A Deceptive Parallel (2.60) Gauss ρ ρ f ɛ 0 E D D E D Coulumb D(r) 1 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
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1 The Electric Displacement Gauss Gauss s Law in the Presence of Dielectrics ρ b = P σ b = P ˆn free chargetrue 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 (2.61) S Q fenc = V ρ f dτ Gauss ρ f ρ b (2.60) σ b ρ b Gauss (2.61) Example 4.4: 2.19 λ a D 2.19 Gauss s L Gauss (2.61) D(2πsL) =λl (2.62) D = λ 2πsŝ (2.63) P =0 2.19: E = 1 D = λ for s>a (2.64) ɛ 0 2πɛ 0 sŝ, P 48
2 A Deceptive Parallel (2.60) Gauss ρ ρ f ɛ 0 E D D E D Coulumb D(r) 1 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 (2.61) D P =0 D D Boundary Conditions D above σ f σ f Dabove D below D below 2.21: 2.22: E Griffiths Sec E above E below = 1 ɛ 0 σ (2.67) E above E below =0 (2.68) D above = ɛe above + P above (2.69) 49
3 D below = ɛe below + P below, (2.70) (2.67)(2.68) D Dabove Dbelow = ɛ(eabove Ebelow)+P above Pbelow (2.71) (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 2.21 Gauss Gauss (2.61) 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 (2.81) Linear Dielectrics Susceptibility, Permittivity, Dielectric Constant D Guass Gauss 50
4 P P = ɛ 0 χ e E (2.82) χ e electric susceptibilityχ 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 (1 + χ e )E (2.83) D E ɛ D = ɛe (2.84) ɛ = ɛ 0 (1 + χ e ) (2.85) (permittivity) χ e =0 ɛ = ɛ 0 ɛ 0 (2.85) ɛ 0 ɛ r =1+χ e = ɛ ɛ 0 (2.86) (relative permittivity) (dielectric constant) Material Dielectric Constant Material Dielectric Constant Vacuum 1 Benzene 2.28 Helium Diamond 5.7 Neon Salt 5.9 Hydrogen Silicon 11.8 Argon Methanol 33.0 Air (dry) Water 80.1 Nitrogen Ice ( 30 C) 99 Water vapor (1100 C) KTaNbO 3 (0 C) 34, C Handbook of Chemistry and Physics, 79th ed. (Boca Raton: CRC Press, Inc., 1997) 51
5 Example 4.5: 2.23 a Q b ɛ 2.23: Gauss (2.61) 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 ( 1 ɛ 0 b + 1 ɛa 1 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 (2.91) ɛ 0 χ e Q 4πɛb 2 = χ e Q 4π(1 + χ e )b 2 σ b = P ˆn = ɛ 0χ e Q 4πɛ 0 a 2 = χ e Q 4π(1 + χ e )a 2 (2.92) 1/4πɛ 0 (Q/r 2 ) 1/4πɛ(Q/r 2 ) a<r<b a<r<b 52
6 2.24: Q b = χ e 1+χ e Q (2.93) Q b Q a Q Q b =1/(1 + 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 = 1 ɛ r E vac (2.97) 53
7 q E = 1 q ˆr (2.98) 4πɛ r Example 4.6 ɛ r 2.26: 1/ɛ r V 1/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 ) (2.100) χ exx,χ exy, susceptibility tensor (2.82) isotropic 54
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