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
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Electromagnetic
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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
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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
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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
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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
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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
=
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Electromagnetic
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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
+ =
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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)
− −
=
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9. Problem 1:
8 5
9
2
(4 10 ) (6 10 )
F 9 10 N
2 (0.1)
− −
=
F 1.08N
=
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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
= + + +
+ + +
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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
=
=
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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°
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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
−
=
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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
=
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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
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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
=
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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
= +
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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
=
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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
=
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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
=
− +
=
=
− +
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21. Electrostatic Potential
r
r r
q 1
4 r
=
=
−
=
q 1 1
4 r
−
= −
q
4 r
=
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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
= − +
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23. Relation between Electric Field and Potential
1
E.dr .dr
= −
Using general notation
E.dr .dr
= −
E .dr 0
+ =
E = −
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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
= − − − − − −
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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
−
= − −
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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
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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
=
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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
=
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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
=
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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
→
= =
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Electromagnetic
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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
=
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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
=
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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