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Chapter 14

Chapter 14. Linear Dielectric Properties. Area under curve. Relative dielectric constant …. Very important. It is a measure of how much charge a solid can store relative to vacuum. In general. D =    P. where D is the displacement (C/m 2 ) E is the applied electric

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Chapter 14

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  1. Chapter 14 Linear Dielectric Properties

  2. Area under curve

  3. Relative dielectric constant …. Very important It is a measure of how much charge a solid can store relative to vacuum

  4. In general D = P where D is the displacement (C/m2) E is the applied electric field, (V/m) and P is the polarization (C/m2) of your material. In vacuum, P = 0 and D = 

  5. Dielectric Properties

  6. Something happens in the solid that allows the parallel plate capacitors to store more charge. That something is called polarization P = qd Thus understanding polarization is the key to understanding dielectric properties.

  7. Electronic polarization Pe Ionic polarization Pi Dipolar or orientation polarization Pdip Convince yourself that when E is applied the center -ve charge is no longer coincident with the center of +ve charge.

  8. N = number of diploes/m3 polarizability, which is an ionic/ atomic property. Only valid for dilute gases or when Eapplied = Elocal Only valid for cubic symmetry but used in many situations

  9. Most Important Equation in This Chapter • Why? • Because it is link between micro and macro… • Always recall that it is an approximate expression and if it agrees with experiment it is because you were born under a lucky star • and your mother loves you ....

  10. 4 Fundamental Polarization Mechanisms in Solids Electronic Polarization Ionic Polarization Dipolar Polarization ( linear) is also referred to as orientational. Space charge - occurs at electrodes and is very important in electrochemistry… Not discussed in this class…

  11. The Effect of Frequency Purple = V Red = q Blue = I When charges are in perfect sync with Eapp you have a perfect dielectric, with no losses. If charges are in phase what is the current doing? Hint: I = dq/dt In a perfect dielectric the current ?? the voltage by ??

  12. Ideal vs. Real Dielectric • Charge in phase with voltage; since i = dQ/dt, then current, Ichg, is 90° out of phase. This current is called a displacive current and does NOT lead to energy loss. • In a real dielectric, there is energy loss. • To take these into account: G = 1/R =conductance Power dissipation, W/m3

  13. A digression on i • It is a shorthand notation that - in the context of dielectric and optical properties you have two components.. • a brilliant tool to solve DE and describe various physical phenomena.. • When you see i, then your first thought should be: there must be some form of energy loss somewhere in this system… • The last jewel:

  14. Vectorial Sum Some charges are in phase with V and result in a charging current - but no loss. Others are 90° out of phase and lead t energy dissipation. Itot = Ichg + I loss Tan = I loss/Ichg

  15. Measuring Dielectric Properties If  is 90° then you have a perfect dielectric with no losses. If  is 0°, then you have a perfect conductor and no capacitance. You use a something called a lock-in analyzer ..

  16. How can you measure k’’ • In principle, one way would be to simply measure the temperature change in the system… • If k” is zero there is no energy loss in the system and thus no heat increase.

  17. Electronic Polarization Assumptions: i) Applied field = local field…. ii) The electrons are collectively attached by a spring to the nucleus, with a natural frequency of vibration of o and a spring constant = So. iii) Recall:

  18. Electronic Polarization Newton’s Law F = ma Zi = atomic number of atom/ion f is a friction factor, and is thus related to k’’

  19. Resonance; a thing of beauty Any examples from real life?

  20. Region 1 DC limit, viz.  = 0 and Region 2 Near resonance:  =  and ke increases dramatically. and would go to infinity, if there was no friction, i.e. if f =0 Region 3  <<  then charges cannot follow the field and drops out. Then ke goes to 1.

  21. Compare:

  22. What determines, ke’ Radius of atom

  23. Very important result… Electronic polarizability is Proportional to the volume of an atom or ion. 1- Size 2 - Charge 3 - Presence of d-electrons which are less shielding.

  24. Very simple model. Assumed an electron jelly around a nucleus attached with a spring…. Life is more complicated. Quantum mechanics tells us: Other simplification that local E = Applied E, does not change the physics, but only resonant frequency.

  25. Ionic Polarization

  26. In DC limit  = 0 and k”= 0 then: Nion = number of ion pairs/m3

  27. What determines kion What is r0???

  28. ro

  29. - + P ABO3 Dipolar Polarization Teams teams… teams ..

  30. Dipolar Polarization Dipole moment = q

  31. In English: What determines k’dip ?? charge on the dipole is a big one density of dipoles is another jump distance… another big one and finally, T ……. and here comes Mr. Entropy!

  32. Debye Model • At high frequency, • As freq goes to 0,

  33. Debye Equations For DC what is k’dip?? How about at very high  For DC what is k’’dip?? How about at very high  For DC what is ?? How about at very high 

  34. Relaxation

  35. Temperature Dependence

  36. Total polarization P = Pe + Pi + Po With increasing frequency you tend to lose the various mechanisms in order shown.

  37. Dielectric Loss

  38. Slight digression:What is Pv for DC conditions?Does anybody recognize theexpression?

  39. Dielectric Breakdown Intrinsic: That’s when the electrons go ballistic or postal. :-) Thermal Breakdown: Lossy dielectric, leads to T increase - leads to more current - leads to more Heat - lead to more current leads to death of capacitor.

  40. Worked Example Note # of ion pairs is also equal to # of each ion individually Consider CsCl: Lattice parameter = 0.412 nm Cs+: e = 3.35x10-40 Fm2 Cl-: e = 3.4x10-40 Fm2 Mean ionic polarizability per ion pair = 6x10-40 Fm2 Estimate the dielectric constant of CsCl at low and optical frequencies. _______________________________________________________ If you solve for k’ you get 7.56.

  41. Insulators and Capacitors • For capacitor functions: • k’ should be maximum of minimum?? • k” should be low or high?? • For insulator functions: • k’ should be maximum of minimum?? • k” should be low or high??

  42. How about at optical frequencies? Solving for k’e gives: 2.71. Experimental values are: 7.2 and 2.62, respectively. Moral of the story: If you want to use CsCl as a window what is k’?? How about if you want to use it as a capacitor, then what?

  43. MP of compound Effect of dipolar polarization on k’.

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