Steady Electric Current Current and current density Continuity Equation & KCL Ohm’s Law Method of Image EE301

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

Current 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 Electrolytic current: charge from ion EE301

Current Density Consider the current per area.  Current density J. Direction is the direction of the current. dS to measure the current Charge moving EE301

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 + + + + + + + + Thus, J =vvA/m2. + + + + Comment: Not necessary to follow Ohm’s law. EE301

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. (2) charge velocity z Theory: EE301

Ex.(1): Current Calculation (2) (1) Find current I. ANS (2) Find v. ANS EE301

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. I EE301

Continuity Equation(2) From 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

Continuity Equation & KCL Steady current case: no charge stored/depleted from the system. Thus, Iin = Iout I1 I2 I3 EE301

Conduction Current From moving free electron Follow Ohm’s law:  : conductivity [Siemen/m, S/m] Current and the current density still related in the same manner as before. EE301

Ohm’s Law Circuit: V = IR Point form: L S s EE301

Resistance Calculation Check whether L,  and S are the same for the entire resistance or not. If yes, use If no, Find the thin surface (along one of the three axes) that will provide the constant L,  and S. Calculate R of the thin surface by dR = Rthin = Integration for the entire object (may need to convert to dG = dR). EE301

Calculation for E, J Consider how the resistance are connected. Serial connection (integration using ) Constant current: Parallel connection (integration using ) Constant voltage: Sometime, it is the same value as the one of the thin R’s. EE301

Ex. (2): Resistance Calculation Theorem: y R=?, J = ? t Cut the resistor into small section.   d x a b I ANS EE301

Ex. (3): Resistance Calculation Theorem: y R=? t Cut the resistor into small section.   d x a b ANS Parallel connection (integration by dG) EE301

Ex. (3): Resistance Calculation (2) y t Parallel connection (integration by dG)  x ANS a b EE301

Ex. (4): Resistance Calculation y b Theorem: R, I, J=? a  = be-x/b x Cut the resistor into small section with constant . z t dx I V0 ANS ANS Uniform current distribution: ANS EE301

Ex. (5): Resistance Calculation y b Theorem: R, I, J=? a  = 0e-z/t x Cut the resistor into small section with constant . z t dz I V0 ANS EE301

Ex. (5): Resistance Calculation (2) ANS y Parallel connection: (Integration by dG) b a  = 0e-z/t ANS x z t I V0 EE301

Power and Joule’s Law Definition: the change in energy per time unit. Consider the power in moving a q-C charge in E. Assume that the charge is moving at v m/s in a small volume, dv. The volume charge density is v C/m3 Integrate for total power: Joule’s law EE301

Power Dissipation in Conductor Consider power loss in a conductor. Assume constant E and J for simplicity. Power loss in R as in circuit theory, EE301

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

Method of Image: Illustration Guarantee to have same E, V above the conductor. Q1 Conductor Q1 d1 d1 Conductor d1 Not so below the conductor -Q1 point charge + conductor point charge + image charge (no conductor) EE301

Ex. (6): Method of Image Conductor z = 0 plane Conductor z = 0 plane Find V0 at (x, y, z); z > 0. 1C (0,0,1nm) Assume 1 nm Conductor z = 0 plane Theorem: Method of image Conductor z = 0 plane 1C 1 nm -1C 1 nm EE301

Ex. (6): Method of Image (2) R1C R-1C 1C Position to find V is very far away. R1C || R-1C RR1C R-1C 2 nm  -1C R-1C – R1C ANS EE301