Chapter 4: Electrostatic Fields in Matter

Transcription

1 Chapter 4: Eectrostatic Fieds in Matter 4. Poarization 4. The Fied of a Poarized Oject 4.3 The Eectric Dispacement 4.4 Sef-Consistance of Eectric Fied and Poarization; Linear Dieectrics

2 4. Poarization 4.. Dieectrics 4.. Induced Dipoes and Poarizaiity 4..3 Aignment of poar moecues 4..4 Poarization and Susceptuiity

3 4. Poarization Dieectrics According to their response to eectrostatic fied, most materias eong to two casses: conductors: charges free to move in the materia insuators (or dieectrics): a charges are attached to specific atoms or moecues. Stretching and rotating are the two principe mechanisms y which eectric fieds can distort the charge distriution of a dieectric atom or moecue.

4 4.. Induced Dipoes and Poarizaiity A neutra atom Atomic poarizaiity A simpe estimate: The fied at a distance d from the center of a uniform charge sphere E At equiirium, E e appied Poarized atom qin q 3 d qd 3 3 4πε d 4πε d a 4πε a = = = e qd p 4πε a 3 4πε a 3 = E = = α = 4πε a = 3ε v 3 Induced dipoe p = qd p = αe (v is the voume of the atom)

5 In case of a moecue it is more compex, their poarization may not e isotropic: E 4 α = c m N Generay: P = α E + α E E α = c m N the most genera inear reation P = α E ( α is poarizaiity tensor) or in (x,y,z) P = α E + α E + α E x xx x xy y xz z P = α E + α E + α E y yx x yy y yz z P = α E + α E + α E z zx x zy y zz z α ij depend on the orientation of the axis you chose. It s possie to seect axes such that α ij = for i j

6 4..3 Aignment of Poar Moecues Poar moecues have uit-in, permanent dipoe moments d Torque appied eectric fied (uniform) ( ) ( N = r ) + F+ + r F ( d ) = ( qe) + ( d ) ( qe) = qd E N = P E qd = P ( )

7 4..3 If the appied eectric fied is nonuniform, the tota force is not zero. F = F + F = q E E = q E + + ( ) ( ) for short dipoe, de = E d = d ( ) x x x de = E d = d ( ) y y y de = E d = d ( ) z z z de = d E ( ) F = P E ( ) Ex x Ey y E z z

8 4..4 Poarization and Susceptiiity The poarization of a poarized dieectric P dipoe moment per unit voume P = ε X E X e is eectric susceptiiity tensor e P = ε ( X E + X E + X E ) x e x e y e z xx xy xz P = ε ( X E + X E + X E ) y e x e y e z yx yy yz P = ε ( X E + X E + X E ) z e x e y e z zx zy zz for inear dieectric P = ε χ E e X e is eectric susceptiiity

9 4. The Fied of a Poarized Oject 4.. Bound charges 4.. Physica Interpretation of Bound Charge 4..3 The Fied Inside a Dieectric

10 4. The Fied of a Poarized Oject Bound charges: Divide the materia into infinitesima dipoes and sum up R V = P d = P ( ) dτ ˆ R τ 4πε V R 4πε V R = [ P dτ ] ( P) dτ ( ) 4πε V R V R = P da ( P) dτ 4πε surf R 4πε V R ound charges: surface charge voume charge σ ρ = P nˆ = P 4 πε surf R 4πε V R V = σ da + ρ d τ

11 4.. Physica Interpretation of Bound Charge ρ and σ represent perfecty genuine accumuations of charge. for uniformy distriuted dipoes for a uniform poarization and perpendicuar cut σ ρ = dipoe moment = P( Ad) = qd q = PA s σ q A = = P

12 4.. () for a uniform poarization and oique cut q = PA σ = q = P cosθ = P nˆ A end σ = P nˆ ρ =? If the poarization is nonuniform, δ P δ x δ q = ρ dτ = δ ( Pdτ ) In 3D this is ρ = P δ x

13 Exampe: Find the fied of a uniformy poarized sphere of radius R ρ =. P = σ = P nˆ = P cosθ r cosθ = z for for Ex.9 of chapter 3 V ( r, θ ) = r R V = 3 P z ε P ˆ E = V = z = P 3ε 3ε 3 r R V 4π = R prˆ 4πε 3 r Simiar to a potentia of a dipoe moment p = cos for 3 P r θ r R ε 3 cos for 3 P R θ r R ε r 4π 3 R P 3

14 4.3 The Eectric Dispacement Gauss Law in The Presence of Dieectrics: Tota charge density ρ = ρ + ρ f ound free Gauss Law ρ E = E = = + = P + ε ( εe + P) = ρ f So in a dieectric medium Gauss aw can e written as D = ρ f ε ρ ρ ρ ρ surface D da = Q f enc. f f D = ε E + P eectric dispacement

15 Exampe: A ong wire with inear charge λis covered with an insuation of radius a. Find the eectrica dispacement. r R D da = Q surface f enc. D ( π rl) = λl D = λ π r rˆ inside E =? need to know P outside λ E = D = rˆ ε πε r for r > R

16 4.3. A Deceptive Parae ρ Rˆ E = E = ρ dτ ε 4πε R E = since = this soution is unique (Hemhortz theorem: V(r) = ( V(r) ) ( V(r) ) But D = ε( E) + P P is not aways zero Rˆ D = ρ f ut D ρ f dτ 4π R And in genera there is no potentia for D (cannot e written as a gradient of a scaar)?

17 Permittivity, Dieectric Constant: for inear dieectrics permittivity P = ε χ E e Eectric susceptiiity D = εe + P = εe + εχee = ε ( + χe ) E D = ε E ε = ε( + χ e ) permittivity of free space dieectric constant or reative permittivity ρ ε r = + χ = εχ ( e χ ) ( e χ = P = D = ) D = ( e ) ρ f ε + χ + χ e e e ε ε χ e ρ f ρ f ρ = ρ + ρ f = ρ f = = + χe + χe εr

18 Exampe: A meta sphere of radius a has a charge Q. It is surrounded y a iner dieectric materia of permittivity ε. Find the potentia at the center, and ound charges of the medium (,. ε dieectric V center =? E = P = D = for r < a Q D = r for r > a ˆ Q 4π r for > r > a r ˆ 4πε r E = Q for r > r ˆ 4πε r a Q a Q Vcenter = Va = E d = dr dr 4πε r 4πε r Q = ( + ) 4π ε εa ε

19 ε χeq P = ε χee = r 4πε r ˆ ρ = P = σ at r = at r = a for > r > a ( P = r P r + = ) r r εxeq ˆ 4πε = P n = εxeq 4πε a

20 Exampe: A parae pate capacitor is fied with an insuating materia of dieectric constant ε. What is the effect of the dieectric on its capacitance? Dieectric without dieectric V = E. d vac E the dieectric reduces E to so V ε r d. e. = E d ε Q since C= C V = ε C d. e. r vac

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22 Boundary Conditions: Fied norma ) to surface: Fied parae ( to surface: surface E da aove Q σ A enc = = ε ε = E A E A eow = E aove E σ eow ε E d = ( ie. E = ) E aove =E eow E aove E eow In case of D aove conditions ecome: D V n Veow n aove aove D eow = σ f ε aove ε eow = σ f D D = P P aove eow aove eow V aove = V eow

23 Exampe: Dieectric sphere in a uniform eectric fied, find the eectric fied inside the sphere. V = V at r = R in out Vin V ε = ε out out at r = R (no free charges at the surface) r r V E z = E cos θ for r >> R out inside the sphere: V ( r, θ ) = A r P (cos θ ) B outside the sphere: V ( r, θ ) = E R cos θ + P (cos θ ) = + = r B oundary condition (i) (cos θ ) cos θ (cos θ ) B so A R = -E R + R A R P = ER + P + = = R B and A R = for R +

24 B oundary condition (i) (cos θ ) cos θ (cos θ ) A R P = ER + P + = = R B B so A R = - ER + and A for R = + R R ( + ) B oundary condition (ii) ε r A R P (cos θ ) = E cos θ P (cos ) θ + = = R B ( + ) B so ε r A = - E and ε - for 3 ra R = + R R ( + ) B ε ra R = ε -( ) for r = + + R this can not e true, so A = B = for ; 3E ε A = r cos θ, A = R E r 3 ε r + ε r + 3E 3E 3E Vin ( r, θ ) = r cos θ = z E( r, θ ) = ε + ε + ε + r r r

25 4.4.3 Energy in a Dieectric System Energy of a charged capacitor W = CV If the capacitor is fied with a dieectric C = ε C r vac So the stored energy shoud increases y the same factor: ε energy stored in any eectrostatic system (vacuum) W = E dτ ε with inear dieectric W = εr E dτ = D Edτ which suggests the energy in the dieectric is = D Edτ

26 4.4.3 Energy in Dieectric media (forma proof) Suppose we are graduay uiding up charge in a dieectric medium δw = ( δρ f ) Vdτ D = ρ f ρ f = D = ( δ D) Vdτ ( δ DV ) = ( δ D) V + δ D ( V ) = ( δ DV ) dτ δ D Vdτ = δ DV da + δ D Edτ s for S for inear dieectric, D = ε E = δ ( D E) dτ = δ[ D Edτ ] W = D Edτ δ D E = δε E E = δ ( ε E ) = δ ( D E)

27 4.4.4 Forces on dieectrics x d Dieectric with Q constant W Q = CV = C dw = F dx ex dw Q dc dc F = Fex = = = V ds C dx dx Q ε A εw dc ε ( ) Xew C = = = εr χex = V d d dx d F εχ = d e w V note we used Q W = not W = CV C

28 If V is kept constant: If W = CV used dw F = Fex = = V dx dc dx wrong sign The reason is to keep V constant the attery must do work. So that has to e taken in to account dw = Fex dx + VdQ W = CV dw dq F = Fex = + V dx dx dc dc = V + V dx dx dc = V now with correct sign/direction dx

29 λ λ' Hint: find the capacitance of tue first