Chapter%205 - Chapter 5 Conductors, Dielectrics, and...

Info iconThis preview shows pages 1–13. Sign up to view the full content.

View Full Document Right Arrow Icon
Dr. Ray Chen© Chapter 5 Conductors, Dielectrics, and Capacitance
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Dr. Ray Chen© 5.1 Current and Current density Fig. 5.1 S J I , S J I , dt dQ I N v v = = = = S S d J I v v , t x S I V = ρ
Background image of page 2
Dr. Ray Chen© 5.1 Current and Current density Fig. 5.1 , x V x v J ρ = , Sv I x V = density. current the is which v J V r = r
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Dr. Ray Chen© 5.2 Continuity of current Current through an enclosed surface = S d J I v v dt dQ S d J I i = = v v Q i : total charge inside the closed surface . From divergence theorem: ∫∫ = S Vol dv ) J ( S d J v v v = vol v vol dv dt d dv ) J ( ρ v = vol v vol dv t dv ) J ( v v t v ) J ( v = v so it is true for an incremental volume continuity of Equation t J v = v
Background image of page 4
Dr. Ray Chen© 5.2 Continuity of current Example 1: theorem divergence and integral surface using 1 z y, x, 0 cube the leaving current total the Find b) 1 z 0, z 1, y 0, y by bounded and 1 x surface the crossing direction a the in current Find a) V 10e V A/m 10 x -x 2 5 < < = = = = = = = ˆ y sin , V J v
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Dr. Ray Chen© 5.2 Continuity of current Solution to Example 1: kA 170 10 10 I 1 z 0 1, y 0 1, x 1 0 1 0 1 6 1 0 1 6 = = = < < < < = ∫∫ ) y cos ( e dydz y sin e ) a x y z (1,0,0) (1,1,1) () y x x x y x x x z y x a ˆ y cos e a ˆ y sin e J a ˆ y cos e a ˆ y sin e a ˆ z V a ˆ y V a ˆ x V V 10 10 10 10 10 5 + = + = + + = v
Background image of page 6
Dr. Ray Chen© 5.2 Continuity of current Solution to Example 1(cont.): () 0 I 1, z 0; I 0, z kA 342 1 1 10 1 10 1 10 J 1, y kA 632 1 10 10 I 0, y kA 460 1 1 10 10 10 I 0, x b) 1 0 1 6 1 0 6 6 1 6 1 0 1 0 6 y 6 1 0 1 0 5 = = = = = = = = = = = = = = = = = ∫∫ ) e ( cos dxdz e cos I cos e ) e ( dz dx e ) cos ( dz ydy sin x x x 0 342 632 460 170 I total = + = 0 0 iny 10 10 have we theory divergence Using = = = vol vol x x vol dv dv s e y sin e dv J v x y z (1,0,0) (1,1,1)
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Dr. Ray Chen© 5.3 Metallic Conductors Fig 5.2
Background image of page 8