1
LOGO
Chapter 5
Electric Field in
Material Space
Part 1
iugaza2010.blogspot.com
Materials:
(1)Conductor σ>>1
(2) Insulator (or dielectric) σ<<1
(3) Semiconductor
Conductivity(σ)
the measure of the ability of material
to conduct current (mho/m) (S/m)
,depends on: temperature and frequency.
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dq
Current : I 
dt
I
2
Current density : J 
(A/m )
S
and I   J.dS
Perfect Conductor(  )
R  0 , E  0 ,  v  0(inside)
Perfect Insulator(  0)
R ,J 0
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Convection Current
doesn' t apply ohm' s law
occurs when current flow in insulating medium
J   vu
u : change velocity
Conduction Current
occurs when current flow in conducting medium
J  E
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A conductor of uniform cross section under
an applied E field
V 
 E.dl
V
E
l
I
V
J 
 E 
S
l
I
V

S
l
V
l

 R
I
 S
R
l
S

c l
S
1
where  c  is the resestivity

I
J
S
6
For non uniform cross section
V
R 

I
 E.dl
 E.dS
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For current density J=10zsin2Ф aρ A/m2
Find the current through the cylindrical surface:
ρ=2 ,1≤ z ≤ 5m
I   J .dS , dS  ddz a 
5 2
I
  10z sin
2
ddz 
z 1  0
5 2
  20z sin
2
ddz
z 1  0
I  240  754 A
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For convective charge transport
A=0.1m2, u=10 m/s , R=1014 Ω , ρs=0.5 μc/m2
Find the potential difference.
V
J   s u  I   s uA    s uA
R
6
14
V   s uAR  (0.5 *10 )(10)(0.1)(10 )  50MV
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The free charge density in copper is 1.81*1010 c/m3
for current density 8*106 A/m2 ,
σ=5.8*107 S/m
Find the electric field intensity and drift velocity .
 v  1.8110 c / m
10
J  8 10 A / m
6
J  v u  u 
3
2
J
4
 4.42 10 m / s
v
J
J   E  E   0.138N / c

10
A lead(σ=5*106 S/m) bar of square cross section
has a hole bored a long its length of 4m and filled
with copper (σ=5.8*106 S/m)
Determine the resistance of the composite bar.
R Lead 
l
 L AL
R copper 
l
 c Ac


4
2

 0.01  
6
2
(5 10 ) (0.03)   
 
 2  

 9.74 104 
4
3

8
.
7

10

2
  0.01  
6
(5.8  10 )  
 
  2  
Rtot  R Lead || R copper 
R Lead R copper
R Lead  R copper
 8.759104 
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Determine the total current in a wire of radius
1.6mm if J=500/ρ az A/m2 placed along Z-axis.
I   J .dS , dS  dd a z
2 1.6 mm
I 




0
0
500

dd  5.026 A
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The charge 10-4 e-3t C is removed from a sphere
through a wire. Find the current in the wire at t=0
and t=2.5s .
dq
 4  3t
I
 (3)10 e
dt
4
I (t  0)  3  10 A
4
I (t  1)  3  10 e
 3 ( 2 .5 )
9
 166 10 A
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R=1MΩ , l=2m , r=4mm (cylinder)
Determine the conductivity.
R
l
 A

l
0.02
4


 3.97810 S / m
6
2
R A (10 )(  0.004 )
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Cylindrical bar of Carbon (σ=3*104 S/m) of radius
5mm and length 8cm are maintained at potential
difference of 9V , find
(a) The resistance of the bar.
(b) The current through the bar.
(c) The power dissipated in the bar.
2
8 10
3
(a) R 

 34 10 S / m
4
2
 A (3 10 )(  0.005 )
V
9
(b) I  
 265A
3
R 34 10
2
2
3
(c) P  I R  (265) (34 10 )  2387.7W
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l
The resistance of a long wire of diameter 3mm is
4.04 Ω/km if I=40A
find (a)σ (b)J
l
1000
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(a)  

 3.7 10 S / m
2
R A (4.04)(  0.0015 )
I
40
6
2
(b) J  
 5.65 10 A / m
2
S   0.0015
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A coil is made of 150 turns of copper wound on a
cylindrical core. If the mean radius of the turn is
6.5mm and the diameter of the wire is 0.4mm
calculate the resistance of the coil.
L  2 (6.5 103 )(150)  6.1261m
L
6.1261
R

 0.84
2
 A
  0.4 103  
 
(5.8 107 )   
2
 
 
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A hollow cylindrical of length 2m has its cross
section as in fig. If the cylinder is made of carbon
(σ=3*104 S/m) .
Determine the resistivity between the ends of the
cylinder.

A   5 10

2 2

  3 10

2 2
 5.02 10 2 m 2
L
2
R

 1.328
4
2
 A (3 10 )(5.02 10 )
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A composite conductor 10m long consist of an
inner core of steel of radius 1.5cm and an outer
sheath of copper whose thickness is 0.5cm
(σcopper=5.8*107 S/m) , (σsteel=8.4*106 S/m) .
(a) Determine the resistance of the conductor.
L1
10
R steel 

 1 A1 (8.4 106 )   1.5 10 2



Cop.
2
St.
 1.67 103 
L2
10
R copper 

 2 A2 (5.8 107 )   2 102 2    1.5 102





2
 3.136104 
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Rtot  R steel || R copper 
R steel R copper
R steel  R copper
 2.64 104 
(b) If the total current in the conductor is 60A what
current flows in each metal?
60( Rs )
Ic 
(current divider)
Rc  Rs
(60)(1.67  103 )

 50.51A
3
4
(1.67  10 )  (3.136  10 )
Is  60  50.51  9.48 A
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LOGO
iugaza2010.blogspot.com
melasmer@gmail.com
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