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

2. 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

3. 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

4. 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

5. 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

6. 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

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

8. 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

9. 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

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

11. 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

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

13. 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

14. 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

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

16. 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

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

18. Ex. (4): Resistance Calculationy 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

19. Ex. (5): Resistance Calculationy 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

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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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