This document discusses Coulomb's law and some key concepts in electrostatics, including: - Coulomb's law describes the electrostatic force between two point charges, being directly proportional to the product of the charges and inversely proportional to the square of the distance between them. - The electric field intensity and electric flux density are introduced. - Properties of the electric field such as it being irrotational are examined. - The electrostatic potential is defined and its relationship to the electric field is explored.

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 –Coulomb's Law
27-10-2021 Arpan Deyasi, EM Theory 1
Arpan Deyasi
Electromagnetic
Theory

2. Coulomb's Law
qi
qj
rij
ij i j
F q q

ij 2
ij
1
F
r

Two stationary point charges qi and qj separated by distance rij exert force Fij on each other
Fij is proportional to the product
of magnitude of charges of the
point charges
Fij is inversely proportional to the square of the
distance (rij ) of separation of point charges
27-10-2021 Arpan Deyasi, EM Theory 2
Arpan Deyasi
Electromagnetic
Theory

3. Coulomb's Law
In vector notation
i j
ij ij
2
ij
q q
ˆ
F r
4 r
=

Statement: two point charges exert forces on each other along the line joining between
them, and this force, repulsive for like charges, and attractive for unlike
charges, are directly proportional to the product of the charges, and
inversely proportional to the square of the distance between them
27-10-2021 Arpan Deyasi, EM Theory 3
Arpan Deyasi
Electromagnetic
Theory

4. Coulomb's Law
Assumptions
Charges are point charges, spherically symmetric
Charges are generally static, but the law is applicable
for moving charges if v<<c
Medium is continuous and conductive
Minimum separation distance between charges is 10-15 m
27-10-2021 Arpan Deyasi, EM Theory 4
Arpan Deyasi
Electromagnetic
Theory

5. Coulomb's Law
Q. Why the factor 4π has been introduced?
Because of spherical nature of point charges, 4π term appears
in the numerator. So in order to simplify the mathematical calculation,
one 4π term is invoked into the denominator. Therefore, final result
will become simple.
Q. Where Coulomb’s law fails?
For nuclear interaction, Coulomb’s law fails
27-10-2021 Arpan Deyasi, EM Theory 5
Arpan Deyasi
Electromagnetic
Theory

6. Coulomb's Law Newton’s Law
i j
ij ij
2
ij
q q
ˆ
F r
4 r
=

Force applied on qj due to qi
Force applied on qi due to qj
j i
ji ji
2
ji
q q
ˆ
F r
4 r
=

27-10-2021 Arpan Deyasi, EM Theory 6
Arpan Deyasi
Electromagnetic
Theory

7. Coulomb's Law Newton’s Law
i j j i
ij ji ij ji
2 2
ij ji
q q q q
ˆ ˆ
F F r r
4 r 4 r
+ = +
 
i j
ij ji ij ji
2
ij
q q
ˆ ˆ
F F r r
4 r
 
+ = +
 

i j
ij ji 2
ij
q q
F F 0
4 r
+ =
 ij ji
F F 0
+ =
27-10-2021 Arpan Deyasi, EM Theory 7
Arpan Deyasi
Electromagnetic
Theory

8. Problem 1:
Calculate the force of interaction between two charges of values 4 ⨯ 10-8 C and 6⨯10-5 C
and spaced 10 cm apart. Assume the medium has permittivity 2.
i j
2
q q
F
4 r
=

Soln:
i j
2
0 r
q q
F
4 r
=
 
8 5
2
0
(4 10 ) (6 10 )
F
4 2 (0.1)
− −
  
=
  
27-10-2021 Arpan Deyasi, EM Theory 8
Arpan Deyasi
Electromagnetic
Theory

9. Problem 1:
8 5
9
2
(4 10 ) (6 10 )
F 9 10 N
2 (0.1)
− −
  
=  

F 1.08N
=
27-10-2021 Arpan Deyasi, EM Theory 9
Arpan Deyasi
Electromagnetic
Theory

10. Superposition Theorem
qt
qi
qj
q1
q2
qn
Fti
Ftj
Ft1
Ft2
Ftn
Assume: [i] charge distribution is continuous
[ii] medium is continuous and conductive
[iii] nature of interaction is non-nuclear
[iv] Forces acting on the test charge
are mutually independent
t t1 t2 ti tj tn
F F F ... F F ... F
= + + + + + +
t 1 t 2
t t1 t2
2 2
t1 t2
t j
t i t n
ti tj tn
2 2 2
ti tn
tj
q q q q
ˆ ˆ
F r r ...
4 r 4 r
q q
q q q q
ˆ ˆ ˆ
r r ... r
4 r 4 r
4 r
= + + +
 
+ + +
 

27-10-2021 Arpan Deyasi, EM Theory 10
Arpan Deyasi
Electromagnetic
Theory

11. Superposition Theorem
j
t 1 2 i n
t t1 t2 ti tj tn
2 2 2 2 2
t1 t2 ti tn
tj
q
q q q q q
ˆ ˆ ˆ ˆ ˆ
F r r ... r r ... r
4 r r r r
r
 
 
= + + + + + +
  
 
n
t i
t ti
2
i 1( t) ti
q q
ˆ
F r
4 r
= 
=


27-10-2021 Arpan Deyasi, EM Theory 11
Arpan Deyasi
Electromagnetic
Theory

12. Problem 2:
Three charges of 0.25 μC are placed at the vertices of an equilateral triangle whose
sides is 100 μm. Determine the magnitude and direction of the resultant force on
one charge due to the other charges.
Soln:
A
B C
F
BA
F
CA
F
i j
BA CA 2
q q
F F
4 r
= =

i j
2
0
q q
F
4 r
=

30° 30°
27-10-2021 Arpan Deyasi, EM Theory 12
Arpan Deyasi
Electromagnetic
Theory

13. Problem 2:
9 6 6
2
9 10 0.25 10 0.25 10
F N
(0.1)
− −
    
=
3
F 56.25 10 N
−
= 
net BA CA BA
F F F 2 F
= + =
3
net
F 2 56.25 10 cos30 N
−
=   
3
net
F 97.425 10 N
−
= 
27-10-2021 Arpan Deyasi, EM Theory 13
Arpan Deyasi
Electromagnetic
Theory

14. 27-10-2021 Arpan Deyasi, EM Theory 14
Electric Field Intensity
Field Intensity (E) at an external point due to a point charge measured at any
continuous conductive medium is the force on an unit positive charge placed
at that point.
+q
P
r
A
Intensity (E) at P will be given by
2
q 1
E
4 r

=

2
q
ˆ
E r
4 r
 =

3
q
E r
4 r
=

Arpan Deyasi
Electromagnetic
Theory

15. 27-10-2021 Arpan Deyasi, EM Theory 15
Electric Flux Density
Flux density (D) is the field Intensity measured at an external point in any
continuous conductive medium multiplied with the permittivity of that
medium.
Flux Density (D) at P will be given by
D E
= 
2
q
D
4 r
=

3
q
D r
4 r
=

+q
P
r
A
It is also called electric displacement
Arpan Deyasi
Electromagnetic
Theory

16. Properties of Electric Field
Q. Show that electric field is irrotational
i j
ij ij
2
ij
q q
ˆ
F r
4 r
=

i j
ij ij
3
ij
q q
F r
4 r
=

In simple notation
According to Coulomb’s law
i j
3
q q
F r
4 r
=

Electric field
3
q
E r
4 r
=

27-10-2021 Arpan Deyasi, EM Theory 16
Arpan Deyasi
Electromagnetic
Theory

17. Properties of Electric Field
3
q
E r
4 r
 = 

3
q r
E
4 r
 
 = 
 
 
 
3 3
q 1 1
E r r
4 r r
 
 
 
 =  +  
 
 
  
 
27-10-2021 Arpan Deyasi, EM Theory 17
Arpan Deyasi
Electromagnetic
Theory

18. Properties of Electric Field
3
q 1
E 0 r
4 r
 
 
 
 = +  
 
 
  
 
( )
( )
5
q
E 0 3 r r r
4
−
 = − 

( )
( 2)
n
n r r
 −
 =
Since So ( )
5
3
1
3 r r
r
−
 
 
 = −
 
 
E 0
 =
27-10-2021 Arpan Deyasi, EM Theory 18
Arpan Deyasi
Electromagnetic
Theory

19. Electrostatic Potential
Electrostatic potential (φ) at an external point due to a point charge is the
work done externally against the field as an unit +ve charge is brought from
infinity to the point.
Alternatively it is the work done by the field as an unit positive charge is
removed to infinity from the point.
+q
P
r
A
Force on unit +ve charge at P
2
q 1
F
4 r

=

27-10-2021 Arpan Deyasi, EM Theory 19
Arpan Deyasi
Electromagnetic
Theory

20. Electrostatic Potential
As the unit +ve charge is shifted by distance ‘dr’, the work done by the field is given by
2
q
dw dr
4 r
=

Net work done by the field as unit +ve charge is shifted from P to infinity = Potential at P
r
2
r r
q dr
4 r
=
=
 =
 
r
2 1
r r
q r
4 2 1
=
− +
=
 
 =  
 − +
 
27-10-2021 Arpan Deyasi, EM Theory 20
Arpan Deyasi
Electromagnetic
Theory

21. Electrostatic Potential
r
r r
q 1
4 r
=
=
−  
 =  
 
q 1 1
4 r
−  
 = −
 
 
 
q
4 r
 =

27-10-2021 Arpan Deyasi, EM Theory 21
Arpan Deyasi
Electromagnetic
Theory

22. Relation between Electric Field and Potential
O
P
Q
r
r dr
+
Potential at P is φ1
Potential at Q is φ2
Work done W E.dr
=
1 2
E.dr =  − 
1 2
E.dr (r) (r dr)
=  −  +
1 1 1
E.dr (r) (r) .dr
 
=  −  + 
 
27-10-2021 Arpan Deyasi, EM Theory 22
Arpan Deyasi
Electromagnetic
Theory

23. Relation between Electric Field and Potential
1
E.dr .dr
= −
Using general notation
E.dr .dr
= −
E .dr 0
 
+  =
 
E = −
27-10-2021 Arpan Deyasi, EM Theory 23
Arpan Deyasi
Electromagnetic
Theory

24. Problem 3:
A potential field is given by V. Calculate electric field at P (2,-1,4).
Soln:
( )
2
3x y yz
 = −
E = −
( )
2
ˆ ˆ ˆ
i j k 3x y yz
x y z
 
  
= − + + −
 
  
 
( ) ( ) ( )
2 2 2
ˆ ˆ ˆ
i 3x y yz j 3x y yz k 3x y yz
x y z
  
= − − − − − −
  
27-10-2021 Arpan Deyasi, EM Theory 24
Arpan Deyasi
Electromagnetic
Theory

25. Problem 3:
( ) ( ) ( )
2
ˆ ˆ ˆ
i 3y 2x j 3x z k y
= −  − − − −
( )
  ( ) ( )
2
2, 1,4
ˆ ˆ ˆ
E i 3( 1) 2 2 j 3 2 4 k 1
−
 = − −   −  − + −
( )
2, 1,4
ˆ ˆ ˆ
E 12i 8j k
−
= − −
27-10-2021 Arpan Deyasi, EM Theory 25
Arpan Deyasi
Electromagnetic
Theory

26. 27-10-2021 Arpan Deyasi, EM Theory 26
Problem 4:
Calculate potential at A(1) due to B(3) where electric field is given by
60
E V/ m
r
=
Soln: A
B
E.dr
 = −
1
3
60
dr
r
= −
1
3
60ln(r)
= −
65.925
= volt
Arpan Deyasi
Electromagnetic
Theory

27. 27-10-2021 Arpan Deyasi, EM Theory 27
Electrostatic Energy
It is defined as the work done per unit positive charge
Work done in moving a small charge dq against potential difference V is
dW Vdq
=
q
dW dq
C
=
If the capacitor is initially uncharged and the process of charging continued till a charge Q
is achieved, then total work done
Q
W
0 0
q
dW dq
C
=
 
Arpan Deyasi
Electromagnetic
Theory

28. 27-10-2021 Arpan Deyasi, EM Theory 28
Electrostatic Energy
2
1
W CV
2
=
2
1 Q
W
2 C
=
2 2
1
W C V
2C
=
Arpan Deyasi
Electromagnetic
Theory

29. 27-10-2021 Arpan Deyasi, EM Theory 29
Then energy ΔW stored in the unit volume ΔV is
( )( )
2
1
W C V
2
 =  
Electrostatic Energy
Small increment in capacitance ( )
C d
 =  
Small increment in potential energy ( )
V E d
 = 
Arpan Deyasi
Electromagnetic
Theory

30. 27-10-2021 Arpan Deyasi, EM Theory 30
Electrostatic Energy
( )( )
2
1
W d E d
2
 =  
( )
3
2
1
W E d
2
 =  
( )
2
1
W E V
2
 =  
Energy density is the energy per unit volume (w) given by
2
V 0
W 1
w Lt E
V 2
 →

= = 

Arpan Deyasi
Electromagnetic
Theory

31. 27-10-2021 Arpan Deyasi, EM Theory 31
Electrostatic Energy
Then total energy stored in the capacitor is given by
E
V
W wdv
= 
2
E
V
1
W E dv
2
= 
 E
V
1
W E.Edv
2
= 

E
V
1
W D.Edv
2
= 
Arpan Deyasi
Electromagnetic
Theory

32. 27-10-2021 Arpan Deyasi, EM Theory 32
Problem 5:
Find the total energy stored in the uniform electric field in a charged spherical shell of
charge Q and radius r
2
E
V
1
W E dv
2
= 

Soln:
2
2
2
V
1 Q
(4 r dr)
2 4 r
 
=  
 

 

2
2
V
1 Q dr
2 4 r
  
=  
 
  
 

Arpan Deyasi
Electromagnetic
Theory

33. 27-10-2021 Arpan Deyasi, EM Theory 33
If the shell has lower radius a and higher radius b,
b 2
E 2
a
1 Q dr
W
2 4 r
  
=  
 
  
 

2
E
Q 1 1
W
8 a b
  
= −
 
 
  
 
Arpan Deyasi
Electromagnetic
Theory