1. Lecture 5 Electric Flux Density and Dielectric Constant Boundary Conditions Electromagnetics Prof. Viviana Vladutescu

2. Electric Flux Density Gauss’s Law: The total outward flux of the dielectric displacement (or simply the outward flux) over any closed surface is equal to the total free charge enclosed in the surface

3. Where- ε is the absolute permittivity (F/m) -ε r is the relative permittivity or the dielectric constant of the medium -ε 0 is the permittivity of free space -χ e is the electric susceptibility (dimensionless)

4. Material Dielectric Constants Vacuum 1 Glass 5-10 Mica 3-6 Mylar 3.1 Neoprene 6.70 Plexiglas 3.40 Polyethylene 2.25 Polyvinyl chloride 3.18 Teflon 2.1 Germanium 16 Strontiun titanate 310 Titanium dioxide (rutile) 173 perp 86 para Water 80.4 Glycerin 42.5 Liquid ammonia(-78°C 25 Benzene 2.284 Air(1 atm) 1.00059 Air(100 atm) 1.0548

5. Homogeneous -ε r independent of position Anisotropic ε r is different for different of the electric field

6. Biaxial, Uniaxial and Isotropic Medium - biaxial - uniaxial -isotropic

7. KDP ADP crystals – Electric field along optic axis Uniaxial crystal becomes biaxial with applied field!

8. GaAs CdTe crystals – Electric field along optic axis Isotropic crystal become biaxial with applied field!

9. Dielectric Strength The maximum electric field intensity that a dielectric material can stand without breakdown MaterialDielectric Strength (V/m) Air3e6 Bakelite24e6 Neoprene rubber12e6 Nylon14e6 Paper16e6 Polystyrene24e6 Pyrex glass14e6 Quartz8e6 Silicone oil15e6 Strontium titanate8e6 Teflon60e6

10. Boundary Conditions

11. Medium 1&2 are dielectrics CONDITION I (tangential components)

14. Condition I (tangential components)

15. CONDITION II (normal components)

17. Condition II (normal components) The normal component of D field is discontinuous across an interface where a surface charge exists, the amount of discontinuity being equal to the surface charge density

18. Example : Two dielectric media with permittivity ε 1 and respectively ε 2, are separated by a charge free boundary. The electric field intensity in medium 1 at point P 1 has a magnitude E 1 and makes an angle α 1. Determine the magnitude and direction of the electric field intensity at point P 2 in medium 2. α1α1 α2α2 P1P1 P2P2

20. The magnitude of E 2 :

21. Boundary conditions at a Dielectric/Conductor Interface - inside a good conductor E=0 E T =0 D=0 D n =ρ s

23. Practice Problems

24. Suppose D = 6rcosf a f C/m 2. (a) Determine the charge density at the point (3m, 90 , -2m). Find the total flux through the surface of a quartered-cylinder defined by 0 ≤ r ≤ 4m, 0 ≤ f ≤ 90 , and -4m ≤ z ≤ 0 by evaluating (b) the left side of the divergence theorem and (c) the right side of the divergence theorem. (a) (b) Divergence Theorem and Gauss’s Law

25. note that the top, bottom and outside integrals yield zero since there is no component of D in the these dS directions. So, (c)

26. The potential field in a material with ε r = 10.2 is V = 12 xy2 (V). Find E, P and D. Electric Potential

27. For z ≤ 0,  r1 = 9.0 and for z > 0,  r2 = 4.0. If E 1 makes a 30  angle with a normal to the surface, what angle does E 2 make with a normal to the surface? Boundary Conditions

28. Therefore and after routine math we find Using this formula we obtain for this problem  2 = 14°. also