Download Steady Electric Current

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.
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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
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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.
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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
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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 dd
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
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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”.
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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
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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
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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
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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).
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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.
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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
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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  2t d
dG



    
d
 2
b 2t d
2t  b 
G

ln   S
a 

 a

ANS
R

b
2t ln  
a
Parallel connection (integration by dG)
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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


2Vo 
J  E  
a A/m 2
ANS

a
b
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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
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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 dvE  v  E  v vdv  E  Jdv
• Integrate for total power: Joule’s law
EE301
 
P   E  Jdv [W]
volume
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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,
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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.”
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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  110 9.
Assume
1 nm
Theorem: Method of image
Conductor
z = 0 plane
V
1C
Q
40 R
1 nm
Conductor
z = 0 plane
1 nm
-1C
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Ex. (6): Method of Image (2)
R1C
1C
RR1C R-1C
2 nm

R-1C
-1C
R-1C – R1C
V
40 R
Position to find V is very far away. R1C || R-1C
Vtotal  V1C  V1C
1
1
1  1
1 







40 R1C 40 R1C 40  R1C R1C 
1  R1C  R1C 
1  2 109 cos  

 



2
40  R1C R1C  40 
R


1  2 109 z / x 2  y 2  z 2

40 
x2  y 2  z 2
 9
10 z

V
2
2
2 1.5
20 x  y  z

EE301
Q




ANS
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