PHYSICS-II (PHY C132)
ELECTRICITY & MAGNETISM
Introduction to Electrodynamics:
by David J. Griffiths (3rd Ed.)
Four basic forces of nature:
1. Strong nuclear force: short range
2.Electromagnetic
3.Weak: short range
4.Gravitational: Long range but 1040 times
weaker than EM force!
Hence the most common force is the
electromagnetic force: It is the force which
holds you together. Friction, normal reaction
are all Electromagnetic in nature
Electromagnetism deals with
electromagnetic force and field.
The unification of forces & phenomena
Electromagnetism
Electricity
Magnetism
Optics
Course Outline:
•Electrostatics: Coulomb’s law, Gauss law
•Magnetostatics: Ampere’s law
•Electrodynamics: Faraday’s law, Lenz’s law
So what’s new?
Use of Vector analysis
VECTOR ANALYSIS
 Differential Calculus (Revisited)
 Integral Calculus (Revisited)
 Curvilinear Coordinates (Revisited)
 The Dirac Delta Function
 Theory of Vector Fields
In Electromagmetism, we deal with difference
in potential between two points, so we require
differential calculus &
When we talk about work done in moving a
(unit) charge, we encounter integral calculus
 
  E.dl
Differential Calculus
Derivative of any function f(x,y,z):
 f
 f 
df  
dx  
 x 
 y

 f 
dy  
dz
 z 


 f
f
f  ˆ
  i dx  jˆdy  kˆdz
df   iˆ
 jˆ
 kˆ
y
z 
 x


 f  dl

  
where

f
f
f
ˆ
ˆ
ˆ
f  i
 j
k
x
y
z
Gradient of function f
Gradient of a function
Change in a scalar function f corresponding
to a change in position dr
 


df  f  dr 
 f is a VECTOR
Geometrical interpretation of Gradient
Z

f
P
X
dr
Q
 
f (x , y , z)  C
Y

change in f : df  f  dr 
=0
=> f  dr
Z
Q
f  C2  C1
dr
P
f  C1
Y
X
 


df  C2  C1  f  dr 
For a given |dr|, the change in scalar function
f(x,y,z) is maximum when:
 
dr || f
=> f is a vector along the direction of
maximum rate of change of the function
Magnitude: slope along this maximal direction
If  f = 0 at some point (x0,y0,z0)
=> df = 0 for small displacements
about the point (x0,y0,z0)
(x0,y0,z0) is a stationary point of f(x,y,z)
Prob. 1.12
The height of a certain hill (in feet) is:
h(x,y) = 10(2xy – 3x2 -4y2 -18x + 28y +12)
where x is distance (in mile) east and y north of
BPHC.
(a) Where is the top located ?
Ans: 3 miles North & 2 miles West
Prob. 1.12 (contd.)
h(x,y) = 10(2xy – 3x2 -4y2 -18x – 28y +12)
(b) How high is the hill ?
Ans: 720 ft
(c) How steep is the slope at 1 mile north and 1
mile east of BPHC? In what direction the slope
is steepest, at that point ?
Ans: 311 ft/mile, direction is Northwest
How do we compute the gradient in other coordinate systems?
We identify a point by its three coordinates:
u,v,w. We assume that the system is
orthogonal, i.e., the unit vectors are mutually
perpendicular.
The infinitesimal displacement vector from
(u,v,w) to (u+du,v+dv,w+dw) can be written

dl  fduuˆ  gdvvˆ  hdwwˆ
f,g and h are the scale factors. These are
needed since we have to get the component
of the displacement along each axis.
For Cartesian system, f = g = h =1
Spherical polar coordinates:
f  1, g  r , h  r sin 
Cylindrical polar coordinates
f  1, g  r , h  1
If you move from point (u,v,w) to point (u+du,
v+dv, w + dw), a scalar function t(u,v,w)
changes by an amount
 t 
 t 
 t 
dt  
du

dv





dw
 u 
 v 
 w 
We can write it as a dot product:
  


df  f  dl
 1 t
1 t
1 t 
   fduuˆ  gdvvˆ  hdwwˆ 
  uˆ
 vˆ
 wˆ
g v
h w 
 f u
The gradient of t then is:
1 t
1 t
1 t
t  uˆ
 vˆ
 wˆ
f u
g v
h w
Prob. 1.13
Let rs be the separation vector from
(x,y,z) to (x,y,z) .

 2
a  rs   2rs
 1
b  
 rs
rˆs

   2
rs

 n
n 1
c  rs   nrs rˆs
The Operator 




  iˆ
 jˆ
 kˆ
x
y
z
 is NOT a VECTOR,
but a VECTOR OPERATOR
Satisfies:
•Vector rules
•Partial differentiation rules
 can act:
 On a scalar function f : f
GRADIENT
 On a vector function F as: . F
DIVERGENCE
 On a vector function F as:  × F
CURL
Divergence of a vector

   

  ˆ
ˆ
ˆ
ˆ
  i Fx  jˆFy  kˆFz
  F   i
 j
k
y
z 
 x

  Fx
Fy
Fz
F 


x
y
z
Divergence of a vector is a scalar.
.F is a measure of how much the vector F
spreads out (diverges) from the point in
question.
Which of these vector fields
have non-zero divergence?

v  constant
 r̂
v 2
r
What about the divergence of
this field?
 yˆ
v
y
Physical interpretation of Divergence
Flow of a compressible fluid:
(x,y,z) -> density of the fluid at a point (x,y,z)
v(x,y,z) -> velocity of the fluid at (x,y,z)
(rate of flow in)EFGH
(rate of flow out)ABCD
Z
 ρv x |x 0 dydz
 ρv x |x dx dydz



 ρv x dx  dydz
  ρv x 
x

 x 0
H
D
dz
A
X
G
C
E
F
dx
dy
B
Y
Net rate of flow out (along- x)




ρv x dxdydz
x
Net rate of flow out through all pairs
of surfaces (per unit time):
 



 ρv x  
 ρv z  dxdydz

ρv y 
y
z
 x

 


    ρv dxdydz
Net rate of flow of the fluid
per unit volume per unit time:


    ρv 
DIVERGENCE
Divergence in curvilinear coordinates:
û
v̂
(u,v,w)
ŵ
Suppose that we have to evaluate the integral of
a vector function

A(u, v, w)  Au uˆ  Av vˆ  Aw wˆ
over the surface of an infinitesimal volume
Volume of the infinitesimal volume element =
d   fghdudvdw
 
For the front surface: A.da  ( ghA )dvdw
u
For the back surface:
  
 ( ghAu ) 
A.da   ghAu 
du  dvdw
u


The front and back surfaces together give:
1 



 u ( ghAu ) dudvdw  fgh  u ( ghAu ) d




Considering all the surfaces we get:
 
1 



 A.da  fgh  u ( ghAu )  v ( fhAv )  w ( fgAw )d
Using the divergence theorem, we obtain:

1 



. A 
( ghAu )  ( fhAv ) 
( fgAw )

fgh  u
v
w

Example:
 
  r f r 
df r 
 3f r   r
dr

n 1
n
  rˆ r  n  2 r
Prob. 1.16

rˆ
v  2
and compute its divergence.
r
Sketch the vector function
 
 v  0
Does the result surprise you? Plot this vector field
Curl
î
 
  F   / x
Fx
fy
 fz
ˆ
 i 

z
 y

 

ĵ
k̂
 / y
 / z
Fy
Fz
ˆj  f x  fz   kˆ  fy  f x
 x
x 
y
 z




Curl of a vector is a vector
×F is a measure of how much the vector F
“curls around” the point in question.
Physical significance of Curl
Circulation of a fluid around a loop:
x 0  dx , y 0  dy 
x , y  dy 
Y
3
0
0
2
4
x 0 , y 0 
x 0  dx , y 0 
1
X
Circulation (1234)   Vx dx   V y d y
1
2
  Vx dx   V y d y
3
4
∂
Vy


 v x ( x 0 , y 0 )dx  v y ( x 0 , y 0 ) 
dx dy
∂
x




∂
Vx
 v x ( x 0 , y 0 ) 
dy  (-dx )  v y ( x 0 , y 0 )( -dy )
∂
y


Vy ∂
∂
Vx 
dxdy
 
x
∂
y 
 ∂
Circulation per unit area = ( × V )|z
z-component of CURL
Curl in curvilinear coordinates:
v̂
To obtain the curl in
curvilinear co-ordinates, we
calculate the line integral
 
 A.dl
û
Along the bottom segment:
 
A.dl  ( fAu )du
Along the top leg, the sign is reversed and fAu
is evaluated at (v + dv). Therefore along the
top leg:
 
A.dl  ( fAu ) v  dv du
Together, the two edges give us:
 
  ( fAu ) 
A.dl   
dudv


v


Similarly, the right and left sides together
yield:

    ( gAv ) 
A.dl  
dudv
d
a

 u 
    ( gAv )  ( fAu ) 
 A.dl   u  v  dudv
  1   ( gAv )  ( fAu )  
 A.dl  fg  u  v  wˆ .da
 ( fg )dudvwˆ
 1  (hAw ) ( gAv ) 
ˆ
 A  

u


gh  v
w 
1  ( fAu ) (hAw )  1  ( gAv ) ( fAu ) 
ˆ
ˆ

v


w
fh  w
u  fg  u
v 
fuˆ

1 
 A 
fgh u
fAu
gvˆ

v
gAv
hwˆ

w
hAw
Sum Rules
For Gradient:



∇f1  f2   ∇f1  ∇f2
For Divergence:

For Curl: 


 
 
  F1  F2    F1    F2

 



 
 
 






  F1  F2    F1    F2
Rules for multiplying by a constant
For Gradient:
For Divergence:
For Curl:


 kf   kf
 


 




 
  kF  k   F


 
  kF  k   F
Product Rules
For Gradients:



∇f1f2   f1∇f2  f2∇f1






     
  
 A B  A   B  B    A
     
 A B  B  A
 


Try to prove the product rules one component
at a time:
 
x component of (A.B) 

 Ax Bx  Ay B y  Az Bz 
x
Expanding this we obtain
Product Rules
For Divergences:

 

 
  f F  f   F  F  f

 





 

  
  
  
  F1  F2  F2    F1  F1.   F2

Product Rules
For Curls:
 


 

 






  f F  f   F  F  f





  
  

  F1  F2  F2   F1  F1   F2
  
  
 F1   F2  F2   F1




Quotient Rules

 F 
   
f 


  f1  f2f1  f1f2
  
2
f2
 f2 
 
 
f   F  F  f

f2


 


 





  F  f   F  F  f
    
2
f
f
 
Second Derivatives
Divergence of
gradient:
 
 
  f   f
Curl of gradient:
2
Laplacian of f
 0


  f
Prob. 1.27: Prove it !
Second Derivatives
Gradient of divergence:
Divergence of Curl:
  
 F


 
 F

 0
Prob. 1.26: Prove it !
Second Derivatives
Curl of Curl:







  
    F     F   2F
Recall  Prob. 1.16
Sketch the vector function and
compute its Divergence

rˆ
v  2
r
 
1   2 1 
 v  2
0
r
2 
r r 
r 
The volume integral of v:


τ

 
  v dτ  0
Surface integral of v over a sphere of
radius R:


 v  dA
S
 4π
From divergence theorem:


τ

 
  v dτ
 4π
Calculation of Divergence =>



 
  v dτ  0
τ
Divergence theorem =>


τ

 
  v dτ  4π
Note: as r  0; v  ∞
 
  v  0, everywhere but;
 0,
at r  0
And integral of v over any volume
containing the point r = 0


τ

 
  v dτ  4π
THE DIRAC DELTA FUNCTION
0
δ x   


and
if x  0 

if x  0
 δ x dx

1
The Dirac Delta Function
An infinitely high,
infinitesimally narrow
“spike” with area 1
(x) NOT a Function
But a Generalized Function
OR distribution
Properties:
f x  δ  x   f 0 δ  x 

 f  x δ x dx


 f 0   δ x  dx  f 0 

Delta function is something that is to be
always used under an integral sign.

If
 f  x  D  x dx
1

then,


 f  x  D  x dx
2

D1 x   D2  x 
[ D1(x) & D2(x) are two expressions involving Delta
functions and f(x) is any ordinary function. ]
One can show:
1
δkx  
δ x 
|k |
 δ  x   δ  x 
………..for a proof, see Example 1.15
The Dirac Delta Function
Shifting the singularity from 0 to a;
0
δ x  a   


and
if x  a 

if x  a 


δ
x

a
dx

1


The Dirac Delta Function
Properties:
f x  δ x  a  f a δ x  a

 f  x δ x  a dx

 f a 
Prob. 1.43:
a  :  3 x 2  2 x  1 δ x  3 dx
6
2
3
c  :  x 3 δ x  1dx
0
 20
0
Prob. 1.45 :
a  Pr ove :
b 
d
δ x    δ x 
x
dx
1
θx   
0
if x  0

if x  0
dθ
To show :
 δ x 
dx
The Dirac Delta Function
(in three dimension)

0
δ r   

3
everywhere but ;

 at 0,0,0 




δ
r
d
τ

1

3
all space
Why such a function ?
Describe very short range forces like
nuclear force
Describe a point particle in terms of a mass
density
Describe a point charge in terms of a
charge density
Prob. 1.46:
Charge density of a point charge q at r :


3 
ρr   qδ r  r  
Charge density of a dipole with -q at 0 and +q at a:




3
3
ρr   qδ r  a   qδ r 
Prob. 1.46: (contd.)
Charge density of a thin spherical shell of radius R
and total charge Q:
Q


ρr  
δ
r

R
2
4 πR
The Paradox of Divergence of
  rˆ
   2
r



From calculation of Divergence:
   rˆ
τ    r 2

  dτ  0

By using the Divergence theorem:
   rˆ
τ    r 2

  dτ  4π

So now we can write:
  rˆ 
3 
   2   4 πδ r 
r 
Such that:
   rˆ




τ   r 2

  dτ  4π

Theory of Vector Fields
By specifying appropriate
boundary conditions, Helmholtz
theorem implies that the field
can be uniquely determined from
its divergence and curl.
Given the values of the divergence and the
curl of a vector field, the field can always be
determined.
But there is a catch!
The fields have to go zero at infinity.
Eg: Can we have a vector field whose
divergence and curl are both zero?
Yes!

v  yxˆ  xyˆ
But then there other fields, for which the curl
and divergence are zero.

v  yzxˆ  zxyˆ  xyzˆ
So in this case the divergence and curl do not
uniquely specify the vector field. That is
because the field does not go to zero at
infinity.
Potentials
THEOREM 1:

( For Curl-less fields )
  F  0 everywhere 
b

 F  dl  path independent
a


 F  dl
 0  closed path
F  V
V
: a scalar potential 
Conclusions from theorem 1

 

  F  0  F  V
If curl of a vector field vanishes,
(everywhere), then the field can always be
written as the gradient of a scalar potential
( not unique )
Potentials
THEOREM 2:
For Divergence-less fields
   F  0 everywhere 

 F  da  surface independen t
s
  F  da  0  closed surface
 F  A
 A : a vector
potential 
Conclusions from theorem 2
  
 
F  0  F   A
If divergence of a vector field vanishes,
(everywhere), then the field can always be
written as the curl of a vector potential
( not unique )
Helmholtz theorem:
Any vector field F with both source and circulation
densities vanishing at infinity may be written as
the sum of two parts: one of which is curl-less
and the other is divergence-less.


 
F  V    A
(Always)