EE330 R
Prof Viviana Vladutescu
Homework 5
1. A semi-infinite slab exists for z > 0 with  = 300 S/m, r = 10.2, and r = 1.0. At the
surface (z = 0),
E(0,t) = 1.0 cos( x 106t) ax V/m.
Find the instantaneous expressions for E and H anywhere in the slab.
The general expression for E is: E( z, t )  1.0e  z cos  x106 t   z  a x
j  j  x106  4 x107   j 3.948
V
m
j  j  x106  10.2  8.854 x1012   j 284 x10 6
Here,  (i.e. it is a good conductor), so
1
   f   24.3  
m

  2 e j 45  0.115e j 45 

So now we have
V
E( z, t )  1.0e 24 z cos  x106 t  24 z  a x
m
To find B we’ll work in phasors.
Es  1e z e j z a x , H s 
1

a P  Es 
1

a z 1e z e j z a x 
1

e z e j z a y
1
A
e 24 z cos  x106 t  24 z  45  a y
0.115
m
A
H( z, t )  8.7e 24 z cos  x106 t  24 z  45  a y
m
2. A 0.1 m layer of copper is deposited atop a very thick slab of nickel. For a field
incident on the copper surface, (a) calculate Rs at 1.0 GHz. Compare this with Rs at 1.0
GHz for (b) a semi-infinite slab of copper and (c) for a 0.1 m thickness of copper by
itself.
H( z, t ) 
Refer to to the figure above
In the copper portion the field is Ex  Exo e Cu z
In the nickel portion, Ex   Exo e Cu t  e  Ni  z t 
The current density in the copper is J xCu   Cu Exo e Cu z , and in the nickel is


J xNi   Ni Exo e Cu t e
 Ni  z t 
. The current is
I    Cu Exo e Cu z dydz    Ni Exo e Cu t e  Ni ( z t ) dydz , or
t

o
t
I  w Cu Exo  e Cu z dz  w Ni Exo e Cu t  e  Ni ( z t ) dz , and upon evaluating



I  wExo  Cu 1  eCut  Ni eCut  , and with V=ExoL,
 Ni
 Cu



1



L
we have  R  Rs , where Rs =  Cu 1  e Cu t  Ni e Cu t  .
w
 Ni
  Cu

Now we’re ready to perform the calculations using the following data:
S
Np
 Cu  5.8 x107 , r  1,  Cu  479 x103
m
m
S
Np
 Ni  1.5 x107 , r  600,  Cu  596 x106
m
m
(a) 0.1m Cu over Ni: Rs = 176 m
(b) Semi-infinite Cu: Rs = 8.3 m
(c) 0.1 m Cu: Rs = 177 m


3. Calculate the DC resistance per meter length of a 4.0 mm diameter copper wire. Now
find the resistance at 1.0 GHz.
DC:
R 1 1
1
1
m


 1.37
2
2
7
L  a
m
 5.8x10    0.002 
R
R
1
 s 
;   1  f   2.09 x106 m
L 2 a  2 a
R
1


 0.66
7
6
L  5.8 x10  2.09 x10  2  0.002 
m
1 GHz:
4. Assume distilled water ( = 10-4 S/m, r = 81, r = 1.0) fills the region z > 0. At the
surface, we have E(0,t) = 8.0cos(2x108t) ax V/m. Determine, for z > 0, (a) E(z,t), (b)
H(z,t), and (c) Pav at z = 1.0 m. (d) Find the power passing through a 10 square meter
surface located at z = 1.0 m.
(a) The general expression for E is: E( z, t )  Eo e  z cos t   z    a x
and we can see from the given information that
V
rad
Eo  8 ,   2 x108
, f  108 Hz,   0 . Also
m
s
   2 x108   81 8.854 x1012   0.45,   10 4 , so

 1 (low loss dielectric).

 104 1
Np

120   0.0021
2 
2 81
m

rad
    
 r  18.8


c

m

1
 120
 41.9

81
so E( z, t )  8e 0.0021z cos  2 x108 t  18.8 z  a x
V
m
(b)
E s  8e 0.0021z e  j18.8 z a x
1
V
,
m
8 0.0021z  j18.8 z
mA
e
e
a y  191e 0.0021z e  j18.8 z a y

41.9
m
mA
so H( z, t )  191e 0.0021z cos  2 x108 t  18.8 z  a y
m
2
1 Exo 2 z
W
(c) Pavg 
e a z  0.764e2(0.0021)(1)a z  0.761 2
2 
m
2
(d) P  Pavg (10m )  7.6W
Hs 
a P  Es 
V
,
m
5. A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0 mV/m
amplitude at the surface. How much power passes through a square meter at the surface?
How much power passes through a square meter area 10. m beneath the surface?
f  200MHz , Eo  1
Cu:   2
mV
1 Eo 2
, Pavg 
m
2 
 j 45
Np
e ,    f   214 x103
, so   5.22e j 45 m

m
3
1 10 
W
Pavg 
 96 2 ; P  Pavg S  96W
3
2 5.22 x10
m
Now at 10 m beneath the surface, we have
2
E ( z  10 m)  Eo e  (10  m )  103 e  (214 x10 )(10  m )  118 x106
3
1 118 x10 
W

 1.3 2 ; P  1.3W
3
2 5.22 x10
m
6 2
Pavg
V
m