K K B below, H below B below,h below 3.5: 3.6: K B Griffiths Sec B 3.5,3.6 Babove Bbelow =0 (3.74) B above B below = µ 0(K ˆn) (3.75) H (3.74) H above H below = ( ) ( ) Babove Mabove Bbelow Mbelow µ 0 µ 0 (3.76) = µ 0 (B above B below) (M above M below) (3.77) = (M above M below) (3.78) 77

81 H above H below = (M above M below) (3.79) (3.75) H above H below = µ 0 (B above B below ) (M above M below ) (3.80) = K ˆn (M above M below ) (3.8) K b =(M below M above ) ˆn (3.82) [ ] K b ˆn = (M above M below ) ˆn ˆn (3.83) A (B C) =B(A C) C(A B) [ ] K b ˆn = (M above M below ) ˆn ˆn +(M above M below )(ˆn ˆn) (3.84) = (Mabove Mbelow)ˆn +(M above M below )=M above M below (3.85) (3.8) H above H below =(K K b) ˆn (3.86) K = K f + K b H above H below = K f ˆn (3.87) 78

82 4 Chapter 7 Electrodynamics 4. Maxwell Maxwell s Equations Maxwell E = ɛ 0 ρ Gauss (4.) B =0 (4.2) E = B t Faraday (4.3) E B = µ 0 J + µ 0 ɛ 0 t Maxwell Ampère (4.4) 4.2 Maxwell Maxwell s Equations in Matter Maxwell ρ b P M ρ b = P (4.5) J b = M (4.6) J p 4. 4.: ±σ b = ±P ˆn P P P + dp σ b σ b + dσ b dσ b = dp ˆn a da dσ b = da dp da = da ˆn di = σ b t da = P t da (4.7) di = J da J p = P t polarization current J p = P t = t ( P) = ρ b t 79 (4.8) (4.9)

83 ρ = ρ f + ρ b = ρ f P (4.0) Gauss J = J f + J b + J p = J f + M + P t (4.) E = ɛ 0 (ρ f P) (4.2) D D ɛ 0 E + P (4.3) D = ρ f (4.4) Ampère ( B = µ 0 J f + M + P ) E + µ 0 ɛ 0 t t (4.5) H (B µ 0 M)=µ 0 J f + µ 0 t (P + ɛ 0E) (4.6) H µ 0 B M (4.7) H = J f + D (4.8) t Faraday B =0 ρ J Maxwell D = ρ f Gauss (4.9) B =0 (4.20) E = B t Faraday (4.2) H = J f + D t Maxwell Ampère (4.22) P = ɛ 0 χ e E, M = χ m H (4.23) D, H E, B D = ɛe, H = µ B (4.24) ɛ = ɛ 0 ( + χ e ),µ= µ 0 ( + χ m ) 80

84 Boundary Conditions σ K E, B, D, H Maxwell (i) D da = Q fenc S S (ii) B da =0 (iii) (iv) S E dl = d P dt S P H dl = I fenc + d dt B da P S D da Gauss 4.2 (i) S D a D 2 a = σ f a (4.25) a D D D 2 = σ f (4.26) (ii) B B 2 =0 (4.27) (iii) 4.3 Amperian loop E l E 2 l = d dt S B da (4.28) Amperian loop E dl E E 2 =0 (4.29) E (iv) H l H 2 l = I fenc + d D = I fenc (4.30) dt I fenc Amperian loop ˆn (ˆn l) Amperian loop l loop I fenc K f I fenc = K f (ˆn l) =(K f ˆn) l (4.3) H H 2 = K f ˆn (4.32) S H 8

85 4.2: 4.3: (4.26),(4.27),(4.29),(4.32) E, B (i) ɛ E ɛ 2 E 2 = σ f (ii) E E 2 =0 (ii) B B 2 =0 (iii) B µ B 2 µ = K f ˆn 2 (i) ɛ E ɛ 2 E 2 =0 (ii) E E 2 =0 (ii) B B 2 =0 (iii) (4.33) B µ (4.34) B 2 µ =