Electromagnetics Quick revision

ELECTROMAGNETISM
Electric Flux:
The number of lines of force passing through a given area is known as electric flux.
The scalar product E.dS is defined as the electric flux for the surface. It is given by
 
 E   E.ds
s


N  m2
Unit
coulomb
Electric flux is a property of an electric field. Electric field lines are usually considered
to start on positive electric charges and to end on negative charges. If there is no given
net charge within a given closed surface then every field line directed into the given surface
continues through the interior and is usually directed outward elsewhere on the surface. The
negative flux just equals in magnitude the positive flux, so that the net or total, electric flux is
zero. If a net charge is contained inside a closed surface, the total flux through the surface is
proportional to the enclosed charge, positive if it is positive, negative if it is negative.


Gauss law in Electrostatics:
The mathematical relation between electric flux and the enclosed charge is known
as Gauss law for the electric field. It is one of the fundamental laws of electromagnetism. Gauss’s
law states that the total electric flux (ϕE) over a closed surface is equal to 1/ε0 times the total
charge Q enclosed within the surface.
  q
 E   E.dS 
S
0
Here S is known as Gaussian surface and ε0 is the permittivity of the free space.
In a dielectric medium Gauss’s law is given by
  q
 E   E.dS  where ε is the permittivity of free space
S

Proof:
Let a charge +Q is placed at O within a closed surface of irregular shape. Consider a
point P on the surface at a distance r from O. Now take a small area ds around P. The normal
to the surface ds is represented by a vector ds which makes an angle θ with the direction of
electric field E along OP. The electric flux dΦ E outwards through the area ds is given by
d E  E.dS  EdS cos …… (1)


Dr. P. Venkata Ramana, AUCE (A)
1

, From coulombs law, the electric intensity E at a point P distance r from a point charge
Q is given by
1 q
E   ……. (2)
40  r 2 
From eqn (1) and (2), we get
q
d E  .dS cos
40 r 2

q  dS cos 
d E   
40  r 2 

 dS cos  
But   is the solid angle dω subtended by dS at O. Hence
 r 
2



q
d E  d …… (3)
40
The total flux ΦE over the entire whole surface is given by
q
E 
40  d
Where  d is the solid angle subtended by whole surface at O. This is equal to 4π.
q
E   4
40

q
E  ….. (4)
0
Let the closed surface encloses several charges say +q1, +q2, +q3 ,….. +qʹ1, +qʹ2, +qʹ3
,….. Now each charge will contribute to the total electric flux. Here it should be remembered
that for positive charges, the electric flux will be outward and hence positive while for the


Dr. P. Venkata Ramana, AUCE (A)
2

, negative charges, the electric flux will be inward and hence negative. Thus the total flux is
given by

E 
1
q1  q2  q3  .......  q1  q2  q3 
0
1
E 
0
q where q is algebraic sum of all charges

So the total normal electric flux over the closed surface is equal to 1/ε0 times the total
charges enclosed within the surface which established the Gauss’s law.
Gauss’s Law in Differential form:
  Q
Gauss’s law is given by  .dS 
E
S
0
For a charge distribution
Q   dv where ρ is the charge density
V

Using Gauss divergence theorem

 E.ds   .E dv
S V


1
 0 V
 dv   .E dv
V


 
  .E  
V 0
dv  0


 .E  0
0

 .E  This is the differential form of Gauss’s law
0
 The charges enclosed by the surface may be point charges or continuous charge
distribution.
 The net electric flux may be outward or inward depending upon the sign of charges.
 Electric flux is independent of shape and size of Gaussian surface.
 The Gaussian surface can be chosen to have a suitable geometrical shape for
evaluation of flux.
Limitation of Gauss’s law:
(a) Since flux is a scalar quantity Gauss’s law enables us to find the magnitude of electric
field only.
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Dr. P. Venkata Ramana, AUCE (A)