ECE 3318 Applied Electricity and Magnetism. Spring Prof. David R. Jackson Dept. Of ECE. Notes 20 Dielectrics

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1 ECE 3318 Applied Electicity and Magnetism Sping 218 Pof. David R. Jackson Dept. Of ECE Notes 2 Dielectics 1

2 Dielectics Single H 2 O molecule: H H Wate ε= εε O 2

3 Dielectics (cont.) H H Wate ε= εε O Vecto dipole moment of molecule p: Dipole model: p = p d p = d 3

4 Dielectics (cont.) The dipoles epesenting the wate molecules ae nomally pointing in andom diections. Note: The molecules ae not floating in wate they make up the wate. 4

5 Dielectics (cont.) The dipoles patially align unde an applied electic field. E x x V 5

6 Dielectics (cont.) Define: P = total dipole moment/volume 1 P V V p i N molecules (dipoles) inside V V We can also wite this as 1 ave i N V N P= p = N p V ( )( ) v p ave = aveage vecto dipole moment Define the electic flux density vecto D: N v = nume of molecules pe unit volume N = N / V v D ε E P 6

7 Dielectics (cont.) P = εχe Linea mateial: e Define: D= εe εχe e ( 1 ) χe = ε ε 1 χ e E Note: χ e > fo most mateials The tem χ e is called the electic susceptiility. Note: A negative value of χ e would mean that the dipoles align against the field. Then we have D = εε E o D = ε E whee ε= εε 7

8 Typical Linea Mateials Teflon Wate Styofoam Quatz ε ε ε ε = 2.2 = 81 = 1.3 = 5 (a vey pola molecule, faily fee to otate) Note: ε > 1 fo most mateials: ε 1 χ, χ > e e 8

9 Dielectics: Bound Chage Assume (hypothetically) that P x inceases as x inceases inside the mateial: v < Note: Fo simplicity, the dipoles ae shown pefectly aligned, with the nume of aligned dipoles inceasing with x. A net volume chage density is ceated inside the mateial, called the ound chage density o polaization chage density. v = ound chage density : it is ound to the molecules. v = fee chage density : this is chage that you place inside the mateial. You can feely place it wheeve you want. 9

10 Dielectics: Bound Chage (cont.) Bound chage densities s < Bound suface chage density s > v < Note: The total chage is zeo since the oject is assumed to e neutal. Total chage density inside the mateial: = total v v v This total chage density may e viewed as eing in fee space (since thee is no mateial left afte the molecules ae emoved). 1

11 Dielectics: Bound Chage (cont.) Fomulas fo the two types of ound chage ( ) P= εχe = ε ε E e 1 E = electic field inside mateial = P nˆ s ε ( xyz,, ) ˆn The unit nomal points outwad fom the dielectic. v = P The deivation of these fomulas is included in the Appendix. 11

12 Dielectics: Gauss s Law Inside a mateial: ε E = = ( ) total v v v This euation is valid, ut we pefe to have only the fee chage density in the euation, since this is what is known. The goal is to calculate (and hopefully eliminate) the oundchage density tem on the ighthand side. 12

13 Dielectics: Gauss s Law (cont.) The feespace fom of Gauss s law is ε E = = = ( ) total v v v v P o ε E P = ( ) v Hence D = v This is the usual Gauss s law: This is why the definition of D is so convenient! 13

14 Dielectics: Gauss s Law (Summay) D = v D = ε E S D nˆ ds = Q encl ε= εε Impotant conclusion: Gauss law woks the same way inside a dielectic as it does in vacuum, with only the fee chage density (i.e., the chage that is actually placed inside the mateial) eing used on the ighthand side. 14

15 Example Point chage inside dielectic shell Find D, E a S Dielectic spheical shell with ε Gauss law: ˆ encl S D n ds = Q = (The point chage is the only fee chage in the polem.) 15

16 Example (cont.) a S Hence D ( 2 4 ) D π = ˆ 4π C/m 2 = 2 We then have E = ˆ 2 4πε [ V/m ] < a, > E = ˆ 2 4πε ε [ V/m] a< < 16

17 Example (cont.) Flux Plot ε Note that thee ae less flux lines inside the dielectic egion (assuming that the flux lines epesent the electic field). 17

18 Homogeneous Dielectics v = P = ( εχe) = εχ e εε 1 = εχ e D εε ε 1 = v ε e e = εχ E D Pactical case: Homogeneous dielectic No fee chage density inside. xyz,, ( ) ε v = ε = constant Hence v = Note: All of the inteio ound chages cancel fom one ow of molecules to the next. 18

19 Homogeneous Dielectics (cont.) Pactical case (homogeneous) v = Blownup view of molecules ε ε ( xyz,, ) = constant v = 19

20 Summay of Dielectic Pinciples Impotant points: A dielectic can always e modeled with a ound volume chage density and a ound suface chage density. A homogeneous dielectic with no fee chage inside can always e modeled with only a ound suface chage density. All of the ound chage is automatically accounted fo when using the D vecto and the total pemittivity ε. When using the D vecto, the polem is then solved using only the (known) feechage density on the ighthand side of Gauss law. 2

21 Example A point chage suounded y a homogeneous dielectic shell. Find the ound suface chages densities and veify that we get the coect answe fo the electic field y using them in fee space. Dielectic shell Fom Gauss s law: ε a E ˆ = 2 4πε E ˆ = 2 4πε ε 21

22 Example (cont.) Dielectic shell ε s v = = v The alignment of the dipoles causes two layes of ound suface chage density. 22

23 Example (cont.) s = P nˆ ( εχ ee) nˆ ε ( ε 1) = ( E) = nˆ D = ε ( ε 1) ˆ n εε ε 1 = D nˆ ε Note: We use the electic field inside the mateial, at the oundaies. Hence ˆn Feespace model ˆn s D ˆ 4π = 2 a s ε 1 = 2 ε 4πa s ε 1 = 2 ε 4π 23

24 Feespace model 24 1 a ε ε = 1 ε ε = 2 ˆ : 4 a E πε < = : a encl a Q < < = 2 ˆ : 4 a E πε ε < < = Fom Gauss s law: Example (cont.) 1 a ε ε ε = = a = Note : 2 ˆ 4 Q encl E πε = 2 4 a a s a π = 2 4 s π = whee a S a < <

25 Feespace model 25 1 a ε ε = 1 ε ε = : a encl Q > = 2 ˆ : 4 E πε > = Fom Gauss s law: Example (cont.) a = a = Note : whee a S >

26 Osevations Fom the pevious example, we have the following osevations: We get the coect answe y accounting fo the ound chages, and putting them in fee space. Howeve, it is much easie to not woy aout ound chages, and simply use the concept of ε and the fee chage. S D nˆ ds = Q encl Fee chage enclosed D = εe = εε E 26

27 Exotic Mateials Plasmas Plasmas have a elative pemittivity that is less than one (and can even e negative) Lossless plasma: 1 ωp ε = ε ω 2 (deived in ECE 634) ω p = plasma esonance feuency Lossy plasma: 2 1 ωp ε = ε ωω ( jυ) υ = plasma collision feuency (loss tem) 27

28 Exotic Mateials (cont.) Plasmas Low feuencies (f < 3 MHz) will eflect off the ionosphee. The ionosphee has a elative pemittivity that is less than one, so the waves will end (efact) away fom the nomal and tavel ack down to the eath, ouncing off of the eath. Shotwave adio signals popagate aound the eath y skipping off the ionosphee. 28

29 Exotic Mateials (cont.) Atificial metamateials that have een designed that have exotic pemittivity and/o pemeaility pefomance. Negative index metamateial aay configuation, which was constucted of coppe spliting esonatos and wies mounted on intelocking sheets of fieglass cicuit oad. The total aay consists of 3 y 2 2 unit cells with oveall dimensions of mm. ε < µ < (ove a cetain andwidth of opeation) 29

30 Exotic Mateials (cont.) The Duke cloaking device masks an oject fom one wavelength at micowaves. Image fom D. David R. Smith. Cloaking of ojects is one aea of eseach in metamateials. 3

31 Appendix In this appendix we deive the fomulas fo the ound change densities: v = P = P nˆ s whee nˆ = n = outwad nomal to the dielectic ounday ˆ 31

32 Appendix (cont.) Model: Dipoles ae aligned in the x diection endtoend. The nume of dipoles that ae aligned (pe unit volume) changes with x. V Q N v = # of aligned dipoles pe unit volume Osevation point 1 v = V Q x d /2 d x x d /2 32

33 Appendix (cont.) 1 v = Q V Q = N V N V v d d x v x 2 2 V Q Hence = N v = v d v d N N x x 2 2 v x d /2 d x x d /2 N v = # of aligned dipoles pe unit volume 33

34 Appendix (cont.) v = N v P x = x pn P = p N v v V Q Hence v 1 = P p x 1 = Px d 1 = Px d = Px x o v = dp x dx x d /2 d x N v = # of aligned dipoles pe unit volume x d /2 34

35 Appendix (cont.) Fo dipole aligned in the x diection, v = dp x dx In geneal, dp dp dp dx dy dz x y v = z o v = P 35

36 Appendix (cont.) Afte applying the divegence theoem, we have the integal fom P nˆ = S Q encl Applying this to a shallow pillox suface at a dielectic ounday, we have ( ) ˆ diel ai encl s P P n S = Q = S Denoting P = P diel nˆ S S ε we have P n = ˆ s 36