Notes 24 Image Theory

Transcription

1 ECE 3318 Applied Electricity and Magnetism Spring 218 Prof. David R. Jackson Dept. of ECE Notes 24 Image Teory 1

2 Uniqueness Teorem S ρ v ( yz),, ( given) Given: Φ=ΦB 2 ρv Φ= ε Φ=Φ B on boundary Inside region (known carge density) on boundary (known B.C.) Φ(, y, z) is unique Note: We can guess te solution, as long as we verify tat te Poisson equation and te B.C. s are correctly satisfied! 2

3 Image Teory z Point carge q Φ ( yz,, ) Infinite PEC ground plane Φ= Note: Te electric field is zero below te ground plane (z < ). Tis can be justified by te uniqueness teorem, just as we did for te Faraday cage discussion (make a closed surface by adding a large emispere in te lower region). 3

4 Image Teory (cont.) Image picture: z Note: Tere is no ground plane in te image picture! q Original carge q Image carge Te original carge and te image carge togeter give te correct electric field in te region z >. Tey do NOT give te correct solution in te region z <. (A proof tat tis is a valid solution is given after te net slide.) 4

5 z > Image Teory (cont.) z R 1 r Here is wat te image solution looks like. #1 q R 2 Φ=Φ 1+Φ2 #2 q q q Φ= + 4πε R 4πε R 1 2 5

6 Image Teory (cont.) To see wy image teory works, we construct a closed surface S tat as a large emisperical cap (te radius goes to infinity). S = S + S z S Original problem q Φ ( yz,, ) S PEC Φ= On S : Φ = (te surface is on a perfect electric conductor at zero volts). On S : Φ = (te surface is at infinity were E =, and Φ = on te bottom). Φ= on S 6

7 Image Teory (cont.) Image picture: S = S + S #1 q S V R 1 S R 2 #2 q q q Φ= + 4πε R 4πε R 1 2 correct carge in V (carge #1) r S : Φ= since R 1 = R 2 r S : Φ= since R 1 = R 2 = (correct B.C. on S) 7

8 Image Teory (cont.) z < #1 q V #2 q S Wrong carge inside! (Te original problem does not ave a carge in region #2.) Image teory does not give te correct result in te lower region (te region were te image carge is). 8

9 Final Solution for z > z > : q Φ= + 4πε 4 q ( ) πε ( ) y + z + y + z+ z R 1 r #1 q R 2 #2 q 9

10 Final Solution (All Regions) z > : q Φ= + 4πε 4 q ( ) πε ( ) y + z + y + z+ z < : Φ= Point carge z q Φ ( yz,, ) Infinite PEC ground plane Φ= 1

11 Hig-Voltage Power Line y V Hig-voltage power line (radius a) Infinite in z Eart (ε, σ) [ V] R Te eart is modeled as a perfect conductor (E = ). Note: Tis is an eact model in statics, since tere will be no current flow in te eart. J = σ E 11

12 Hig-Voltage Power Line (cont.) Te loss tangent of a material sows ow good of a conductor it is at any frequency. (Tis is discussed in ECE 3317.) ω = 2π f ε= εε r 12 ε = [F/m] tanδ σ ωε Table sowing tanδ for eart 1 [Hz] [Hz] [khz] tanδ >> 1 ( good conductor) 1 [khz] tanδ << 1 ( good dielectric) 1 [khz] [MHz] 225 Assume te following parameters for eart: 1 [MHz] 22.5 σ =.1 [S/m] ε = 8 r 1 [MHz] [GHz] [GHz]

13 Hig-Voltage Power Line (cont.) Line-carge approimation: Note: ρ = 2πaρ l s ρ l We need to find te correct value of ρ l. Eart R Te power line is modeled as an effective line carge at te center of te line. Tis is valid (from Gauss s law) as long as te carge density on te surface of te wire is approimately uniform. Tis gives te correct field outside te wire. 13

14 Hig-Voltage Power Line (cont.) Image teory: R 1 r ρ l R 2 ρ l b ρ l b Φ= ln + ln R1 R2 or ρ R l 2 Φ= ln R1 ρ l Line carge formula: ρ l ln b Φ= ρ Note: In te formula b is te distance to te arbitrary reference point for te single line-carge solution. Te line-carge density ρ l can be found by forcing te voltage to be V at te surface of te wire (wit respect to te eart). 14

15 Hig-Voltage Power Line (cont.) Wire (radius a) ρ l Set VAB = V A B ρ l Te point A is selected as te point on te bottom surface of te power line. 15

16 Hig-Voltage Power Line (cont.) Original wire (radius a) #1 ρ l A Set VAB = V B ρl 2 a Φ ( A) = ln a ρl Φ ( B) = ln = #2 ρ l l 2 Φ= ln R1 AB ρ ρ ( ) ( ) V = V =Φ A Φ B l = ln R 2 a a 16

17 Hig-Voltage Power Line (cont.) ρ ρ Hence ρ = V 2 a ln a ρ V 2 ln a 17

18 Hig-Voltage Power Line (cont.) y ρ l Alternative calculation of ρ l AB B V = V = E dr = E dy V A Hence ρ a l = + a y 1 1 y + y A B Line carge formula: ρ l ρ l ρ l E = yˆ yˆ + 2 dy ρ l E = ˆ ρ ρ Along te vertical line we ave: ( y) πε ( + y) From top carge From bottom carge 18

19 Hig-Voltage Power Line (cont.) y Performing te integration, we ave: ρ l V ρ l = + ρ ( ln ( y) ln ( y) ) l = + + ρ ( ln ( ) ln ( a) ln ( ) ln ( 2 a) ) l = + + ρ ( ln ( 2 a) ln ( a) ) l = ρ l = a ln 1 1 y + y 2 a a dy a V Hence ρl 2 a = ln a A B ρ l 19

20 Hig-Voltage Power Line (cont.) y Final solution r = ( y,,) ρ R 1 R 2 y > ρ ρ R 2 Φ ( y, ) = ln R1 ρ = V 2 a ln a ( ) ( ) = + 2 = + + R y R y 2

21 Hig-Voltage Power Line (cont.) Electric Field y R 1 ρ r= ( y,,) Line carge formula: ρ l E = ˆ ρ ρ R y > 2 ρ E ( y ρ, ) ˆ ˆ 2 R ρ = + πε 2 R 1 2 R1 πεr2 V ρ = 2 a ln a ( ) ( ) ( ) ( ) R = ˆ + yˆ y R = ˆ + yˆ y+ 1 2 ( ) ( ) = + 2 = + + R y R y 21

22 Hig-Voltage Power Line (cont.) Electric Field on Eart y ρ R 1 (,) y = R R1 = R2 = R = + ρ ρ ρ ρ E R R R R R R R ( ) ˆ ˆ, = ( ˆ ˆ ) 1+ 2 = ( ) ( ) ( ) ( ) R = ˆ + yˆ R = ˆ + yˆ Rˆ 1 Rˆ ˆ 2 = R1 R2 = y 2 R R ( ) ( ) ( ) 22

23 Hig-Voltage Power Line (cont.) Hence, we ave on te surface on te Eart: ρ πε R (,) = yˆ E 2 y ρ = V 2 a ln a ρ y = ρ R R (, ) 2 2 R = + 23

24 Hig-Voltage Power Line (cont.) Final result: (,) E 1 2V 1 ˆ 2 a ln 1 + ( / ) a = y 2 y V (,) Eart 24

25 Hig-Voltage Power Line (cont.) Eample Find te electric field at te surface of te eart. Find te electric field at te bottom surface of te line. Parameters: V = 23 [kv] = 5 [m] a = 1 [cm] In MKS units: V = [V] = 5 [m] a =.1 [m] 25

26 Hig-Voltage Power Line (cont.) Part (a) At te surface of te Eart we ave (,) E 1 2V 1 ˆ 2 a ln 1 + ( / ) a = y 2 Tis gives us : 1 E(, ) = yˆ ( ) V/m ( / ) [ ] 26

27 Hig-Voltage Power Line (cont.) Part (b) Original wire (radius a) y Along te vertical line we ave: ρ l ρ l ρ l E(, y) = yˆ yˆ + 2 ( y) πε ( + y) A B ρ l Hence, at te bottom of te line (y = -a) we ave: bot ρ l ρ l ρ l E = E(, a) = yˆ + yˆ yˆ a ( 2 a) a 27

28 Hig-Voltage Power Line (cont.) E so tat E bot bot Tis gives us: bot E ρ l yˆ a ˆ 1 a ln y 2 6 ( )[ V/m] = yˆ V a a Original wire (radius a) Recall: V ρ = 2 a ln a y ρ l A B ρ l Recall: E c = [V/m] 28

29 Image Teory for Dielectric Region Point carge over a semi-infinite dielectric region (Please see te tetbook for a derivation.) y q Note: We are now interested in bot regions (z > and z < ). ε r Note: A line carge could also be done te same way. 29

30 Image Teory for Dielectric Region (cont.) Solution for air region (z > ): q Observation point ε q ε r 1 q q ε = ε r + 1 Note: A very large permittivity acts like a PEC (q' = - q). 3

31 Image Teory for Dielectric Region (cont.) Solution for dielectric region (z < ): q εε r Observation point εε r 2ε r q = q ε r

32 Image Teory for Dielectric Region (cont.) Sketc of electric field Note: Te flu lines are straigt lines in te dielectric region. q ε εε r Te flu lines are bent away from te normal (as proven earlier from B.C.s). 32