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Electric fields in Material Space Sandra Cruz-Pol, Ph. D. INEL 4151 ch 5 Electromagnetics I ECE UPRM Mayagüez, PR.

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Presentation on theme: "Electric fields in Material Space Sandra Cruz-Pol, Ph. D. INEL 4151 ch 5 Electromagnetics I ECE UPRM Mayagüez, PR."— Presentation transcript:

1 Electric fields in Material Space Sandra Cruz-Pol, Ph. D. INEL 4151 ch 5 Electromagnetics I ECE UPRM Mayagüez, PR

2 Last Chapter: free space NOW: different materials

3 Some applications  superconductors  High permittivity dielectrics  Transistors  Electromagnets

4 We will study Electric charges:  Conductors or Insulators Depends on Frequency and Temperature…  Boundary conditions Conductors (metals) Insulators (dielectrics) Semiconductors

5 Material @ 20 o C Low frequency Conductivity (S/m) Silver6.1 x 10 7 Copper5.8 x 10 7 Gold4.1 x 10 7 Aluminum3.5 x 10 7 Carbon3 x 10 4 Sea water4 Silicon4.4 x 10 -4 Pure water10 -4 Dry Earth10 -5 Glass, Quartz10 -12, 10 -17 Colder metals conduct better. (superconductivity) Insulators at most lower frequencies. Conductors- have many free electrons available. semiconductor Appendix B

6 Current Units: Amperes [A] Definition: is the electric charge passing through an area per unit time. Current Density, [A/m 2 ] Is the current thru a perpendicular surface:

7 Depending on how I is produced: There are different types of currents.  Convection- I flows thru isolator: liquid, gas, vacuum. Doesn’t involve conductors, doesn’t satisfies Ohm’s Law  Conduction- flows thru a conductor  Displacement (ch9)

8 Current in a filament  Convection current, [A]  Convection density, A/m 2 SS vv u ll

9 Conduction Current  Requires free electrons, it’s inside conductor.  Suffers collisions, drifts from atom to atom  Conduction current density is: Newton’s Law where  v = ne

10 A Perfect conductor Has many charges that are free to move.  Therefore it can’t have an E field inside which would not let the charges move freely.  So, inside a conductor Charges move to the surface to make E=0

11 Resistance  If you force a Voltage across a conductor:  Then E is not 0  The e encounter resistance to move I E V + - S l  c = resistivity of the material

12 Power in Watts =Rate of change of energy or force x velocity Joule’s Law

13 PE 5.1 Find the current thru the cylindrical surface  For the current density

14 PE 5.2 In a Van de Graaff generator, w=0.1m, u=10m/s and the leakage paths have resistance 10 14 .  If the belt carries charge 0.5 C/m 2, find the potential difference between the dome and the base. w= width of the belt u= speed of the belt

15 PE 5.3 The free charge density in Cu is 1.81 x 10 10 C/m 3..  For a current density of 8 x 10 6 A/m 2, find the electric field intensity and the drift velocity.

16 Polarization in dielectrics The effect of polarization on a dielectric is to have a surface bound charge of: and leave within it an accumulation of volume bound charge:  ps and  pv are the polarization (bounded) surface and volume charge densities

17 Permittivity and Strength  Not really a constant!

18 Dielectric properties  Linear =  doesn’t change with E  Isotropic=  doesn’t change with direction  Homogeneous=  doesn’t change from point to point. Coulomb’s Law for any material:

19 PE 5.6. A parallel plate capacitor with plate separation of 2mm has a 1kV voltage applied to its plane.  If the space between its plates is filled with polystyrene, find E and P.

20 PE 5.7. In a dielectric material, E x = 5V/m and  Find:

21 Continuity Equation  Charge is conserved.

22 For steady currents:  Change= output current –input current = 0

23 Substituting in: where T r = is called the Relaxation time

24 What is Relaxation Time? [s]

25 Is the time it takes a charge placed in the interior of a material to drop to e -1 of its initial value.  Find T r for silver  Find T r for rubber:

26  We have two materials  How the fields behave @ interface? Boundary Conditions

27  We have two materials  How do the fields behave @ interface? We look at the tangential and the perpendicular component of the fields.

28 Cases for Boundary Conditions: 1.Dielectric- dielectric 2.Conductor- Dielectric 3.Conductor-Free Space

29 Dielectric-dielectric B.C.  Consider the figure below: E1E1 E2E2 E 1t E 1n E 2t E 2n ab c d  w hh 11 22 11

30 Dielectric-dielectric B.C.  Consider the figure below: D1D1 D2D2 D 1t D 1n D 2t D 2n hh  S 22 11 SS

31 Dielectric-Dielectric B.C. E1E1 D2D2 E 1t E 1n D 2t D 2n hh 22 11 In summary: 11

32 Conductor-dielectric B.C.  Consider the figure below: E EtEt EnEn a b c d  w hh 11  2 =∞ E 2 =0 11 dielectric conductor

33 Conductor-dielectric B.C.  Consider the figure below: E EtEt EnEn hh 11  2 =∞ E 2 =0 11 dielectric conductor  S SS

34 Conductor-Free Space B.C.  Consider the figure below: E EtEt EnEn a b c d  w hh 11  2 =∞ E 2 =0 oo Free space conductor

35 PE 5.9 A homogeneous dielectric ( r =2.5) fills region 1 ( x 0) is free space.  Find

36 5.29 Lightning strikes a dielectric sphere of radius 2-mm for which  r =2.5,  =5x10 -6 S/m and deposits uniformly a charge of 1C.  Determine the initial volume charge density and the volume charge density 2s later. Answer: 29.84KC/m3, 18.98 kC/m3


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