This document discusses Gauss's law and related electromagnetic theory concepts taught in a course. It provides: 1) An overview of Gauss's law, which states that the total outward electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of the medium. 2) Derivations and proofs of Gauss's law using calculus theorems like divergence theorem. 3) Applications of Gauss's law to problems involving charge distributions and electric field calculations. 4) Discussions of related concepts like Gauss's law in polarized media, current continuity equation, relaxation time, and Poisson's equation. 5) Example problems demonstrating the application of these electromagnetic theory principles.

1. Course: Electromagnetic Theory
paper code: EI 503
Course Coordinator: Arpan Deyasi
Department of Electronics and Communication Engineering
RCC Institute of Information Technology
Kolkata, India
Topics: Electrostatics – Gauss' Law
27-10-2021 Arpan Deyasi, EM Theory 1
Arpan Deyasi
Electromagnetic
Theory

2. Gauss’ Law
It states that in S.I unit, for a closed surface, surface integral of normal component of
electric field is equal to the total charge enclosed by the surface divided by
permittivity of the medium.
For a closed surface enclosing a charge ‘qenc’
dS
E
n̂
enc
S
q
E.ds =

 S
This is integral form
27-10-2021 Arpan Deyasi, EM Theory 2
Arpan Deyasi
Electromagnetic
Theory

3. Gauss’ Law
is the electric flux through the elementary area ds
E.ds
So the alternative statement is
In S.I unit, total outward electric flux through a closed surface is equal to the total
charge enclosed by the surface divided by permittivity of the medium.
dS
E
27-10-2021 Arpan Deyasi, EM Theory 3
Arpan Deyasi
Electromagnetic
Theory

4. Gauss’ Law
Applying Divergence theorem
S
q
E.ds =


V
q
.Edv
 =

 V V
1
.Edv dv
 = 

 
.E

 =

This is differential form
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Arpan Deyasi
Electromagnetic
Theory

5. Gauss’ Law
Assumptions
Charge distribution is continuous
Electric field is continuous over the closed surface
Medium should be continuous and conductive
Permittivity is constant throughout the medium where potential is defined
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Arpan Deyasi
Electromagnetic
Theory

6. Gauss’ Law
Proof
dS
E
n̂
S
qi
r
i
3
q
E r
4 r
=

Total outward flux of that electric field through an elemental
area dS is
i
3
S S
q
E.ds r.ds
4 r
=

 
27-10-2021 Arpan Deyasi, EM Theory 6
Arpan Deyasi
Electromagnetic
Theory

7. Gauss’ Law
Proof
r.ds
r
represents the projection of area ds on a plane perpendicular to r
3
r.ds
r
 is the ratio of projected area over ‘r2’, which gives the solid angle dω
subtended by dS at origin
dω
i
S S
q
E.ds d
4
= 

 
i
S
q
E.ds 4
4
=  


i
S
q
E.ds =


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Arpan Deyasi
Electromagnetic
Theory

8. 27-10-2021 Arpan Deyasi, EM Theory 8
Problem 1:
Calculate total charge enclosed in a tiny sphere of radius 1 μm centered at (5,8,1) at free
space when electric field is given by
2 2
2 2 2
2xy 3x 2x yz
ˆ ˆ ˆ
E i j k MV/m
1 z 1 z 1 z
= + +
+ + +
Soln:
2 2
2 2 2
2xy 3x 2x yz
ˆ ˆ ˆ
E i j k
1 z 1 z 1 z
= + +
+ + +
(5,8,1)
E 207.38 MV/m
=
Arpan Deyasi
Electromagnetic
Theory

9. 27-10-2021 Arpan Deyasi, EM Theory 9
Applying Gauss’ law
enc
S
q E.ds
=  
enc 0
S
q E.ds
=  
2
enc 0
q E .(4 r )
=  
12 6 6 2
8.854 10 207.38 10 (4 3.14) (10 ) C
− −
=      
enc
q 0.0231pC
=
Arpan Deyasi
Electromagnetic
Theory

10. Difference between Coulomb’s Law and Gauss’ Law
Coulomb’s law deals with attractive or repulsive force between two point charges
Gauss’ law gives the net amount of electric flux associated due to continuous
charge distribution over a closed surface
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Arpan Deyasi
Electromagnetic
Theory

11. Gauss’ Law Coulomb’s Law
E
qi
r
i
S
q
E.ds =


2 i
q
4 r E
 =

i
2
q
E
4 r
=
 
i j
j 2
q q
F Eq
4 r
= =
 
i j
3
q q
F r
4 r
=
 
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Arpan Deyasi
Electromagnetic
Theory

12. Gauss’ Law in presence of Polarized Charge
QP: amount of polarized charge in the closed surface
Qf: amount of free charge in the closed surface
: polarization vector due to polarized charges
P
Total charge contained within the closed surface = f P
(Q Q )
+
By Gauss’ Law
( )
f P
S
1
E.ds Q Q
= +


It states that in S.I unit, for a closed surface, surface integral of normal component of
electric field is equal to the sum of the free and bound charges enclosed by the
surface divided by permittivity of the medium.
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Arpan Deyasi
Electromagnetic
Theory

13. Gauss’ Law in presence of Polarized Charge
P
S
1
P.ds Q
= −


Since
f
S S
Q 1
E.ds P.ds
 = −
 
 
f
S S
1 Q
E.ds P.ds
+ =
 
 
f
S S
E.ds P.ds Q
 + =
 
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Arpan Deyasi
Electromagnetic
Theory

14. Gauss’ Law in presence of Polarized Charge
( ) f
S
E P .ds Q
 + =

f
S
D.ds Q
=

D E P
=  +
where
D is called electric displacement
In S.I unit, total outward electric displacement through a closed surface is equal to
he total free charge enclosed by the surface.
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Arpan Deyasi
Electromagnetic
Theory

15. 27-10-2021 Arpan Deyasi, EM Theory 15
Gauss’ Law in presence of Polarized Charge
Now
S V
D.ds .Ddv
= 
 
f
V
.Ddv Q
  =

f
V V
.Ddv dv
 = 
 
f
.D
 = 
Arpan Deyasi
Electromagnetic
Theory

16. 27-10-2021 Arpan Deyasi, EM Theory 16
Problem 2:
A positive charge ‘Qv’ C/m3 occupies a solid sphere. At a point in the interior at a distance ‘r’
from the center, small charge ‘q’ is inserted. What is the force acting on the small charge?
Soln
f
S
D.ds Q
=
 ( )
2
f
D. 4 r Q
 =
f
V
.Ddv Q
 =
 v f
V
Q dv Q
=

3
v f
4
Q r Q
3
 
 =
 
 
Arpan Deyasi
Electromagnetic
Theory

17. 27-10-2021 Arpan Deyasi, EM Theory 17
Comparing
( )
2 3
v
4
D. 4 r Q r
3
 
 = 
 
 
v
1
D Q r
3
=
v
1
E Q r
3
=

v
q
F Q r
3
=

Arpan Deyasi
Electromagnetic
Theory

18. Current Continuity Equation
Statement: Due to the principle of charge conversation, the time rate of decrease of
charge within a given volume must be equal to the net outward flow
through the closed surface of the volume
Proof:
Current coming out of the closed surface
out
S
I J.ds
= 
out enc
d
I Q
dt
= −
Again
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Arpan Deyasi
Electromagnetic
Theory

19. Current Continuity Equation
enc
S
d
Q J.ds
dt
− = 
Applying Divergence Theorem
S V
J.ds .Jdv
= 
 
enc
V
d
.Jdv Q
dt
 = −

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Arpan Deyasi
Electromagnetic
Theory

20. Current Continuity Equation
V V
d
.Jdv dv
dt
 = − 
 
V V
.Jdv dv
t

 = −

 
.J
t

 = −

.J 0
t

 + =

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Arpan Deyasi
Electromagnetic
Theory

21. 27-10-2021 Arpan Deyasi, EM Theory 21
Problem 3
For a given current density
Calculate [i] total outward current at 1 sec for 5 cm radius,
[ii] Charge density, [iii] velocity of charge carriers
t
1
ˆ
J e r
r
−
=
Soln
t
1
ˆ
J e r
r
−
=
Total outward current ( )
1 2
out
1
I J.S e 4 5 amp
5
−
 
= = 
 
 
out
I 23.1 amp
=
Arpan Deyasi
Electromagnetic
Theory

22. 27-10-2021 Arpan Deyasi, EM Theory 22
.J
t

− = 

t
1
ˆ
. e r
r
−
 
=   
 
2 t
2
1 1
r e
r r r
−
  
=  
  
t
2
1
e
r
−
=
t
2
1
e dt
r
−
 = −
t
2
1
e K
r
−
= +
Arpan Deyasi
Electromagnetic
Theory

23. 27-10-2021 Arpan Deyasi, EM Theory 23
t , 0
→  →
As
t 3
2
1
e C/m
r
−
 =
Velocity is given by
J v
= 
J
v =

t
t
2
1
e
r
v
1
e
r
−
−
=
v r m/sec
=
Arpan Deyasi
Electromagnetic
Theory

24. 27-10-2021 Arpan Deyasi, EM Theory 24
Relaxation Time
Continuity equation states .J 0
t

 + =

. E 0
t

   + =

Ohm’s law gives J E
= 
D
. 0
t

  + =
 
Arpan Deyasi
Electromagnetic
Theory

25. 27-10-2021 Arpan Deyasi, EM Theory 25
Relaxation Time
.D
t
 
 = −
 
f
t
 
 = −
 
At t 0,
= f 0
 = 
f 0 exp t

 
  =  −
 

 
The term (ε/σ) is
called relaxation time
Arpan Deyasi
Electromagnetic
Theory

26. 27-10-2021 Arpan Deyasi, EM Theory 26
Problem 4
Calculate relaxation time for distilled water having relative permittivity 80 and
conductivity 2⨯10-4 mho
Soln
Relaxation time is given by
12
4
80 8.854 10
sec
2 10
−
−
  
 = =
 
3.54 s
 = 
Arpan Deyasi
Electromagnetic
Theory

27. Poisson’s Equation
Differential form of Gauss’ law gives
.E

 =

Relation between electric field and potential gives
E = −
.

  = −

2 
  = −

27-10-2021 Arpan Deyasi, EM Theory 27
Arpan Deyasi
Electromagnetic
Theory

28. 27-10-2021 Arpan Deyasi, EM Theory 28
Problem 5
In free space, . Calculate electric field at point P(1, 2, -5).
Also calculate charge density at the same point
2
6xy z 8
 = +
Soln
E = −
ˆ ˆ ˆ
i j k
x y z
  
= − − −
  
2 2
ˆ ˆ ˆ
6y zi 12xyzj 6xy k
= − − −
Arpan Deyasi
Electromagnetic
Theory

29. 27-10-2021 Arpan Deyasi, EM Theory 29
2 2
(1,2, 5) (1,2, 5)
ˆ ˆ ˆ
E 6y zi 12xyzj 6xy k
− −
 = − − −
(1,2, 5)
ˆ ˆ ˆ
E 120i 120j 24k
−
= + −
2 2 2
2
2 2 2
x y z
     
  = + +
  
12xz
=
Arpan Deyasi
Electromagnetic
Theory

30. 27-10-2021 Arpan Deyasi, EM Theory 30
2
0
 = −  
0
12 xz
= − 
12 3
(1,2, 5)
12 8.854 10 1 5 C/m
−
−
 =    
3
(1,2, 5)
531.24 pC/m
−
 =
Arpan Deyasi
Electromagnetic
Theory

31. Laplace’s Equation
For charge-free region
0
 =
2
0
  =
27-10-2021 Arpan Deyasi, EM Theory 31
Poisson’s equation is given by
2 
  = −

Arpan Deyasi
Electromagnetic
Theory

32. 27-10-2021 Arpan Deyasi, EM Theory 32
Show the potential field exist in a dielectric medium
having ϵ=2ϵ0 does not satisfy Laplace's equation. Also calculate the total charge
within the unit cube 0<x, y, z<1 m by this potential field
Problem 6
2 3
2x yz y z
 = −
Soln
2 2 2
2
2 2 2
x y z
     
  = + +
  
2
2yz
  = − 2
0
  
Hence the potential does not satisfy Laplace’s equation
Arpan Deyasi
Electromagnetic
Theory

33. 27-10-2021 Arpan Deyasi, EM Theory 33
2
0
 = −  
0
2 yz
 = 
V
Q dv
= 

1 1 1
0
0 0 0
Q 2 yzdxdydz
= 

Q 8.854pC
=
Arpan Deyasi
Electromagnetic
Theory

34. Significance of Poisson’s & Laplace’s Equation
Any electrostatic problem can be solved by using Poisson’s equation or Laplace’s
Equation subject to the appropriate boundary conditions. The obtained solutions
are physically reasonable and potential can be determined at all the points inside
the region where the problem is specified.
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Arpan Deyasi
Electromagnetic
Theory

35. Uniqueness Theorem
1st Uniqueness Theorem:
Solution of Laplace’s equation in some region is uniquely determined if the value of
potential is a specified function on all boundaries of that region
Proof
We consider a Laplacian region and its boundary.
We further consider that two different solutions φ1 & φ2 satisfy Laplace’s equation
2
1 0
  =
2
2 0
  =
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Arpan Deyasi
Electromagnetic
Theory

36. Uniqueness Theorem
Let
3 1 2
 =  
Therefore, φ3 should obey Laplace’s equation
2
3 0
  =
2 2
1 2 0
    =
1 2 K
  =
So, potential can uniquely be determined
27-10-2021 Arpan Deyasi, EM Theory 36
Arpan Deyasi
Electromagnetic
Theory

37. Uniqueness Theorem
2nd Uniqueness Theorem:
In a region containing conductors and filled with specified charge density, electric field is
uniquely determined if the total charge on each conductor is given
Proof
We consider two electric fields & satisfy the Gauss’ Law
1
E 2
E
1
.E

 =

2
.E

 =

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Arpan Deyasi
Electromagnetic
Theory

38. Uniqueness Theorem
Let
3 1 2
E E E
=
Therefore, should obey Laplace’s equation
3
E
( ) ( ) ( )
3 3 3 3 3 3
. E .E E .
  =   + 
( ) ( )
3 3 3 3 3
. E 0 E .
  =   + 
( ) 2
3 3 3
. E E
  = −
27-10-2021 Arpan Deyasi, EM Theory 38
Arpan Deyasi
Electromagnetic
Theory

39. Uniqueness Theorem
( ) 2
3 3 3
V V
. E dv E dv
  = −
 
2
3 3 3
V V
.E dv E dv
  = −
 
2
3
V
E dv 0
− =
 3
E 0
=
1 2
E E
=
27-10-2021 Arpan Deyasi, EM Theory 39
Arpan Deyasi
Electromagnetic
Theory

40. Equipotential Surface
Relation between electric field and potential gives
E = −
0
 =
Equipotential surface gives
E 0
=
Therefore, for equipotential surface, parallel electric field is zero,
i.e., conductor has equipotential surface
27-10-2021 Arpan Deyasi, EM Theory 40
Arpan Deyasi
Electromagnetic
Theory

41. Equipotential Surface
.E 0
 =
0

=

0
 =
Therefore, all charges must reside entirely on the surface of conductor
27-10-2021 Arpan Deyasi, EM Theory 41
Arpan Deyasi
Electromagnetic
Theory

42. 28-10-2021 Arpan Deyasi, EM Theory 42
Problem 7
Show the electric field is perpendicular to equipotential surface
Soln
Force acting on charge ‘q’ F qE
=
Work done by the force W F.ds
=
W Fdscos( )
= 
W qEdscos( )
= 
Arpan Deyasi
Electromagnetic
Theory

43. 28-10-2021 Arpan Deyasi, EM Theory 43
For equipotential surface, net work done is zero
W 0
=
qEdscos( ) 0
 =
2

 =
Arpan Deyasi
Electromagnetic
Theory