Local Field. Estimate for E indiv?-some sort of average?

Transcription

1 &M Lecture 6 Topics: (1) Local Field in a dielectric medium (2) Clausius-Mossotti equation (3) non-uniform polarisation (4) lectric displacement (5) Origins of volume bound charge density

2 Local Field the field creating dipole (called the local field or loc ) is not the same as the actual field because the latter contains the effect of all the dipoles created! Recall the Capacitor where = vac - pol = vac / The key word above is all! In fact loc is the actual field less the effect of one dipole (essentially the one being created ) - think about one solitary individual dipole being created within a capacitor, the creating field ( vac ) is larger than the actual field ( = vac - indiv ) by the contribution of this one individual dipole. Vectorially, obtain loc by subtracting indiv from : = loc indiv stimate for indiv?-some sort of average?

3 Local Field continued stimate for indiv comes from L4: recall the average field of a point charge inside a spherical volume. Or, more generally, the average field of a spherical volume containing a specific dipole moment p: p indiv = = 4π R = p πr 3 3 loc = indiv P + P 3 3 loc + χ e 1+ χ e 3ε 3 o ( ) P Approximation arises from attempt to fill space with spheres, more correct to replace 3 by b 3 3

4 Clausius-Mossotti equation A practical equation relating to mass density ρ m : χ e Nα = 1+ χ e loc 3 = =1+ χ e 3 Nα = χ e 1+ χ e 3 = 3χ e ( χ e when χ e small) 3 + χ e (or when loc : eg gases) Substituting χ e = -1 Nα 3 = N Aα 3 = M ρ m but N = N Aρ m M where N A is Avogadro s number, M is Molecular Weight

5 Non-uniform polarisation Uniform polarisation results in surface bound charge density only; non-uniform polarisation results in surface and volume bound charge densities. Focus on latter (ρ b ), origin later. Consider box, cornered on (x,y,z) volume Δx by Δy by Δz, small but still containing large number of molecules and exhibiting non-uniform polarisation: Focus on x-component of P, and assume it is independent of y or z value at x is From L5, charge crossing area = P.da ˆ z (x,y,z) P x value at x + Δx is ˆ y (x+δx,y,z) ˆ x P x + P x x Δx

6 Non-uniform polarisation continued Charge entering LH yz face is P x ΔyΔz Charge exiting RH yz face is P x + P x x Δx ΔyΔz Net charge entering box is z ˆ (x,y,z) P x x ΔxΔyΔz y ˆ (x+δx,y,z) Total charge on box (including zx and xy pairs of faces) is P x x + P y y + P z ΔxΔyΔz =.P z ρ b =.P ( )( volume) ˆ x looks a bit like.=ρ t /?

7 lectric displacement (D).a convenience vector? 2 distinct types of charge, free and bound, are combined as total Gauss s Law (differential form) gives. = ρ t = 1 ( ρ f + ρ ) b = 1 ( ρ f.p) ρ f =. +.P =. + P ( ) write D = + P ρ t = ρ f + ρ b Displacement: a vector whose div equals free charge density Units: C m -2 (same as P). xpress D in terms of? D = + P = + χ ε e o = ε ( o 1+ χ e ) = ε r

8 Gauss s Law for, P and D. = ρ t.da = 1 ( encl. total ch. ).P = ρ b P.da = encl. bound ch. ( ).D = ρ f D.da = encl. free ch. ( ) use D as a route to? Field in capacitor? A cond. Gauss :.A = 1 ( σ f A) = σ f D Gauss : = D = σ f D.A = ( σ f A) D = σ f as =1in vacuum ( ) σ f

9 Gauss s Law for, P and D continued Trivial example but field in non-vacuum capacitor? Gauss :.A = 1 ( σ f σ b )A dielectric σ b conductor =? as σ b is not easily known σ f D Gauss : D.A = ( σ f A) D = σ f = D = σ f Still a relatively trivial example perhaps but one that shows how bound charge gets hidden in the relative permittivity!

10 Origins of volume bound charge density Recall uniform dielectric has no volume bound charge density (although it has dipoles), only surface bound charge density D = + P P = D ρ b =.P =. D If no longer treat as a spatial constant: ρ b =.D D. 1 1.D ( ) =. D D =.D origins: + D. 1 ε = 1 1 ε ρ + D. 1 f ε r r r (1) embedded ρ f induces ρ b of opposite sign (2) non-uniform medium, i.e. varying : either via dipole number density or magnitude of dipole moment