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e-Notes by Prof.T.Basavaraj Sri Revanna Siddeshwara Institute of Technology, Bangalore
FIELD THEORY
Sub Code : EC44
Hrs/Week : 04
Total Hrs. : 52
IA Marks : 25
Exam Hrs. : 03
Exam Marks : 100
1. Electric Fields
a. Coulomb’s law and Electric field intensity
b. Electric flux density, Gauss law and divergence
c. Energy and potential
d. Conductors, dielectrics and capacitance
e. Poisson’s and Laplace’s equations
18 hours
2. Magnetic fields
a. The steady magnetic field
b. Magnetic forces, materials and inductance
14 hours
3. Time varying fields and Maxwell’s equations
5 hours
4. Electromagnetic waves
15 hours
Text Books :
William H Hayt Jr and John A Buck, “Engineering Electromagnetics”, Tata McGraw-Hill,
6th Edition, 2001
Reference books :
John Krauss and Daniel A Fleisch, “Electromagnetics with Application”, McGraw-Hill,
5th Edition, 1999
Guru and Hiziroglu, Electromagnetics Field theory fundamentals, Thomson Asia Pvt. Ltd
I Edition, 2001
Joseph Edminster, “Electromagnetics”, Schaum Outline Series, McGraw-Hill
Edward C Jordan and Keith G Balmain, “Electromagnetic Waves and Radiating Systems”,
Prentice-Hall of India, II Edition, 1968, Reprint 2002.
David K Cheng, “Field and Wave Electromagnetics”, Pearson Education Ais II Edition, 1989,
Indian Repr-01
1
Introduction to Field Theory
The behavior of a physical device subjected to electric field can be studied either by Field
approach or by Circuit approach. The Circuit approach uses discrete circuit parameters like
RLCM, voltage and current sources. At higher frequencies (MHz or GHz) parameters would no
longer be discrete. They may become non linear also depending on material property and
strength of v and i associated. This makes circuit approach to be difficult and may not give very
accurate results.
Thus at high frequencies, Field approach is necessary to get a better understanding of
performance of the device.
FIELD THEORY
The ‘Vector approach’ provides better insight into the various aspects of Electromagnetic
phenomenon. Vector analysis is therefore an essential tool for the study of Field Theory.
The ‘Vector Analysis’ comprises of ‘Vector Algebra’ and ‘Vector Calculus’.
Any physical quantity may be ‘Scalar quantity’ or ‘Vector quantity’. A ‘Scalar quantity’ is
specified by magnitude only while for a ‘Vector quantity’ requires both magnitude and direction
to be specified.
Examples :
Scalar quantity : Mass, Time, Charge, Density, Potential, Energy etc.,
Represented by alphabets – A, B, q, t etc
Vector quantity : Electric field, force, velocity, acceleration, weight etc., represented by alphabets
with arrow on top.
   
A, B, E, B etc.,
  
Vector algebra : If A, B, C are vectors and m, n are scalars then
(1) Addition



A  B  B 



A  ( B  C) 

A



(A  B)  C
Commutativ e law
Associativ e law
(2) Subtraction
 


A - B  A  (- B)
(3) Multiplication by a scalar


mA  Am


m (n A)  n (m A)



(m  n) A  m A  n A
 


m (A  B)  m A  m B
Commutativ e law
Associativ e law
Distributi ve law
Distributi ve law
2
A ‘vector’ is represented graphically by a directed line segment.
A ‘Unit vector’ is a vector of unit magnitude and directed along ‘that vector’.

â A is a Unit vector along the direction of A .

Thus, the graphical representation of A and â A are

A

Vector A
Unit ve ctor â A


Also â A  A / A


or A  â A A
Product of two or more vectors :
(1) Dot Product ( . )
 


A . B  A ( B COS θ OR {


A COS θ } B , 0  θ  π

B

B

A Cos θ


B Cos θ

A
A.B = B.A


A
(A Scalar quantity)
(2) CROSS PRODUCT (X)
C=AxB=
Ex.,
A B SIN θ n̂


where ' θ ' is angle between A and B ( 0  θ  π )


and n̂ is unit vecto r perpendicu lar to plane of A and B
  
directed such that A B C form a right handed system of vectors




A x B  - B x A







A x ( B  C)  A x B  A x C
3
CO-ORDINATE SYSTEMS :
For an explicit representation of a vector quantity, a ‘co-ordinate system’ is essential.
Different systems used :
Sl.No.
1.
2.
3.
System
Rectangular
Cylindrical
Spherical
Co-ordinate variables
x, y, z
ρ, , z
r, , 
Unit vectors
ax , ay , az
aρ , a , az
ar , a , a
These are ‘ORTHOGONAL‘ i.e., unit vectors in such system of co-ordinates are mutually
perpendicular in the right circular way.
i.e., x y z ,   z , r  
RECTANGULAR CO-ORDINATE SYSTEM :
Z
x=0 plane
az
p
y=0
plane
ax
X
Y
ay
z=0 plane
ax . ay  ay . az  az . ax  0
ax x ay  az
ay x az  ax
az x ax  ay
az is in direction of ‘advance’ of a right circular screw as it is turned from ax to ay
Co-ordinate variable ‘x’ is intersection of planes OYX and OXZ i.e, z = 0 & y = 0
Location of point P :
If the point P is at a distance of r from O, then
If the components of r along X, Y, Z are x, y, z then

r  x ax  y ay  z az 
r ar
4
Equation of Vector AB :

If OA  A  A x a x  A y a y  A z a z

and OB  B  Bx a x  By a y  Bz a z then


 
A  AB  B or AB  B - A
B

B
0
AB

A
A
where A s , A y & A z are components of A along X, Y and Z
and Bs , By & Bz are components of B along X, Y and Z
Dot and Cross Products :
 
A . B  (A x a x  A y a y  A z a z ) . (B x a x  By a y  Bz a z )  A x Bx  A y By  A z Cz
 
A x B  (A x a x  A y a y  A z a z ) x (B x a x  By a y  Bz a z )
Taking ' Cross products' term by term and grouping, we get
ax


A x B  Ax
ay
az
Ay
Az
Bx
By
Bz
5
Ax
 

A . (B x C )  Bx
Ay
Az
By
Bz
Cx
Cy
Cz
 

If A, B and C are non zero vectors,
 
(i) A . B  0 then Cos θ  0 i.e., θ  900 A and B are perpendicu lar
 
A x B  0 then Sin θ  0
θ  0 A and B are parallel
 

 

(ii) A . ( B x C) represents the volume of a parallelop oid of sides A , B and C
Unit Vector along AB
a AB 
AB
AB
where
Vector length AB  AB

(AB . AB )
Differential length, surface and volume elements in rectangular co-ordinate systems

r  x â x  y â y  z â z




r
r
r
dr 
dx 
dy 
dz
x
y
z

d r  dx â x  dy â y  dz â z

Differential length dr  [ dx 2  dy2  dz2 ]1/2
- - - - -1

Differential surface element, ds
1. r to z : dxdy â z
2.  r to z : dxdy â z
3.  r to z : dxdy â z
------ 2
Differential Volume element
dv = dx dy dz
------ 3
6
z
dx
p’
p
dz
dy

r


r d r
0
y
x
Other Co-ordinate systems :Depending on the geometry of problem it is easier if we use the appropriate co-ordinate system
than to use the Cartesian co-ordinate system always. For problems having cylindrical symmetry
cylindrical co-ordinate system is to be used while for applications having spherical symmetry
spherical co-ordinate system is preferred.
Cylindrical Co-ordiante systems :z
P(ρ, , z)
az

r
x = ρ Cos 
y = ρ Sin 
z=z
ρ
ρ  x 2  y2
0
 ap

r
y
φ  tan -1 y / x
zz

ρ
x

r  x â x  y â y  z â z

r  ρ Cos  â x  ρ Sin  â y  z â z




r
r
r
dr 
dρ 
d 
dz
- - - - - -1
ρ

z


r
r
 Cos  â x  Sin  â y 
â   h ρ â ρ ; h ρ 
ρ
ρ


r
r
 - ρ Sin  â x  ρ Cos  â y 
â   ρ â  ; h 



r
 â z
hz 
z

r
1
ρ

r
 ρ


r
1
z
Thus unit vectors in (ρ, , z) systems can be expressed in (x,y,z) system as
7
a ρ  Cos  a x  Sin  a y
a x  Cos  a   Sin  a 
a   - Sin  a x  Cos  a y
a y  Sin  a   Cos  a 
az  az
;
a  , a  and a z are orthogonal

Further , d r  d ρ â   ρ d â   dz â z
 2
and d r  d ρ 2  (ρ d ) 2  (dz) 2
------2
Differential areas :
ds â z  (d ρ) (ρ d ) . â z
ds â   (dz) (ρ d ) . â 
-------3
ds â  (d ρ dz) â
Differential volume :
d  (d ρ) (ρ d ) (dz)
or d  ρ d ρ d dz
----- 4
8
Spherical Co-ordinate Systems :Z
z
X = r Sin  Cos 
Y = r Sin  Sin 
Z = r Cos 
p
R

0
x
r
y

Y
r Sin
X

R  r Sin  Cos  â x  r Sin  Sin  â y  r Cos  â z


R R
â r 
/
 Sin  Cos  â x  Sin  Sin  â y  Cos  â z
r
r


R R
â  
/
 Cos  Cos  â x  Cos  Sin  â y  Sin  â z
 


R R
â  
/
 - Sin  â x  Cos â y
 



 R
R
R
dR 
dr 
d 
d
r



dR  dr â r  r d â   r Sin  d â 
d Sr  r 2 Sin  d d
d S  r 2 Sin  dr d
d S  r dr d
d v  r 2 Sin  dr d d
9
General Orthogonal Curvilinear Co-ordinates :z
u1 a3
u3
a1
u2
a2
y
x
Co-ordinate Variables : (u1 , u2, u3) ;
Here
u1 is Intersection of surfaces u2 = C & u3 = C
u2 is Intersection of surfaces u1 = C & u3 = C
u3 is Intersection of surfaces u1 = C & u2 = C
â1 , â 2 , â 3 are ubnit vect ors tangentia l to u1 , u 2 & u 3
System is Orthogonal if â1 . â 2  0 , â 2 . â 3  0 & â 3 . â1  0

If R  x â x  y â y  z â z & x, y, z are functions of u1 , u 2 & u 3



R
R
R
then d R 
du1 
du 2 
du 3
 u1
 u2
 u3
 h1 du1 â1  h 2 du 2 â 2  h 3 du 3 â 3
where h1 , h 2 , h 3 are scale factors ;


R
R
h1 
, h2 
 u1
 u2

R
, h3 
 u3
10
Co-ordinate Variables, unit Vectors and Scale factors in different systems
Systems
Co-ordinate Variables
Unit Vector
Scale factors
General
u1
u2
u3
a1
a2
a3
h1
h2
h3
Rectangular
x
y
z
ax
ay
az
1
1
1
Cylindrical
ρ

z
aρ
a
az
1
ρ
1
Spherical
r


ar
a
a
1
r
r sin 
Transformation equations (x,y,z interms of cylindrical and spherical co-ordinate system
variables)
Cylindrical : x = ρ Cos  , y = ρ Sin  , z = z ;
ρ  0, 0    2 - < z < 
Spherical
x = r Sin Cos , y = r Sin Cos , z = r Sin
r 0 , 0     , 0    2
 V 

. A 
1 v
1 v
â 1 
â
h1 u1
h 2 u2
1
h 1 h 2 h3

xA 
1
h1h 2h
 
(h

  u1
3
2
h
3
2

A1) 
h 1 â 1 h 2 â 2


 u1
u2
h1 A1 h 2 A2
1 v
â 3
h 3 u 3

(h 1 h
 u2
h
3
â
3
A2) 

(h 1 h
 u3
2

A3) 

3

u 3
h3 A3
where V  V ( u 1 , u 2 , u 3 ) a Scalar field

& A  A 1 â 1  A 2 â 2  A 3 â 3 is a Vector field where A1  A1 (u1 , u 2 , u 3 )
A 2  A 2 (u1 , u 2 , u 3 ) and A 3  A 3 (u1 , u 2 , u 3 )
11
Vector Transformation from Rectangular to Spherical :
Rectangula r : A R  A x â x  A y â y  A z â z



Spherical : AS  ( A R â r ) â r  (A R â ) â  (A R . â  ) â 
 A r â r  A â  A â 
where A r , A , A are related to A x , A y , A z as
 â x . â r
A r 

 
A    â x . â 
â x . â 
A  
 

â y . â
r
â y . â 
â y . â 
â z . â r 

â z . â  
â z . â  
A x 
A 
 y
 A z 
12
Field Theory
A ‘field’ is a region where any object experiences a force. The study of performance in the

presence of Electric field (E) , Magnetic field () is the essence of EM Theory.
P1 : Obtain the equation for the line between the points P(1,2,3) and Q (2,-2,1)
PQ  a x - 4 a y - 2 a z
P2 : Obtain unit vector from the origin to G (2, -2, 1)
13
Problems on Vector Analysis
Examples :1. Obtain the vector equation for the line PQ between the points P (1,2,3)m and Q (2, -2,
1) m
Z
PQ
P (1,2,3)
Q(2,-2,-1)
0
Y
X
The vector PQ  (x q - x p ) â x  (y q - y p ) â y  (z q - z p ) â z
 (2 - 1) â x  (-2 - 2) â y  (-1 - 3) â z
 (â x - 4 â y - 2 â z )
2. Obtain unit vector from origin to G (2,-2,-1)
G

G
0
The vector G  (x g - 0) â x  (y g - 0) â y  (z g - 0) â z
 (2 â x - 2 â y - â z )

G
The unit vect or , â g  
G

G  2 2  (-2)2  (-1)2  3
 â g  (0.667 â x - 0.667 â y - 0.333 â z )
3. Given
A  2 â x - 3 â y  â z

B  - 4 â x - 2 â y  5â z
 


find (1) A . B and (2) A x B
Solution :
 
(1) A . B  (2 a x - 3 a y  a z ) . (-4 a x - 2 a y  5 a z )
=-8+6+5= 3
Since ax . ax = ay . ay = az . az = 0 and ax ay = ay az = az ax = 0
14
ax
 
(2) A x B  2
ay
az
3
1
4 2
5
= (-13 ax -14 ay - 16 az)
4. Find the distance between A( 2, /6, 0) and B = ( 1, /2, 2)
Soln : The points are given in Cylindrical Co-ordinate (ρ,, z). To find the distance between
two points, the co-ordinates are to be in Cartesian (rectangular). The corresponding
rectangular co-ordinates are (ρ Cos, ρ Sin, z)



 A  2 Cos â x  2 Sin
â y  1.73 â x  â y
6
6



& B  Cos â x  Sin
â y  2 â z  â y  2 â z
6
2
 AB  (B x - A x ) â x  (B y - A y ) â y  (B z - A z ) â z
 - 1.73 â x  (1 - 1) â y  (2 - 0) â z
 - 1.73 â x  2 â z
 (AB) 
1.732  2 2  2.64
5. Find the distance between A( 1, /4, 0) and B = ( 1, 3/4, )
Soln : The specified co-ordinates (r, , ) are spherical. Writing in rectangular, they are (r
Sin  Cos , r Sin  Sin , r Cos ).
Therefore, A & B in rectangular co-ordinates,




A  (1 Sin Cos 0 â x  1 Sin Sin 0 â y  1 Cos â z )
4
4
4
 ( 0.707 â x  0.707 â y )

3
3
3
B  ( Sin
Cos  â x  Sin
Sin  â y  Cos
â z )
4
4
4
 ( 0.707 â x  0.707 â y )
AB  (B x - A x ) â x  (B y - A y ) â y  (B z - A z ) â z
 - 1.414 â x  (- 0.707) â y  (-0.707) â z
AB  ( AB . AB )1/2
 (2  0.5  0.5)1/2  1.732
6. Find a unit vector along AB in Problem 5 above.
1
AB
â AB 
= [ - 1.414 ax + (-0.707) ay + (-0.707) az]
1.732
AB
= ( - 0.816 â x - 0.408 â y  0.408 â z )


7. Transform F  (10 â x - 8 â y  6 â z ) into F in Cylindrica l Co - ordinates.
Soln :
15




FCyl  (F . â p ) â p  (F . â ) â  (F . â z ) â z
 [(10 â x - 8 â y  6 â z ) . (Cos  â x  Sin  â y )] â 
 [ (10 â x - 8 â y  6 â z ) . (- Sin  â x  Cos â y )] â 
 [ (10 â x - 8 â y  6 â z ) . (â x )] â x
ρ
 (10 Cos  - 8 Sin  ) â   (-10 Sin  - 8 Cos  ) â   6 â z

x   Cos 
y   Sin 
  x 2  y 2  12.81
  tan -1
y
x
y
 - 38.660
x
FCyl  [ 10 Cos (- 38.66) - 8 Sin ( - 38.66) ] â p  [- 10 Sin (- 38.66) - 8 Cos (- 38.66)] â  6 â z
 (12.8 â   6 â z )

8. Transform B  y â x - x â y  z â z into Cylindrical Co-ordinates.
x   Cos  , y   Sin 

B   Sin  â x -  Cos  â y  z â z

BCyl  (B. â  ) â   (B. â  ) â   (B . â z ) â z
 [ (  Sin  â x -  Cos  â y  z â z ). (Cos  â x  Sin  â y )] â 
 [ (  Sin  â x -  Cos  â y  z â z ). (- Sin  â x  Cos  â y )] â   z â z
 [  Sin  Cos  -  Sin  Sin  ] â   [ -  Sin 2  -  Cos 2  ] â   z â z
 -  â   z â z
9. Transform 5 â x into Spherical Co-ordinates.
ASph  (A. â r ) â r  (A. â ) â   (A. â  ) â 
 [ 5 â x . (Sin  Cos  â x  Sin  Sin  â y  Cos  â z )] â r 
[ 5 â x . (Cos  Cos  â x  Cos  Sin  â y - Sin  â z ] â  
[ 5 â x . (- Sin  â x  Cos  â y )] â 
 5 Sin  Cos  â r  5 Cos  Sin  â  5 Sin  â 
10. Transform to Cylindrical Co-ordinates G  (2 x  y) â x - (y - 4x) â y at Q (  ,  , z)
Soln :
16
G Cyl  (G. a  ) â   (G. a  ) â   (G. a z ) â z
G Cyl  [ (2 x  y) â x - (y - 4x) â y ] . [ Cos  â x  Sin  â  ] â  
[ (2 x  y) â x - (y - 4x) â y ] . [ - Sin  â x  Cos  â y ] â   0
 [ ( 2x  y) Cos  - (y - 4x) Sin  ] â 
 [ - (2 x  y) Sin  - (y - 4x ) Cos  ] â 
x   Cos  , y   Sin 
G Cyl  [ ( 2  Cos    Sin  ) Cos  - (  Sin  - 4  Cos  ) Sin  ] â 
 [ - ( 2  Cos    Sin  ) Sin  - (  Sin  - 4  Cos  ) Cos  ] â 
 [ 2  Cos 2    Sin  Cos  -  Sin 2   4  Sin  Cos  ] â 
 [ - 2  Sin  Cos  -  Sin 2  -  Sin  Cos   4 Cos 2  ] â 
 ( 2  Cos 2   5  Sin  Cos  -  Sin 2  ) â 
 ( 4  Cos 2    Sin 2  - 3  Sin  Cos  ) â 
11. Find a unit vector from ( 10, 3/4, /6) to (5, /4, )
Soln :
A(r, , ) expressed in rectangular co-ordinates
OA  r Sin  Cos  â x  r Sin  Sin  â y  r Cos  â z

3

3

3
 A  10 Sin
Cos
â x  10 Sin
Sin
â y  10 Cos
â z
4
6
4
6
4




B  5 Sin Cos  â x  5 Sin Sin  â y  5 Cos â z
4
4
4


A  6.12 â x  3.53 â y - 7.07 â z B  - 3.53 â x  3.53 â z
 
AB  B - A  - 9.65 â x - 3.53 â y  10.6 â z
AB  9.652  3.532  10.6 2  14.77

AB
â AB 
 (- 0.65 â x - 0.24 â y  0.72 â z )
AB


12. Transform F  10 â x - 8 â y  6 â z into F in Spherical Co-orindates.
17
â r  Sin  Cos  â x  Sin  Sin  â y  Cos  â z
â   Cos  Cos  â x  Cos  Sin  â y - Sin  â z
â   - Sin  â x  Cos  â y




FSph  (F . â r ) â r  (F . â  ) â   (F . â  ) â 
 (10 Sin  Cos  - 8 Sin  Sin   6 Cos  ) â r
 (10 Cos  Cos  - 8 Cos  Sin  - 6 Sin  ) â 
 (- 10 Sin  - 8 Cos  ) â 
r  10 2  8 2  6 2  200 ;   Cos -1
-8
 - 38.66 0
10
Sin   Sin 64.69  0.9
z
 Cos -1
r
6
 64.89 0
200
  tan -1
Sin   Sin (-38.66)  - 0.625
Cos   Cos 64.69  0.42 Cos   Cos (-38.66)  0.781

F  (10 x 0.9 x 0.781 - 8 x 0.9 x (-0.625)) â r
 ( 10 x 0.42 x 0.781 - 8 x 0.42 x (0.625)) â 
 (-10 x - 0.625 - 8 x 0.781) â 

F  (11.529 â r  5.38 â   0.783 â  )
Line Integrals
In general orthogonal Curvilinear Co-ordinate system
dl  h1 du1 â1  h 2 du 2 â 2  h 3 du 3 â 3

F  F1 â1  F2 â 2  F3 â 3

 F . dl   h1 F1 du1   h 2 F2 du 2   h 3 F3 du 3
C
C
C
C
Conservative Field A field  is said to be conservative if it is such that    . dl  0
C
b
   . dl   d
  (b) -  (a) (does not depend on the path !). If  is electrosta tic flux, then
a

E  -   represent the electric field intensity and
b
   . dl represent the potential between b and a and is zero if it is taken around a closed contour.
a
i.e.,
   . dl  0
Therefore ES flux field is ‘Conservative’.
EXAMPLES :
18
13. Evaluate
line

I   a . dl
integral
where

a  (x  y) â x  (y - x) â y
along y2  x from A (1,1) to B (4,2)
Soln : dl  dx â x  dy â y

a . dl  (x  y) dx  (y - x ) dy
y 2  x or 2 dy dy  dx
2
2

2
  a . dl   (y  y) 2y dy   (y - y 2 ) dy
1
1
2
  (2 y3  2 y 2  y - y 2 ) dy
1
2
  (2 y 3  y 2  y) dy
1
2
 2 y 4
y3
y2 

 


3
2  1
 4 2
 24
23
 


3
 2
8


 8   2 3



1
1
1
 -  
 
3
2
2

2
1
1
4
- 1  11
   12
3
3
3
3
22
2
19


14. Evaluate the Integral I   E . ds where E  x â x and S is hunisphere of radius a
S
Soln:
If S is hemisphere of radius a, then S is defined by
x 2  y2  z2  a 2 , z  0 ;
ds  (a d ) (a Sin  ) d â r
ds  a 2 Sin  d d â r

E  (E . â r ) â r  (E. â  ) â   (E. â  ) a 

E r  x Sin  Cos  â r ; x  a Sin  Cos 


E . ds  E r . ds  a ( Sin  Cos  ) 2 â r . a 2 Sin  d d

E . ds  a Sin 3  Cos 2  d d
0     / 2 , 0    2
/2
2

2
2  a3
3
3
2
3
E
.
ds

a
Sin

d

Cos

d


a
x
x



0
0
3
3
20
where r, r1 , r2 ….. rm are the vector distances of q, q1 , …… qm from origin, 0.
 
r - rm is distance between charge qm and q.
â m is unit vector in the direction of line joining qm to q.
Electric field is the region or vicinity of a charged body where a test charge experiences a
force. It is expressed as a scalar function of co-ordinates variables. This can be illustrated by
drawing ‘force lines’ and these may be termed as ‘Electric Flux’ represented by  and unit is
coulomb (C).

Electric Flux Density (D) is the measure of cluster of ‘electric lines of force’. It is the
number of lines of force per unit area of cross section.


ψ
i.e., D 
c/m 2 or ψ   D n̂ ds C where n̂ is unit vect or normal to surface
A
S

Electric Field Intensity (E) at any point is the electric force on a unit +ve charge at that
point.


F
q1
i.e., E 

â1 N / c
q
4  0 r12



1  q1 
D



â
N
/
c

N
/
c
or
D


E
C in vacuum
1
0
0  4  r12 
0
In any medium other than vacuum, the field Intensity at a point distant r m from + Q C is

Q
E 
â r N / c ( or V / m)
4  0 r r 2



Q
and D  0 r E C or D 
â r C
4  r2


Thus D is independent of medium, while E depends on the property of medium.

E
+QC
q = 1 C (Test Charge)
Source charge

E

r
E 
21
0
r,m

Electric Field Intensity E for different charge configurations

1. E due to Array of Discrete charges

Let Q, Q1 , Q2 , ……… Qn be +ve charges at P, P1 , P2 , ……….. Pn . It is required to find E
at P.
Q1
P1

r2
Q2
Pr

P

r1
P2
Qn

Er

En

r1
1
4  0


E1

E2
0

rn
Qm
r - rm
2
â m V / m

2. E due to continuous volume charge distribution
â R

R
P
ρv C / m 3
The charge is uniformly distributed within in a closed surface with a volume charge density
dQ
of ρv C / m3 i.e, Q   V dv
and V 
dv
V

E 

Er 
V  V
Q
â R 
â R
2
4  0 R
4  0 R 2
V (r 1 )
 4 
V1
0
(r - r 1 ) 2
â R N / C
â R is unit vector directed from ‘source’ to ‘filed point’.

3. Electric field intensity E due to a line charge of infinite length with a line charge
density of ρl C / m
â R
22

R
dl
P
ρl C / m
L

Ep

1
4  0

l dl
R2
L
â R N / C

4. E due to a surface charge with density of ρS C / m2

R
ds
â R
P (Field point)
(Source charge)

Ep

1
4  0

S
 S ds
R2
â R N / C
Electrical Potential (V) The work done in moving a unit +ve charge from Infinity to that is
called the Electric Potential at that point. Its unit is volt (V).
Electric Potential Difference (V12) is the work done in moving a unit +ve charge from one
point to (1) another (2) in an electric field.

Relation between E and V
If the electric potential at a point is expressed as a Scalar function of co-ordinate variables
(say x,y,z) then V = V(x,y,z)


f
dV  - dl  - E . dl
- - - - - - - - (1)
q
V
V
V
Also, dV 
dx 
dy 
dz
x
y
z
dV   V . dl
- - - - - - - - - (2)

From (1) and (2) E  -  V
Determination of electric potential V at a point P due to a point charge of + Q C
â l
23

R  dR
0
+Q

At point P, E 


R
P
â R
Q
â R N / C
4  0 R 2

Therefore, the force f on a unit charge at P.


 f  1 x Ep 
Q
â R N
4  0 R 2
The work done in moving a unit charge over a distance dl in the electric field is


dV  - f . dl  - E . dl
R
 Vp  VP 
Q dl
 4  0 R 2 (â R . â l )  -
Q
Volt
4  0 R 2
R
Q

0
 4 
R2
dR
(a scalar field)
Electric Potential Difference between two points P & Q distant Rp and Rq from 0 is
Vpq  (Vp - Vq ) 
Q
4  0
 1
1 

 volt
 R p R q 
Electric Potential at a point due to different charge configurations.
1. Discrete charges
. Q1
.
Q2
Qm
P
Rm
V1P 
n
1
4  0
R
V2P 
1
4  0
1
Qm
V
m
2. Line charge
xP
ll
R
dl V
l
24
ρl C / m
3. Surface charge
V3P 
xP
S ds
1
4  0

V4P 
1
4  0
S
R
V
ρs C / m 2
4. Volume charge
xP
ρv C/ m3
R

V
V dv
R
V
5. Combination of above V5P = V1P + V2P + V3P + V4P
Equipotential Surface : All the points in space at which the potential has same value lie on a
surface called as ‘Equipotential Surface’.
Thus for a point change Q at origin the spherical surface with the centre of sphere at the
origin, is the equipotential surface.
Sphere of
Radius , R
R
P
0
+Q
equipotential surfaces
Q
V
0
R
Potential at every point on the spherical surface is
VR 
Q
volt
4  0 R
VPQ is difference of potential two equipotent ial surface potential

Gauss’s law : The surface integral of normal component of D emerging from a closed
surface is equal to the charge contained in the space bounded by the surface.

i.e.,  D . n̂ ds  Q C
(1)
S
where ‘S’ is called the ‘Gaussian Surface’.
25
By Divergence Theorem,


----------- (2)
D
.
n̂
ds


.
D
dv


S
V
Also, Q    V dv
---------- (3)
V
From 1, 2 & 3,

----------- (4) is point form (or differential form) of Gauss’s law while
.D  
equation (1) is Integral form of Gauss law.
Poisson’s equation and Laplace equation


In equation 4, D  0 E

  . E   / 0 or  . (-  V)   / 0
2 V -

0
Poisson equation
If   0,  2 V  0 Laplace equation
Till now, we have discussed (1) Colulomb’s law (2) Gauss law and (3) Laplace equation.

The determination of E and V can be carried out by using any one of the above relations.
However, the method of Coulomb’s law is fundamental in approach while the other two use
the physical concepts involved in the problem.


(1) Coulomb’s law : Here E is found as force f per unit charge. Thus for the simple case of
point charge of Q C,

1
Q
E 

 V/ M
4  0  R 2 

V   E dl Volt
l
(2) Gauss’s law : An appropriate Gaussian surface S is chosen. The charge enclosed is
determined. Then

 D n̂
ds  Qenc
S


Then D and hence E are determined

Also V   E dl volt
l
(3) Laplace equation : The Laplace equation  2 V  0

boundary conditions to get V. Then, E  -  V
is solved subjecting to different
26
Solutions to Problems on Electrostatics :1. Data : Q1 = 12 C , Q2 = 2 C , Q3 = 3 C at the corners of equilateral triangle d m.

To find : F on Q3
Solution :
Let Q1 , Q 2 and Q3 lie at P1 , P2 and P3 the corners of equilatera l triangle of side d meter.
If P1 , P2 and P3 lie in YZ plane, with P1
at origin the n
P1  (0,0,0) m
Z
P3
P2  (0, d, 0) m
P3  (0, 0.5 d, 0.866 d) m

r1  0

r2  d â y

r3  0.5 d â y  0.866 â z




The force F3 is F3  F13  F23

Q3  Q1
Q

F3 
â13  22 â 23 
2

4  0  d
d

 
0.5 d â y  0.866 d â z
r - r
â13  3 1 
r3 - r1
d
 
r - r
â 23  3 2  - 0.5 â y  0.866 â z
r3 - r2
d
d
d
P1
Y
P2
X
 0.5 â y  0.866 â z
Substituting,

12 x 10-6
F3  (3 x 10- 6 ) 9 x 109 
( 0.5 â y  0.866 â z ) 
2
 d
27 x 10-3  5 â y  12.12 â z 


 13.11
2
2
d2
 5  12.12 

F3  0.354 â F N where â F  (0.38 â y  0.924 â z )

2 x 10-6
( - 0.5 â y  0.866 â z )
2
d

2. Data : At the point P, the potential is Vp  (x 2  y2  z2 ) V
To find :

(1) E p (2) VPQ given P(1,0.2) and Q (1,1,2) (3) VPQ by using general expression for V
Solution :

Vp
Vp 
 Vp
(1) E p  -  Vp  - 
â x 
â y 
â z 
y
z
 x

2
 - [ 2 x â x  2 y â y  3z â z ] V /m
27
P 
(2) VPQ  -  E p . dl 
Q
1
 2x dx 
0
2
 2y dy   3z dz
2
1
1
2
0 y

2 0
 0  -1 V
(3) VPQ  VQ - VP  - 1 V
3. Data : Q = 64.4 nC at A (-4, 2, -3) m

To find : E at 0 (0,0,0) m
Solution :

E0 

A

E0
0
Q
â AO N / C
4  0 (AO) 2
64.4 x 10-9
[ â AO ] N/ C
10-9
2
4 x
(AO)
36 
AO  (0  4) â x  (0 - 2) â y  (0  3) â z  4â x - 2 â y  3 â z
â AO 
AO
AO

1
(AO)  (0.743 â x - 0.37 â y  0.56 â z )
29

64.4 x 9
E0 
â AO  20 â AO N / C
29
4. Q1 = 100 C at P1 (0.03 , 0.08 , - 0.02) m
Q2 = 0.12 C at P2 (- 0.03 , 0.01 , 0.04) m
F12 = Force on Q2 due to Q1 = ?
Solution :

Q1 Q 2
F12 
â12
2
4  0 R12
 
R12  R 2 - R1  (-0.03 â x  0.01 â y  0.04 â z ) - (0.03 â x  0.08 â y - 0.02â z )
 ( - 0.06 â x - 0.07 â y  0.06 â z ) ; R12  0.11 m
â12  ( - 0.545 â x - 0.636 â y  0.545 â z )

100 x 10- 6 x 0.121 x 10- 6
F12 
x 9 x 109 â12
2
0.11

F12  9 â12 N
28
5. Q1 = 2 x 10-9 C , Q2 = - 0.5 x 10-9 C C

(1) R12 = 4 x 10-2 m , F12  ?

(2) Q1 & Q2 are brought in contact and separated by R12 = 4 x 10-2 m F12`  ?
Solution :

F12 
(1)
2 x 10-9 x - 0.5 x 10-9
10-9
4 x
x ( 4 x 10- 2 ) 2
36 
â12 
-9
x 10-5 â12   5.63  N (attractiv e)
16
1
(Q1  Q 2 )  1.5 x 10-9 C
2
1.52

x 9 x 10-18  13 â12  12.66  N â12
16
(2) When brought into contact Q1`  Q`2 

F12` 
F12`
6.
( 1.5 x 10-9 ) 2
10-9
4 x
x ( 4 x 10- 2 ) 2
36 
 12.66  N (repulsive )
â12
Y
x
P3
x
x
P2
x
P1
0
X
Q1 = Q2 = Q3 = Q4 = 20  C
QP = 200  C at P(0,0,3) m
P1 = (0, 0 , 0) m P2 = (4, 0, 0) m
P3 = (4, 4, 0) m P4 = (0, 4, 0) m
FP = ?
Solution :
29





Fp  F1p  F2p  F3p  F4p

R 1p  3 â z R 1p  3 m â1p  â z

R 2p  - 4 â x  3 â z ; R 2p  5 m â 2p  - 0.8 â x  0.6 â z

R 3p  - 4 â x - 4 â y  3 â z ; R 3p  6.4 m ; â 3p  - 0.625 â x - 0.625 â y  0.47 â z

R 4p  - 4 â y  3 â z ; R 4p  5 m ; â 4p  - 0.8 â y  0.6 â z

Fp 
Qp
4
-9
10
36 
 Q1

Q2
Q3
Q4
 2 â1p  2 â 2p  3 â 3p  2 â 4p 
R 2p
R 3p
R 4p
 R 1p

1
1
1

â

(
0.8
â

0.6
â
)

(-0.625 â x - 0.625 â y  0.47 â z ) 
z
x
z
2
2
2

3
5
6.4
 200 x 10- 6 x 9 x 109 
 20 x 10- 6
 1 ( - 0.8 â  0.6 â )

y
z
 52

100
100
100

â

(0.8
â

0.6
â
)

(-0.625 â x - 0.625 â y  0.47 â z )
z
x
z

9
25
40.96
 200x10 - 6 x9x10 9 x10 9 x10 - 6 x 10- 2 

 100 ( - 0.8 â  0.6 â )

y
z
 25

1


 0.36 (3.2  1.526) â x 
(-1.526 - 3.2) â y  (11.11  2.4  1.15  2.4) â z ) 
2
6.4


 (- 1.7 â x - 1.7 â y  17 â z ) N  17.23 â p N
7. Data : Q1 , Q2 & Q3 at the corners of equilateral triangle of side 1 m.
Q1 = - 1C, Q2 = -2 C , Q3 = - 3 C

To find : E at the bisecting point between Q2 & Q3 .
Solution :
Z
P1 Q1
P1 : (0, 0.5, 0.866) m
P2 : (0, 0, 0) m
P3 : (0, 1, 0) m
P : (0, 0.5, 0) m
Q2
P E1P
Q3
Y
P2
E2P
E3P
P3
30




E P  E1P  E 2P  E 3P
 Q1

1
Q2
Q3
â

â
 2 â1P 

2P
3P
2
2
4  0  R1P
R 2P
R 3P

 - 0.866 â z
R 1P  0.866
â1P  - â z

R1P
R 2P   0.5 â y
R 2P  0.5
â 2P  â y
R 3P  - 0.5 â y
R 3P  0.5
â 3P  - â y
 - 1 x 10- 6

1
- 2 x 10- 6
- 3 x 10- 6
(
â
)

(
â
)

( - â y )
z
y

-9
2
2
2
10
0.5
0.5
 0.866

4
36 
 9 x 103 1.33 â z - 8 â y  12 â y

EP 
 9 x 103

4 â
y
 


 1.33 â z  36 â y  12 â z  03 V / m  37.9 180 k V/m
Z
E1P
EP
( EP ) = 37.9 k V / m
Y
E2P
(E3P – E2P)
E3P
8. Data Pl = 25 n C /m on (-3, y, 4) line in free space and P : (2,15,3) m
To find : EP
Solution :
Z
ρl = 25 n C / m
A

R
ρ (2, 15, 3) m
P
Y
X
The line charge is parallel to Y axis. Therefore EPY = 0

R  AP  (2 - (-3)) â x  (3 - 4) â z  (5 â x - â z ) ;  R  5.1 m

R
â R 
 (0.834 â x - 0.167 â z )
R

l
25 x
EP 
â R 
â R
10- 9
2  0 R
2
x 5.1
36 

E P  88.23 â R V / m
31
9. Data : P1 (2, 2, 0) m ; P2 (0, 1, 2) m ; P3 (1, 0, 2) m
Q2 = 10 C ; Q3 = - 10 C
To find : E1 , V1
Solution :
 


1  Q2
Q3
E1  E 21  E 21 
â
 2 â 21 

31
2
4  0  R 21
R 31



R 21  (2 â x  â y - 2 â z )
R 21  3 â 21  0.67 â x  0.33 â y - 0.67 â z


R 31  â x  2 â y  2 â z
R 31  3 â 31  0.33 â x  0.67 â y  0.67 â z
-6


10- 6
9 10
E1  9 x 10 
(0.67 â x  0.33 â y - 0.67 â z ) 
(0.33 â x  0.67 â y  0.67 â z )
9
 9

3
 10 [ â x  â y ]  14.14 (0.707 â x  0.707 â y ) V / m
V1 
1
4  0
-6
 Q2
Q3 
10- 6 
9 10


9
x
10




  3000 V
3 
 3
 R 21 R 31 
E1  14.14 V / m
V1  3000 V
32
10. Data : Q1 = 10 C at P1 (0, 1, 2) m ; Q2 = - 5 C at P2 (-1, 1, 3) m
P3 (0, 2, 0) m

To find : (1) E3 (2) Q at (0, 0, 0) for E3x  0
Solution :


1  Q1
Q2
(1) E 3 
â 23 
 2 â13 
2
4  0  R 13
R 23


R 13  (2 - 1) â y  (0 - 2) â z  â y - 2 â z
R 13  5

R 23  (0  1) â x  (2 - 1) â y  (0 - 3) â z  â x  â y - 3 â z
R 23  11

R
â13  13  ( 0.447 â y - 0.894 â z )
R 13

R
â 23   23  0.3 â x  0.3 â y - 0.9 â z
R 23

 10 x 10- 6

- 5 x 10- 6
E 3  9 x 109 
(0.447
â
0.894
â
)

(0.3 â x  0.3 â y - 0.9 â z )
y
z
2
2
( 11)
 ( 5)

 (8 â y - 16 â z )  (-1.23 â x - 1.23 â y  3.68 â z )

 - 1.23 â
x


 6.77 â y - 12.32 â z 103 V / m

 Q
 
Q
Q
(2) E 3  9 x 109  12 â13  22 â 23 
â 03  ; R 03  2 â y
2
R 23
R 03
 R 13


E 3x  - 1.23 â x

E 3x cannot be zero
11. Data : Q2 = 121 x 10-9 C at P2 (-0.02, 0.01, 0.04) m
Q1 = 110 x 10-9 C at P1 (0.03, 0.08, 0.02) m
P3 (0, 2, 0) m

To find : F12
Solution :

Q1 Q 2
F12 
â12 N ; R 12  - 0.05 â x - 0.07 â y  0.02 â z
2
4  0 R 12

121 x 10- 9 x 110 x 10- 9
F12 
[â12 ]
R 12  0.088
10-9
-3
4
x 7.8 x 10
36 

F12  0.015 aˆ 12 N
33
12. Given V = (50 x2yz + 20y2) volt in free space

Find VP , E P and â np at P (1, 2, - 3) m
Solution :


VP  50 (1) 2 (2) (-3)  20 (2) 2  - 220 V




E - V V â x V â y V â z
x
y
z

E  - 100 x y z â x - 50 x 2 z â y - 50 x 2 y â z

E P  - 100 (2) (-3) â x - 50 (-3) â y - 50 (2) â z
 600 â x  150 â y - 100 â z
 62 6.5 â P V / m ; â P  0.957 â x  0.234 â y - 0.16 â z
34
Additional Problems

A1. Find the electric field intensity E at P (0, -h, 0) due to an infinite line charge of density
ρl C / m along Z axis.
+
Z
A dz

R AP
z
dEPy

P
Y
dEPz
h
0

d EP
X
â P
-
Solution :
Source : Line charge ρl C / m. Field point : P (0, -h, 0)

dQ
ρ l dz
â R 
â R V / m ; R  AP  - z â z - h â y
2
2
4  0 R
4  0 R

R  AP  z 2  h 2

R
1
  
- h â y - z â z
R
R
dE P 
â R

z
 h

- R â y - R â z   d E Py â y  d E Pz â z


ρ l dz h
ρ l dz z
â y dE Pz  â z
2
4  0 R R
4  0 R 2 R
dE P 
dE Py

ρ l dz
4  0 R 2
Expressing all distances in terms of fixed distance h,
h = R Cos  or R = h Sec  ; z = h tan  , dz = h sec2  d
35
dE P y
ρl h Sec 2 d
x Cos 
4  0 h 2 Sec 2
EPy  -
E -
ρl
Cos  d
4  0 h
ρl
ρl
ρl
[  Sin  ]- / 2/ 2  x2  â y
4  0 h
4  0 h
2  0 h
dE P z  
EPz 
-
ρl h Sec 2 d
4  0 h 2 Sec 2
x
h tan 
ρl
 Sin  d
h Sec 
4  0 h
ρl
[ Cos ]- / 2/ 2  0
4  0 h
ρl
aˆ y V / m
2 π 0 h
An alternate approach uses cylindrical co-ordinate system since this yields a more general
insight into the problem.
Z +
A
dz

R
z

0
P (ρ ,  / 2, 0)
Y
P
/2
AP
X
-
36
dQ  ρ l dz is the elemental change at Z.
The field intensity dE P due to dQ is
dE P 
dQ
 2 â R V / m
4  0 R

where R  ρ â ρ - z â z
and â R 
1
( ρ â ρ - z â z )
R
dQ  ρ l dz C
z 
ρ
 R â ρ - R â z   dE Pρ â ρ  dE P z â z


ρl
ρl
(i) dE Pρ 
ρ dz ; (ii) dE P z  z dz
2
4  0 R
4  0 R 2
dE P 
ρ l dz
4  0 R 2
Taking   OP̂A as integratio n variable , and expressing all distances in terms of ρ and 
ρ
z  ρ tan  , dz  ρ Sec 2  d and R 
 ρ Sec 
Cos 
ρ x ρ x ρ Sec 2 
ρl
(i) dE Pρ  l
d 
Cos  d
3
3
4  0 ρ Sec 
4  0 ρ
E Pρ 
(ii) dE P z
ρl
ρl
ρl
[ Sin  ]- / /22 
x 2 
4  0 ρ
4  0 ρ
2  0 ρ
 ρ l x ρ tan  x ρ Sec 2 
ρl

d 
(- Sin  ) d
3
3
4  0 ρ Sec 
4  0 ρ
ρl
[ Cos ]- / /22  0
4  0 ρ

ρl
 EP 
â ρ V / m
2  0 ρ

Thus, E is radial in direction
EP z 

A2. Find the electric field intensity E at (0, -h, 0) due to a line charge of finite length along Z
axis between A (0, 0, z1) and B(0, 0, z2)
Z
B (0, 0, z2)
2
1
P
dz
A(0, 0, z1)
Y
X
37
Solution :
z 
 h
 R â y - R â z 


ρ l dz
4  0 R 2
dE P 

z2


EP   d EP  -
2
ρl
ρl
Cos  d â y 
4  0 h 1
4  0 h
z1
 Sin  d â
z
1
ρl
ρl
(- Sin  )12 â y 
( Cos )12 â z
4  0 h
4  0 h
 

EP  
2


ρl
(Sin 1 - Sin  2 ) â y  (Cos 1 - Cos  2 ) â z V / m
4  0 h
If the line is extending from -  to  ,
2 

2

EP 
, 1  -

2
- ρl
aˆ y V / m
2  0 h
A3. Two wires AB and CD each 1 m length carry a total charge of 0.2  C and are disposed

as shown. Given BC = 1 m, find E at P, midpoint of BC.
A
B
P
.
C
1m
1m
D
Solution :
(1)
1 = 1800 2 = 1800
A
B
P
1m

E PAB 
ρl
 - (Sin  2 - Sin 1 )  â y  Cos  2 - Cos 1  â z
4  0 h

0
(Indetermi nate)
0
az
(2)
Pay
C
1
1 = - tan-1
2 = 0
1
= - 63.430
0 .5
D
38

E PCD 
ρl
4  0 h
 - (Sin 
- Sin 1 ) â y  (Cos  2 - Cos 1 ) â z



0.2 x 10- 6
- (Sin (-63.43)) â y  (Cos 0 - Cos 63.43) â z
10-9
4
0.5
36 
 3.6 x 103 - 0.894 â y  (1 - 0.447) â z  (-3218 â y  1989.75 â z )


E PCD
2



Since Eρ AB is indeterminate, an alternate method is to be used as under :
Z
dEPz
d
dy
y
B
Y
P
dEPy
A
L
dE P 
R
ρ l dy
â R V / m
4  0 R 2

1
R  (L  d - y) â R ; â R  (-â y )
R
 ρ l â y
dE Py 
dy
4  0 (L  d - y) 2
1
Ld
1
y L;t 
d
Let L  d - y  - t ; - dy  - dt ; y  0 , t 
dE P 
- ρl
dt
4  0 t 2
 ρl
EP   
 4  0

 EP 
1
d

t 1
Ld

ρl  1
1 

4  0  d L  d 
ρl  1
1 
V/ m

4  0  d L  d 
39

0.2 x 10-6  1
1 
E PAB 
â y
-9 
10  0.5 1.5 
4
36 

E PAB  1800 [ 2 - 0.67] â y  2400 â y V/ m



 E P  E PAB  E PCD
 2400 â y - 3218 â y  1990 â z
 (-820 â y  1990 â z )
 2152 â P V / m
where â P  (- 0.381 â y  0.925 â z )

A4. Develop an expression for E due to a charge uniformly distributed over an infinite plane
with a surface charge density of ρS C / m2.
Solution :
Let the plane be perpendicular to Z axis and we shall use Cylindrical Co-ordinates. The
source charge is an infinite plane charge with ρS C / m2 .
dEP Z

AP  R

z
P
0
Y
d

X
A
ρ
AP  AO  OP  - OA  OP

R  ( - ρ â ρ  z â z )
â R 
1
( - ρ â ρ  z â z )
R
The field intensity dE P due to dQ = ρS ds = ρS (d dρ) is along AP and given by
ρ ρ d dρ
ρS
dE P  S
â R 
( - ρ â ρ  z â z ) d ρ dρ
2
4  0 R
4  0 R 3
Since radial components cancel because of symmetry, only z components exist
 dE P 
ρS z
d ρ d ρ
4  0 R 3

ρS
E P   dE P 
4  0
S
2

z ρ dρ
ρS
0 d 0 R 3  4  0 x 2 

zρ
R
3
dρ
0
40
‘z’ is fixed height of ρ above plane and let OP̂A   be integration variable. All distances
are expressed in terms of z and 
ρ = z tan  , d ρ = z Sec2  d ; R = z Sec  ; ρ = 0,  = 0 ; ρ =  ,  =  / 2

ρ
EP  S
2 0

z z tan 
ρS
2
0 z3 Sec3  z Sec  d  2 0
 /2
 Sin  d

0

ρS
[- Cos  ]0 / 2 â z
2 0
ρS
â z (normal to plane)
2 0
A5. Find the force on a point charge of 50 C at P (0, 0, 5) m due to a charge of 500 C that
is uniformly distributed over the circular disc of radius 5 m.
Z
P
h =5 m
0
Y
 ρ
X
Solution :
Given : ρ = 5 m, h = 5 m and Q = 500 C
To find : fp & qp = 50 C
 

ρ
f P  E P x q P where E P  S â z
2 0
Q
 A â z 
2 0

f P  1131 x 103 â z x 50 x 10- 6

f P  56.55 â z N
500  x 10- 6
â z
10-9
2
2 ( 5 ) x
36 
500

x 36  x 103 â z
2 x 25
 1131 x 103 â z N / C
41