4.4 Linear F r stable magnetic θ stable superconductor F image dipole

4.4.1 Susceptibility, Permitivility, Constant r If E is not too strong, the 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 .

ε ε r = = 1+ χe is called the , 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 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 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?