Electromagnetic Wave Theory

1. Series: EMF Theory
Lecture: #1.14
Dr R S Rao
Professor, ECE
ELECTROSTATICS
Passionate
Teaching
Joyful
Learning
Electric field intensity due to various charge distributions like infinite straight line charge,
infinite plane sheet charge, spherical shell charge and solid spherical charge.

2. Expressions of field intensity of various charge distributions
Electric Field Intensity, E
Electrostatics
Electrostatic
Fields

3. Expressions of field intensity for standard charge distributions
Electric Field Intensity, E
Electrostatics
Electrostatic
Fields

4. Electrostatics
Electrostatic
Fields
4
Example I
Consider a circular loop of radius a m, carrying a uniform line charge density, λ C/m.
Assuming free space medium, find the field intensity, E at a distance of h m above its
center.
Solution:
To find field intensity, locate the loop over xy-plane with its center exactly over the origin.
Observation point P now is over positive z-axis at (0,, h), at a height of h from xy-plane.
The source distribution arrangement in the coordinate system is shown in Figure.
The field intensity at field point P(0,,h)due to differential line charge dQ=λdl =λad
at(a,, 0) is
2 2 2 2 2 1 2
ˆ ˆ
1 1 λ ( )
ˆ
4 4 ( ) ( )
o o
dQ ad a h
d
R a h a h

 
 

 
ρ z
E = R
Figure Field intensity due to a circular loop.(2.13)
Radial components of the field get cancelled due to symmetry
of the source, leaving behind only the vertical component.
2
2 2 3 2 2 2 3 2
0
ˆ ˆ
1 λ 1 2 λ
4 ( ) 4 ( )
o o
ad h ah
a h a h

 
 
 

z z
E = =
2 2 3 2
ˆ
1 λ
2 ( )
o
ah
a h
 
z
=

5. Electrostatics
Electrostatic
Fields
5
Three identical and mutually perpendicular infinite charge sheets, each with a charge density,
σ= 17.708 pC/m, are lying in free space over x=0, y=0 and z=0 planes. Determine the field, E
at (5,6,7)m and express it in cylindrical system.
Solution:
Given three mutually perpendicular charge distributions are depicted in Figure. Unit vector of
z=0 plane to point at (5,6,7)m is 𝐳 . Field due to z=0 plane charge over the field point is,
12
12
σ 17.708 10
ˆ ˆ ˆ
= = = V m
2 2 8.854 10
o




 
E n z z
Unit vector of x=0 and y=0 planes to point at (5,6,7)m are 𝐱 and 𝐲 respectively. Fields due to
these two planes are,
12
12
12
12
σ 17.708 10
ˆ ˆ ˆ
= = = V m &
2 2 8.854 10
σ 17.708 10
ˆ ˆ ˆ
= = = V m
2 2 8.854 10
o
o







 

 
E n x x
E n y y
Total field due to the three infinite charges then is,
ˆ ˆ ˆ
( )V/m
t   
E x y z
Figure Three mutually perpendicular sheet charges.
Example II

6. Example II
Electrostatics
Electrostatic
Fields
6
The result is in Cartesian coordinates and it requires to be converted into cylindrical system.
1
tan (6 5) 0.876 rad

 
Substituting Cartesian components of the given field in cylindrical component expressions,
cos sin 1cos 1sin 1.408
x y
E E E
    
    
sin cos 1sin 1cos 0.128
x y
E E E
    
       
1
z
E 
Thus, the field in cylindrical coordinates is,
ˆ ˆ
ˆ ˆ
ˆ ˆ
1.408 0.128 V/m
z
E E E
 
    
E z z
    
 
1
tan y x
 


7. Example III
Electrostatics
Electrostatic
Fields
7
A line charge distribution, in free space, with density λ=5nC/m, is located at y=2m, z=5m
parallel to x-axis. Determine field intensity, E at field point (4,1,3)m. If an additional surface
charge, with density, σ= 9 nC/m2
is placed over plane y=4m, determine the location where total
field is zero.
Solution:
Given line and surface charge distributions are depicted in Figure. Length of radial vector, ρ
joining line charge at (x,2,5)m to (4,1,3)m is 2.24m. Unit vector along radial line joining line
charge to (4,1,3) m is 𝛒=‒(𝐲+2𝐳)/2.24. Hence, field intensity El due to line charge is,
9
12 2
ˆ ˆ
λ 5 10 ( 2 )
ˆ ˆ ˆ
= (17.91 35.82 )V/m
2 2 8.854 10 2.24
l
o
  


   
  
  
y z
E = ρ y z
With the surface charge in place, cancellation of individual field intensities can happen only in
between the charge distributions

8. Example III
Electrostatics
Electrostatic
Fields
8
Let us suppose at (x,y,z)m the fields due to line and surface charges become equal and
opposite, cancelling each other, giving a zero net field intensity. At this point, field intensity
due to line charge becomes,
9
12 2 2
ˆ ˆ
λ 5 10 ( 2) ( 5)
ˆ=
2 2 8.854 10 [( 2) ( 5) ]
l
o
y z
y z
  


   
    
y z
E = ρ
2 2
ˆ ˆ
89.88[( 2) ( 5) ]
= V/m
[( 2) ( 5) ]
y z
y z
  
  
y z
Normal vector to sheet charge is, 𝐧=±𝐲 , and in the present
case, appropriate one is‒ sign.
Figure Line and surface charge distribution.
Hence, field intensity Es due to sheet charge is,
9
12
σ 9 10
ˆ ˆ ˆ
( ) 508.24 V/m
2 2 8.854 10
s
o




 
 
E = n = y = y
For cancellation of individual field intensities,
2 2
ˆ ˆ
89.88[( 2) ( 5) ]
ˆ
+ 0 508.24 =0
[( 2) ( 5) ]
l s
y z
y z
  
 
  
y z
E E = y
Equating z-components gives z=5m. Equating y-components with z=5m gives,
89.88 508.24( 2) =2.18m
y y
  

9. Example IV
Electrostatics
Electrostatic
Fields
9
A point charge, Q= 25 nC is located at y=−4m and a line charge distribution, with density
λ=10nC/m is located at y=4m parallel to z-axis. When the entire charge distribution is in free
space, determine the field intensity, E at (a)P1 (0,−2,0)m, (b) P2 (0,6,0)m and (c) P3 (0,0,4)m.
Solution:
Given charge distribution is illustrated in Figure.
(a) At P1 (0,−2,0)m: At this point the fields due to both charge distributions are entirely in y-
direction. Distances of point and line charges, respectively, from field point are 2m and 6m.
Fields due to point and line charges, respectively, are,
9
2 12 2
ˆ
25 10
ˆ ˆ
= = 56.17 V/m
4 4 8.854 10 2
p
Q
R
 




  
y
E R y
9
12
λ 10 10
ˆ ˆ ˆ
= ( ) 29.96 V/m
2 2 8.854 10 6
l
 



  
  
E = ρ y y
Figure A point and line charge distributions.
Total field is
ˆ ˆ
(56.17 29.96) 26.21 V/m
p l
   
E E E y = y

10. Example IV
Electrostatics
Electrostatic
Fields
10
(b) At P2 (0,6,0)m: Here also fields due to both charge distributions are entirely in y-direction.
Distances of point and line charges, respectively, from field point are 6m and 2m. Fields due to
point and line charges, respectively, are ,
9
2 12 2
ˆ
25 10
ˆ ˆ
= = 2.25 V/m
4 4 8.854 10 10
p
Q
R
 




  
y
E R y
9
12
ˆ
λ 10 10
ˆ ˆ
= 89.88 V/m
2 2 8.854 10 2
l
 




  
y
E = ρ y
Total field is,
ˆ ˆ
(2.25 89.88) 92.13 V/m
p l
   
E E E y = y
(c) At P3 (0,0,4)m: Here the field due to point charge is in y- and z- directions. Distance of
field point from source is 5.66m. The field then can be found as,
9
2 12 3
ˆ ˆ
25 10 (4 4 )
ˆ ˆ ˆ
= = (4.96 4.96 )V/m
4 4 8.854 10 5.66
p
Q
R
 


 
 
  
y z
E R y z
Distance of field point from line charge is 4m. Hence, field due to line charge is,
9
12
λ 10 10
ˆ ˆ ˆ
= ( ) 44.94 V/m
2 2 8.854 10 4
l
 



  
  
E = ρ y y
Total field then is,
ˆ ˆ ˆ ˆ ˆ
(4.96 4.96 ) 44.94 ( 39.98 4.96 )V/m
p l
      
E E E y z y = y z

11. Example V
Electrostatics
Electrostatic
Fields
11
A line charge distribution, with density λ=8nC/m is located at y=−2m parallel to z-axis, and
another sheet charge, with density, σ= 6nC/m2
is located at y=2m in free space. Determine the
field intensity E at points (a)P1 (0,−1,0)m, (b) P2 (0,4,0)m and (c) P3 (0,0,2)m.
Solution:
Given charge distribution is illustrated in Figure.
(a) At P1 (0,−1,0)m: At this point the fields due to both charge distributions are entirely in y-
direction. Distances of line and surface charges,
respectively, are 1m and 5m. Their fields are,
9
12
λ 8 10
ˆ ˆ ˆ
= 143.80 V/m
2 2 8.854 10 1
l
o
  




  
E = ρ y y
9
12
σ 6 10
ˆ ˆ ˆ
( ) V/m
2 2 8.854 10
338.83
s
o




 
 
E = n = y = y
Total field is,
m
338. 3
83 ˆ ˆ
(143.80 ) 195.0 V/
l s
   

E E E y = y
Figure A line and sheet charge distributions.

12. Example V
Electrostatics
Electrostatic
Fields
12
(b) At P2 (0,4,0)m: Here also fields due to both charge distributions are entirely in y-
direction. Distances of line and surface charges, respectively, are 6m and 2m. Field due to line
charge is,
9
12
λ 8 10
ˆ ˆ ˆ
= 23.98 V/m
2 2 8.854 10 6
l
o
  




  
E = ρ y y
Field due to surface charge remains same as it is in earlier case magnitude wise and opposite in
direction. Total field then is,
ˆ ˆ
(23.98 338.83) 362.81 V/m
l s
   
E E E y = y
(c) At P3 (0,0,2)m: Here also fields due to both charge distributions are entirely in y-direction.
Distances of line and surface charges, respectively, are 2m and 2m. Field due to line charge is
9
12
λ 8 10
ˆ ˆ ˆ
= 71.90 V/m
2 2 8.854 10 2
l
o
  




  
E = ρ y y
Field due to surface charge remains same as it was in earlier cases. Total field then is
ˆ ˆ
(71.90 338.83) 266.93 V/m
l s
    
E E E y = y

13. Example VI
Electrostatics
Electrostatic
Fields
13
A spherical shell, with a radius, r=1.5m, carrying a total charge, Q= 30 nC, is located at y=−2m
and a sheet charge, with a density, σ=10nC/m2
is located at y=4m parallel to xz-plane. When
the entire charge distribution is in free space, determine the field intensity, E at points (a)P1
(0,−1,0)m, (b) P2 (0,2,0)m and (c) P3 (0,0,4)m.
Solution:
Given charge distribution is illustrated in Figure .
(a) At P1 (0,−1,0)m: The given point is located within the
shell, and hence, field there due to shell is zero, Esl = 0. Total
field there is only due to the sheet charge.
9
12
σ 10 10
ˆ ˆ ˆ
( ) V/m
2 2 8.854 10
564.72
s
o




 
 
E = n = y = y
Total field is,
m
564.72 2
ˆ ˆ
0 564. V
7 /
sl s  
  
E E E y= y
Figure A shell and sheet charge distributions.

14. Example VI
Electrostatics
Electrostatic
Fields
14
(b) At P2 (0,2,0)m: The given point is located outside the shell, and hence, field there due to
shell plus sheet charge. Shell charge now can be considered as a point charge, located at the
center of shell. Fields due to both charge distributions are entirely in y-direction. Distances of
point and sheet charges, respectively, from field point are 4m and 2m. Fields due to these
charges are,
9
2 12 2
ˆ
30 10
ˆ ˆ
= = 16.85 V/m
4 4 8.854 10 4
sl
o
Q
R
 




  
y
E R y
Field due to sheet charge remains same as in previous case. Total field now is,
m
564 5
ˆ .72 47.8
ˆ ˆ
16.85 7 V/
sl s  
  
E E E y y= y
(c) At P3 (0,0,4)m: The given point is outside the shell, and hence, field there is sum of those
due to both shell and sheet charges. Shell can be replaced by a point charge, located at the
center of shell. Distances of point and sheet charges, respectively, from field point are 4.47m
and 4m. Fields due to these charges are
9
2 12 3
ˆ ˆ
30 10 (2 4 )
ˆ ˆ ˆ
= = (6.04 12.08 )V/m
4 4 8.854 10 4.47
sl
Q
R
 


 
 
  
y z
E R y z
Field due to sheet charge remains same as in previous case. Total field now is
m
564
ˆ ˆ ˆ ˆ ˆ
6.04 12.08 55
7 8.68 12.
. 2 08 V/
sl s 
    

E E E y z y= y z

15. ENOUGH
FOR
TODAY
ENOUGH
FOR
TODAY
ENOUGH
FOR
TODAY
ENOUGH
FOR
TODAY
ENOUGH
FOR
TODAY
15