4.4 Linear Dielectrics F r stable magnetic dipole θ stable superconductor F image dipole 4.4.1 Susceptibility, Permitivility, Dielectric Constant r If E is not too strong, the polarization is proportional to the field. r r r r r r P = χeε 0E (since D = ε 0E + P , D is electric displacement from ρ free ) r r Æ ∇ × D = ∇ × P = 0 P M χe is called the electric susceptibility. χe = , χm = ε0E H r r r r r D = ε 0E + P = ε 0 ()1+ χe E = εE , ε is called the permittivity. ε ε r = = 1+ χe is called the relative permittivity, or dielectric constant. ε 0 Material Dielectric Constant Material Dielectric Constant Vacuum 1 Benzene 2.28 Helium 1.000065 Diamond 5.7 Air 1.00054 Water 80.1 Nitrogen 1.00055 KTaNbO3 34,000 What is high K material? +++++++++ +++++++++ +++++++++ Vacuum E ε r = 2 E - - - - - - - - - - - - - - - - - - - - - - - - - - - It may discharge for an electric field higher than E. a Example: A metal sphere of radius a carries a charge Q. It is b surrounded, out to radius b, by linear dielectric material of permittivity ε . Find the potential at the center. r Q r Q r Q D = 2 rˆ Æ r > b , E = 2 rˆ and b > r > a , E = 2 rˆ 4πr 4πε0r 4πεr Q b 1 a 1 Q 1 1 1 V = − dr + − dr = − + ∫ 2 ∫ 2 4π ∞ ε 0r b εr 4π ε 0b εb εa r r χ ε Q b > r > a , P = χ ε E = e 0 rˆ e 0 4πεr 2 r ρb = −∇ ⋅ P = 0 r χ ε Q r χ ε Q r = a , σ = P ⋅ ()− rˆ = − e 0 and r = b , σ = P ⋅ ()rˆ = e 0 b 4πεa2 b 4πεb2 If the space is entirely filled with a homogeneous linear dielectric, then r r E ∇ ⋅ D = ρ f and ∇ × D = 0 vacuum When crossing the boundary: P r r ∇ × D = ∇ × P ≠ 0 Example: A parallel-plate capacitor is filled with insulating - - - - - - - - - - material of dielectric constant ε r . What effect does this have on its capacitance? ++++++++++ σ + σ + σ + σ + 2AD = Aσ + Æ D+ = , E+,up = , E+,up,out = , E+,down = − 2 2ε 2ε 0 2ε 0 σ − σ + σ + σ + σ + 2AD = Aσ − Æ D− = = − , E−,up = − , E−,down = , E−,doun,out = 2 2 2ε 0 2ε 2ε 0 σ σ Q Q Aε E = Æ ∆V = d = d , C = = = ε C in ε ε Aε ∆V d r vacuum 4.4.2 Boundary Value Problem with Linear Dielectrics If we place free charge inside a linear dielectric, r + r r ε D χ P E 0 e ρb = −∇ ⋅ = −∇ ⋅()χeε 0 = −∇ ⋅ χe = − ρ f ε 1+ χe If we place the free charge on the boundary: D⊥,above − D⊥,below = σ f Æ ε aboveE⊥,above − εbelowE⊥,below = σ f ∂Vabove ∂Vbelow ε above − − εbelow − = σ f , continuity: Vabove = Vbelow ∂n ∂n Example: A sphere of homogeneous linear dielectric material is r placed in an otherwise uniform electric field E0 . Find the E0 electric field inside the sphere. Method 1: ∂Vout ∂Vin The free charge is on the boundary Æ ε out − − ε in − = σ f = 0 ∂r ∂r Only bound charge exists, no free charge. r >> R , Vout → −E0r cosθ ∂V ∂V r = R , ε in = ε out and V = V ∂r 0 ∂r in out A V = −E r cosθ + cosθ , V = Br cosθ out 0 r 2 in A 2A 3 3 ε − ε 0 − E0R + 2 = BR , εB = ε 0 − E0 − 3 Æ B = − E0 , A = R R R 2 + ε /ε 0 2ε 0 + ε 3 3 r ∂V 3 Vin = − E0r cosθ = − E0 z Æ E = − zˆ = E0 zˆ 2 + ε r 2 + ε r ∂z 2 + ε r Method 2: r Uniformly polarized sphere with polarization P = Pzˆ may produce electric field r P E = − zˆ inside. 3ε 0 r P ˆ Total field inside: E = E0 − z inducing the polarization: 3ε 0 r P 3χ r 3 3 ˆ ˆ e ˆ ˆ Pz = χeε 0E = χeε 0 E0 − z Æ P = ε 0E0 Æ E = E0 z = E0 z 3ε 0 3 + χe 3 + χe 2 + ε r Example: Suppose the entire region below the plane z = 0 is filled with uniform linear dielectric material of susceptibility χe . Calculate the force on a point charge q situated a distance d above the origin. Considering q without dielectric material: q 1 d E = − z,below 2 2 2 2 4πε0 ()r + d r + d r σ b Considering dielectric material without q: σ b = P ⋅ zˆ Æ Ez,above = and 2ε 0 σ b Ez,below = − 2ε 0 r qd σ b Total effect: P ⋅ zˆ = σ b = χeε 0Ez,below,total = χeε 0 − 3 − 2 2 2ε 0 4πε0 r + d 1 χ qd σ = − e b 2 2 3 / 2 2π χe + 2 ()r + d Total bound charge: χ Q = σ da = − e q b ∫ b χe + 2 χ e Use image charge Qb = − q at z = -d to solve the problem. χe + 2 1 q Q V = + b z > 0 4πε 2 2 2 2 2 2 0 x + y + ()z − d x + y + ()z + d 1 q + Q V = b z < 0 4πε 2 2 2 0 x + y + ()z − d r 1 qQb attractive force on q: F = 2 zˆ 4πε0 ()2d 4.4.3 Energy in Dielectric Systems To charge up a capactor, it takes energy of Q Q2 1 dW = VdQ Æ W = dQ = = CV 2 ∫ C 2C 2 Change of capacitance in dielectric materials: Cdielectric = ε rCvacuum Æ increase the energy because of an increase of charge r r r ∆W = ∆ρ Vdτ , ρ = ∇ ⋅ D Æ ∆W = ∆ ∇ ⋅ D Vdτ = ∇ ⋅ ∆D Vdτ ∫ f f ∫ ( ) ∫ ( ) r r r r r r r ∇ ⋅ (∆DV )= ∇ ⋅ (∆D)V + ∆D ⋅∇V Æ ∇ ⋅ (∆D)V = ∇ ⋅ (∆DV )+ ∆D ⋅ E Choose V=0 at r → ∞ . r r r r 1 r r ∆W = (∆D)⋅ Edτ for a linear dielectric ∆D ⋅ E = ∆(D ⋅ E) ∫ 2 1 r r W = D ⋅ Edτ 2 ∫ l 4.4.4 Forces on Dielectrics fringing field x Q wlσ ε wl C = = = 0 vacuum σ V d d ε 0 ε w ε w ε w C = ε C Æ C = 0 x + ε 0 ()l − x = 0 ()ε l − χ x dielectric r vacuum d r d d r e 1 Q2 Assume that the total charge on the plate is constant (Q = CV ), W = 2 C dW Q2 dC 1 dC ε χ w F = − = = V 2 = − 0 e V 2 attractive force dx 2C 2 dx 2 dx 2d If you start from constant voltage, the work supplied by a battery must be included. d 1 dQ F = − CV 2 +V dx 2 dx If the voltage is constant and the charge is varying, you must include the force due to the battery for maintaining a constant voltage. ++++++++ + + + + ++++ Exercise: 4.18, 4.23, 4.28 1. Dielectric material Æ increase the capacitance of a capacitor Æ store much more charges 2. P P P E + − 3ε 3ε 0 0 E P P E E + P /ε P 0 − ε 0 E 3. Ferroelectricity: BaTiO3 ferroelectric antiferroelectric Transition temperature? Curie-Weiss law? .