electric

1. Contents:
Introduction
Properties of electric charges
Coulomb’s law
Electric field
Electric field lines
1

2. » An intrinsic property of protons
and electrons, which make up all
matter, is electric charge.
» A proton has a positive charge,
and an electron has a negative
charge.
2
What is Electric Charge?

3. •Basic element of electricity is
electric charge, which is a
fundamental property of matter.
•Lightning: example of an electrical
effect in nature.
•Electric charge: source of electric
force of attraction or repulsion.

4. Properties of electric charge
Benjamin Franklin(1706-1790)
Positive charge – possessed by protons
Negative charge – possessed by electrons
Charges of same sign repel
Charges of opposite signs attract

5. Cont….
Electric charges and conservation of
charges
Principle of conservation of charges state the total
charge in an isolated system is constant (conserved).
C
10
6
1
charge,
of
amount
l
fundamenta
: 19
.
e -
´
charge
electric
:
q
1,2,...
number
integer
positive
: =
n
ne
q =

6. Cont..
» Electric charge unit is coulomb, C
» An electron,e have the electric charge value in
about = - 1.6 x 10-19
» A proton,p have the electric charge value is
about 1.6 x 10-19
6

7. ne
q =
7
1. How many electrons are there in one
coulomb of negative charge?
Solution: From equation,
2. How many electrons make up a charge of
-30.0 C? (Ans:1.875 x 1014 electrons)
electrons
C
C
e
q
n 18
19
10
25
.
6
10
6
.
1
00
.
1
´
=
´
=
= -

8. » An object that has an excess of electrons
is negatively charged, and an object that
has a deficiency of electrons is positively
charged.
» First law of electrostatic:
˃ Like charges repel each other and unlike charges attract each other.
» Conservation of charge:
˃ Single charge can be neither created nor destroyed

9. STATIC ELECTRICITY
Objects can be charged by rubbing
a) Rub a plastic ruler b) bring it close to some tiny pieces of paper

10. A charged comb attracts bits of paper because
charges in molecules in the paper are realigned.

11. » It is possible to transfer electric charge from one object to
another. Usually electrons are transferred, and the body that
gains electrons acquires an excess of negative charge. The
body that loses electrons has an excess of positive charge.
11
• During any process, the net electric charge of an entire
isolated system remains constant (is conserved). This is
referred to as the law of conservation of electric
charge.

12. Not only can electric charge exist on an object, but it
can also move through an object.
» Substances that readily conduct electric charge are
called electrical conductors.
» Materials that conduct electric charge poorly are
known as electrical insulators.
» Semiconductors are materials that are intermediate
between conductors and insulators
12

13. Good
conductors
Poor
conductors
Semiconductors Insulators
Silver
Copper
Aluminium
Tungsten
Iron
Water
Human
body
Carbon
Germanium
silicon
Plastic
Glass
Rubber
Nylon
Paper
13

14. a) A charged metal
sphere and a neutral
metal sphere
b) The two spheres
connected by a metal nail,
which conducts charge from
one sphere to the other
c) The two spheres
connected by an
insulator, almost no
charge is conducted

15. » By Contact:
15
• By Induction :

16. Charging a metallic object by induction
(that is, the two objects never touch each
other).
(a) A neutral metallic sphere, with equal
numbers of positive and negative charges.
(b) The electrons on the neutral sphere are
redistributed when a charged rubber rod is
placed near the sphere.
(c) When the sphere is grounded, some of its
electrons leave through the ground wire.
(d) When the ground connection is removed, the
sphere has excess positive charge that is
nonuniformly distributed.
(e) When the rod is removed, the remaining
electrons redistribute uniformly and there is
a net uniform distribution of positive charge
on the sphere.

17. 17
Coulomb’s Law
states that the magnitude of the electrostatic (Coulomb or
electric) force between two point charges is proportional
to the product of the charges and inversely proportional
to the square of the distance between them.
r
12
F
!
21
F
!
+
2
Q
1
Q
+
2
2
1
r
Q
Q
F µ 2
2
1
r
Q
kQ
F =

18. 18
Coulomb’s Law equation:
This equation gives the magnitude of the force.
Unit : Newton
F : magnitude of electrostatic force
charges
point
o
between tw
distance
:
r
charges
of
magnitude
:
, 2
1 Q
Q
2
2
9
C
m
N
10
0
9
constant
(Coulomb)
tic
electrosta
: -
´
= .
k
2
2
1
r
Q
Q
k
F =

19. 0
4
1
pe
=
k
, hence the Coulomb’s law can be
written as
÷
ø
ö
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
= 2
2
1
0
4
1
r
Q
Q
F
pe
where air)
or
(vacuum
space
free
of
ty
permittivi
:
0
e
)
m
N
C
10
85
.
8
( 2
1
2
12
0
-
-
-
´
=
e

20. 20
The force is along the line connecting the charges,
and is attractive if the charges are opposite, and
repulsive if they are the same.

21. 21
The proportionality constant in Coulomb’s law is
then:
Charges produced by rubbing are typically
around a microcoulomb:

22. 22

23. 1. Calculate the magnitude of the force between
two 3.60 µC point charges 9.3 cm apart.
Solution: From equation,
2. Two positive point charge, Q1 = 50 µC and Q2 =
1 µC, are separated by a distance x. Which is
larger in magnitude, the force that Q1 exerts on
Q2, or the force that Q2 exerts on Q1.(Ans:F21
= F12)
23
2
2
1
r
Q
kQ
F =
N
47
.
13
m)
10
3
.
9
(
C)
10
60
.
3
(
/C
m
.
N
10
99
.
8 2
2
2
6
2
2
9
=
´
´
´
= -
-

24. Solution: From Coulomb's law,
˃ the force that Q1 exerts on Q2 is,
˃ the force that Q2 exerts on Q1 is
˃ Which is the same magnitude. The equation is
symmetric with respect to the two charges, So F21
= F12. 24
2
1
2
21
x
Q
Q
k
F =
2
2
1
12
x
Q
Q
k
F =
x
+
Q2
+ 1 μC
+
Q1
+50 μC

25. EXAMPLE 3
Two isolated small objects have charges of
0.04 µC and –0.06 µC and are 5cm apart
as Figure. What will be the magnitude
of electrostatic force acting on each object?
50.0 cm
Q1 = +0.04µC Q2 = -0.06µC

26. Step 1
Draw the direction of the electric force on
each object
r = 0.05 m
Q1 = +0.04 µC Q2 = -0.06 µC
F12 F21
Step 2
Find the magnitude of force on
Q1 F12
Q1
r
Q
Q
2
2
1
k
F12 =
Q1 = +0.04 µC
Q2 =I-0.06I µC
(0.05 m)2
(9 x 109 Nm2/C2)
F12 = 0.00864 N

27. Step 3
Write the vector component
F12
Q1
F12 = 0.00864 N
the force direction
is to +x
Hence, the vector
component:-
F12 = 0.00864 ˆ
i
Step 4
Calculate Force on Q2
Q2 = -0.06 µC
F21 IF12I = IF21I
F21 = -0.00864 ˆ
i

28. 1. At what separation will two charges, each of
magnitude 6.0 µC, exert a force of 1.4 N on
each other? (Ans: 0.48 m)
2. Calculate the magnitude of force between two
8.5 µC point charges that are 5.4 cm apart.
(Ans: 2.23 x 102 N)
3. Two point charges have charge values of -2.0 µC
and -4.0 µC, respectively. They are separated
by distance of 2.0 cm. Find the magnitude of
the electric force between them. (Ans: 179.8 N) 28

29. Solving Problems Involving Coulomb’s Law and Vectors
» Superposition: for multiple point charges, the forces on each
charge from every other charge can be calculated and then added
as vectors.
» The total force acting that charges is the vector sum of the forces
that the two charges would exert individually.
net 1 2 3
F = F + F + F + ....
! ! ! !

30. 30
Vector addition review:

31. EXAMPLE 4
Calculate the resultant force on the charge
Q3 due to other two charges located as
shown in figure below. Identify the
direction of resultant / net force.
1.0 cm
2.0 cm
Q1 = -4.2µC Q2 = +1.3µC Q3 = +1.1µC

32. Step 1
Draw the direction of the electric force on Q3 due to Q1
and Q2
Step 2
Find the magnitude of force on Q3 due to Q1 and Q2
Q3 = +1.1 µC
+
F31
-
Q1 = -4.2 µC
+
Q2 = +1.3 µC
F32
Force Calculation Method Value
F31
F32
2
3 1
k
Q Q
r
2
6
6
9
02
.
0
)
10
2
.
4
)(
10
1
.
1
(
10
9 -
-
´
´
´
2
3 2
k
Q Q
r
2
6
6
9
01
.
0
)
10
3
.
1
)(
10
1
.
1
(
10
9 -
-
´
´
´
103.95 N
128.70 N

33. Step 3
Write the vector component
Step 4
F31 F32
+
-103.95 i 128.27 i
Total up all the
forces Fnet = F31 + F32
= (-103.95 i) + ( 128.27i)
= 24.32 i
Hence, the MAGNITUDE
of total Force = 24.75 N
the DIRECTION of total
Force = +x direction

34. PROBLEM SOLVING HINTS
1. Determine the directions of the individual forces on Q3
due to charge Q1 and Q2 and indicate it in a sketch.
1. Find the magnitude of forces on Q3 due to charge Q1
& Q2 that is & . Make sure these values
always POSITIVE.
1. Find the vector components for & by using
form.
4. Find the resultant forces vector
5. Identify the direction and magnitude of resultant forces
from
31
F 32
F
31
F
!
32
F
! ˆ
ˆ ˆ
, ,
i j k
net 31 32
F = F + F
! ! !
net
F
!

35. SOLUTION
(1) 31 32
(2) = = = 103.95N
= = = 128.70 N
(3) 31= - 103.95 , 32 = + 128.70
(4) net = 31 + 32 = + 24.75
The magnitude of electric force are 24.75 N and
direction in positive x axis direction.
F
!
31
F
r
Q
Q
2
1
3
k 2
6
6
9
02
.
0
)
10
2
.
4
)(
10
1
.
1
(
10
9 -
-
´
´
´
32
F
r
Q
Q
2
2
3
k 2
6
6
9
01
.
0
)
10
3
.
1
)(
10
1
.
1
(
10
9 -
-
´
´
´
F
! i
ˆ
F
!
i
ˆ
F
!
F
!
F
!
i
ˆ
F
!

36. EXAMPLE 5
Three point
charges QA, QB and
QC of +12μ, -16μ and
+20μ respectively, are
arranged as shown
below. Find the
magnitude and direction
of the net force
on charge QA.
QC = +20µC
QB = -16µC
QA =+12µC
4m
3m

37. Step 1
Draw the direction of the electric force on QA due to QB
and QC
4 m
3 m
QA=+12 µC
QC=+20 µC
QB=-16 µC
FAC
+
+
-
FAB

38. QA=+12 µC
FAC
+
FAB
Step 2
Force Calculation Method Value
FAC
FAB
2
k
A C
Q Q
r
2
k
A B
Q Q
r
9 6 6
2
9 10 (12 10 )(20 10 )
4
- -
´ ´ ´
9 6 6
2
9 10 (12 10 )(16 10 )
5
- -
´ ´ ´
Find the magnitude of force
on QA due to QB and QC
0.135 N
0.069 N

39. Step 3
Write the vector component
Magnitude X - component Y - component
FAC = 0.135 N
FAB = 0.069 N
Total
QA
FAC = 0.135 N
+
FAB = 0.069 N
53.13o
0.135 cos 90o
= 0 i
0.135 sin 90o
= 0.135 j
0.069 cos 233.13o
= -0.04 i
0.069 sin 233.13o
= - 0.055 j
= - 0.04 i = 0.08 j
Step 4
Total up all the
forces

40. Step 5
Find the magnitude from the vector component
calculated
Magnitude = 2 2
(_ ) (_ )
i j
+
Hence,
2 2
(0.04) (0.08)
+
Vector component = - 0.04 i +
0.08 j
Magnitude =
= 0.089 N

41. Step 6
Calculate the direction of the total
force
Direction =
1 ( )
tan
( )
j
i
q -
=
Vector component = - 0.04 i + 0.08 j
Hence,
1 (0.08)
tan 63.43
(0.04)
o
q -
= =
63.43o
0.089 N
Ans : 0.089 N, (116.570)
Check the quadrant for direction of the force.
Step 7

42. •electric field exist around charges
•Electric field is a vector quantity .
•the electric field can be refer as the
arrow known as electric field line

43. The Electric Field
» The electric field vector E at a point is
defined as the electric force Fe acting on
a positive test charge q0 placed at that
point divided by the test charge:
0
e
q
F
º
E

44. 44
cont…
A region where force acts on
a charged body, or
The electric field is the force
on a small charge, divided by
the charge:

45. 45
For a point charge:
Superposition principle for electric fields:
Force on a point charge in an electric field:

46. 46
Field Lines
•The electric field is a vector and can be represented by field lines.
•The field lines indicate the direction of electric field.
•The lines are drawn to that the magnitude of electric field, E is
proportional to the number of lines crossing unit area perpendicular to the
lines.
The direction of electric field always tangent to the electric field
line at each point.

47. RULES FOR DRAWING ELECTRIC FIELD LINES
» The lines must begin on a positive charge and
terminate on a negative charge. In the case of an
excess of one type of charge, some line will begin or
end infinitely far away.
» The number of lines drawn leaving a positive charge or
approaching a negative charge is proportional to the
magnitude of the charge
» No two field lines can cross.

48. The electric field
lines for a point
charge.
(a) For a positive
point charge, the
lines are directed
radially outward.

49. The electric field
lines for a point
charge.
(b) For a negative
point charge, the
lines are directed
radially inward. Note
that the figures show
only those field lines
that lie in the plane
of the page.

50. The electric field
lines for a point
charge.
(c) The dark areas
are small pieces of
thread suspended in
oil, which align with
the electric field
produced by a small
charged conductor at
the center.

51. (a) The electric field
lines for two point
charges of equal
magnitude and
opposite sign (an
electric dipole). The
number of lines leaving
the positive charge
equals the number
terminating at the
negative charge.

52. (b) The dark lines are small pieces of thread suspended in
oil, which align with the electric field of a dipole.

53. (a) The electric field lines for two positive point
charges.

54. (b) Pieces of thread suspended in oil, which align with the
electric field created by two equal-magnitude positive
charges. Courtesy of Harold M. Waage, Princeton
University

55. The electric field between two
closely spaced, oppositely
charged parallel plates is
constant.

56. 56

57. 57

58. » The strength of the
electric field is indicated
by how close the field
lines are to each other.
» The closer the field lines,
the stronger the electric
field in that region.
58
Strength of Electric Field

59. Electric field lines penetrating two surfaces. The
magnitude of the field is greater on surface A than on
surface B.
•The density of lines
through surface A is greater
than through surface B
•The magnitude of the
electric field is greater on
surface A than B
•The lines at different
locations point in different
directions
- This indicates the field
is non-uniform

60. Lets say if you
want to calculate
the E value at P,
you can use
+
P
r
2
kQ
E
r
=
Electric Field
Magnitude at
point(N/C)
Proportionality
constant
The
value of
charge
Distance
between the
charge and the

61. You can also calculate the
Electric Field strength/intensity
at any point by using
0
q
F
E
!
!
=
+
r
q
Force
experienced
by q
Electric Field
Magnitude at
point (N/C)
Charge
value

62. 2
kQ
E
r
=
0
q
F
E
!
!
=
Both equations
can be used to
calculate Electric
Field
strength/intensit
y at any point

63. EXAMPLE 1
Calculate the magnitude and direction of Electric Field at point P
in Figure below which is 30 cm to the right of a point charge Q =
-3.0 x 10-6 C.
P
_
Q = -3.0 x 10-6 C
EXAMPLE 6

64. Step 1
Draw the direction of the electric field on P due to Q
P
_
Q = -3.0 x 10-6 C
Imagine there is a +ve
charge here
+
EP
Step 2
Calculate the magnitude of the electric field on P
2
=
p
kQ
E
r
P
+
EP
+
9 x 109 Nm2 / C2
Q = -3.0 x 10-6
C
0.3 m Ep = 3.0 x 105 N/C

65. Step 3
Write the vector component
P
EP
+
+
The direction
of E is to -X
Ep = 3.0 x 105 N/C
Ep = -3.0 x 105 i N/C
Hence, the
vector
component is
Final Answer
Magnitude = 3.0 x 105 N/C and
direction = to –X axis (to the left)

66. 1. Determine the electric field intensity at a
distance 1.50 mm from a point charge +90 nC.
(Past year; APR2009) (Ans:3.6 x 108 N/C)
2. Two charge +2µC and +5µC are separated by a
distance 40.0 cm. Determine the electric field
at a point midway between the two charges.
(Ans: 3.375 x 105 N/C)
66

67. 3. Two point charges are separated by a distance of
10 cm. One has a change of -25 µC and the
other +50 µC.
a. Determine the direction and magnitude of the
electric field at a point P between the two
charges that is 2.0 cm from the negative
charge. (Ans: 6.3 x 108 N/C)
b. If an electron mass = 9.11 x 10-31 kg) is
placed at rest at P and then released, what
will be its initial acceleration? (Ans: 1.1 x
1020 m/s2).
67

68. 4. Positive charges are situated at three corners of
a rectangle, as shown in below. Find the electric
field at the fourth corner. (Ans: 756.13 N/C)
68

69. 5. What is the electric field at the center of the
triangle in figure below? (Ans: 5.46 x 106 N/C)
69
20 cm
q1 = +4.0 µC
q3 = -4.0 µC
q2 = +4.0 µC
20 cm
20 cm