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Steady Electric Current • Current and current density • Continuity Equation & KCL • Ohm’s Law • Method of Image EE301 1 Objectives • To know the convection and conduction current. • In a conductor, to be able to find the electric field intensity from the current density and vice versa. • To be able to find the resistance, the current density and the electric field intensity of the conductor. • To be able to apply the method of image to find the electric field of the charge system with a very large conductor plate. EE301 2 Current dQ I dt • Definition: rate of the charge changing in time • Cause: – Change of bound charge density displacement current – Movement of free charge • Movement of free charge: – Convection current: free charge in vacuum – Conduction current: charge under E V – Electrolytic current: charge from ion EE301 IR 3 Current Density • Consider the current per area. Current density J. • Direction is the direction of the current. Charge moving dS to measure the current I J dS surface EE301 4 Convection Current Density • Charge moving inside an object at v m/s. Amount of charge flowing through the area in1s + + + + + + + + + + + + Amount of charge flowing in 1s = charge inside the red cylinder = vvdS C I JdS v vdS Thus, J =vvA/m2. J v v Comment: Not necessary to follow Ohm’s law. EE301 5 Ex(1): Current Calculation Given a system having v = –0.2 nC/mm3 and J = 2az A/mm2, find (1) the current flowing through the hemisphere with the radius of 5mm, 0< < /2, 0 < <2. z (2) charge velocity Theory: I J dS J v v surface EE301 6 Ex.(1): Current Calculation (2) (1) Find current I. I J dS (2a z ) (ar dS) S 2 2 0 0 S (2 cos )r sin dd 2 2 2(2 )(5 ) 2 0 50 A (2) Find v. a z ar cos J v v v ANS J v 2a z 0.2 10 9 10 a z 10 EE301 2 sin sin d (sin ) 100 2 0 2 C mm 3 2 s mm C mm ( ) ANS 7 s Continuity Equation • Conservation of charge: charge cannot be created nor destroyed.rate of changing charge is due to the charge being taken out by current. q I I out _ net t v dv J dS volume t surface v dv Jdv volume t volume v J t EE301 8 Continuity Equation(2) • From I out _ net v dv J dS volume t surface • This equation is true only when the current can flow continuously through the closed surface (otherwise, the charge will be depleted.) • Thus, the name “continuity equation”. EE301 9 Continuity Equation & KCL • Steady current case: no charge stored/depleted from the system. • Thus, v J 0 J dS 0 t surface I 0 I2 • Iin = Iout I1 EE301 I3 10 Conduction Current • From moving free electron • Follow Ohm’s law: J E : conductivity [Siemen/m, S/m] • Current and the current density still related in the same manner as before. I J dS surface EE301 11 Ohm’s Law V E dL E L Circuit: V = IR • • Point form: J E EL V R I ES I J dS E S surface L R S S EE301 L 12 Resistance Calculation • Check whether L, and S are the same for the entire resistance or not. – If yes, use – If no, R L S • Find the thin surface (along one of the three axes) that will provide the constant L, and S. Lthin • Calculate R of the thin surface by dR = Rthin = thinS thin • Integration for the entire object (may need to convert to dG = dR). EE301 13 Calculation for E, J • Consider how the resistance are connected. – Serial connection (integration using R dR ) Constant current: J I a current S – Parallel connection (integration using G dG ) V Constant voltage: E a L current Sometime, it is the same value as the one of the thin R’s. EE301 14 Ex. (2): Resistance Calculation t R=?, J = ? y Theorem: R L S Cut the resistor into small section. dR x a d b R b a d t / 2 2 d 2 b ln t t a ANS I EE301 15 Ex. (3): Resistance Calculation y R=? t x a b EE301 Theorem: R L S Cut the resistor into small section. 2 dR td td 2t d dG d 2 b 2t d 2t b G ln S a a ANS R b 2t ln a Parallel connection (integration by dG) 16 Ex. (3): Resistance Calculation (2) y t Parallel connection (integration by dG) x V E a current V0 (a ) 2V0 a V/m L 2 2Vo J E a A/m 2 ANS a b EE301 17 Ex. (4): Resistance Calculation y b R, I, J=? a = be-x/b x z t I Theorem: L S V IR R V E dL, J E Cut the resistor into small section with constant . dR dx be x / b at b 1 1 R e x / b dx 1 e 1 0 bat at V0 ANS V0 at V IR I A ANS 1 dx 1 e V0 2 Uniform current distribution: I JA J ANS a A/m x 1 18 1 e EE301 Ex. (5): Resistance Calculation y b R, I, J=? a = 0e-z/t x z Theorem: L S V IR I J dS , V E dL R Cut the resistor into small section with constant . t I dR dG V0 0 e z / t adz 0e z / t adz b a 0 z / t e dz dz t / 2 b a 0 0.5 0.5 t e e S b b R ANS 19 0.5 0.5 a 0t e e G t/2 EE301 b Ex. (5): Resistance Calculation (2) R a 0t e y b 0.5 e 0.5 ANS Parallel connection: (Integration by dG) b a = 0e-z/t x z V0 a 0t e0.5 e 0.5 V IR I A b V V0 E a current a x V/m L b 0e z / tVo J E a x A/m 2 b ANS t I V0 EE301 20 Power and Joule’s Law • Definition: the change in energy per time unit. • Consider the power in moving a q-C charge in E. w qE L p lim lim qE v t 0 t t 0 t • Assume that the charge is moving at v m/s in a small volume, dv. The volume charge density is v C/m3 dp v dvE v E v vdv E Jdv • Integrate for total power: Joule’s law EE301 P E Jdv [W] volume 21 Power Dissipation in Conductor • Consider power loss in a conductor. Assume constant E and J for simplicity. Pconductor E Jdv E JdSdL volume line surface EdL JdS VI line surface ( IR) I I 2 R • Power loss in R as in circuit theory, EE301 22 Method of Image • Case: conductor in a system. – Conductor is an equipotential surface. – Not conform with Coulomb’s law!! • Method: Add the image charge to the opposite side. “The electric property above the conducting plane behaves as if there were the opposite changes configuring in the other side of the conducting plane.” EE301 23 Method of Image: Illustration Q1 Guarantee to have same E, V above the conductor. Q1 d1 d1 Conductor Conductor Not so below the conductor d1 -Q1 point charge + conductor EE301 point charge + image charge (no conductor) 24 Ex. (6): Method of Image 1C (0,0,1nm) Find V0 at (x, y, z); z > 0. x 2 y 2 z 2 110 9. Assume 1 nm Theorem: Method of image Conductor z = 0 plane V 1C Q 40 R 1 nm Conductor z = 0 plane 1 nm -1C EE301 25 Ex. (6): Method of Image (2) R1C 1C RR1C R-1C 2 nm R-1C -1C R-1C – R1C V 40 R Position to find V is very far away. R1C || R-1C Vtotal V1C V1C 1 1 1 1 1 40 R1C 40 R1C 40 R1C R1C 1 R1C R1C 1 2 109 cos 2 40 R1C R1C 40 R 1 2 109 z / x 2 y 2 z 2 40 x2 y 2 z 2 9 10 z V 2 2 2 1.5 20 x y z EE301 Q ANS 26