1. COULOMB’S LAW AND ELECTRIC FIELD
Lecture 2
COULOMB’S LAW AND ELECTRIC FIELD

2. Learning Objectives You will learn:
How to utilise Coulomb’s law to calculate the force between electric charges.
How to calculate the electric field caused by discrete electric charges.

3. PROPERTIES OF ELECTRIC CHARGES
Electric charge is an intrinsic property of matter, inherent in its atomic structure.
Electronic charge: q = - e
Proton charge: q = + e
with elementary charge: e = 1.60 x Coulomb
Experiments demonstrate that there are two kinds of electric charges:
Positive and negative
(Benyamin Franklin [ ])

5. CONDUCTORS AND INSULATORS

6. CHARGING BY INDUCTION
A conductor connected to earth by means of a conducting wire is said to be grounded.
Charging by induction:
Bring negatively charged rubber rod near uncharged conducting sphere
Ground sphere
Remove wire  sphere is now positively charged
Note: Charging an object by induction requires no contact with the object inducing the charge.

7. ELECTROSTATIC INDUCTION
It is possible to obtain charges without any contact from another charge. They are known as induced charges and the phenomenon of producing induced charges is known as electrostatic induction. It is used in electrostatic machines like Van de Graaff generator and capacitors.
Figure on the right shows the steps involved in charging a metal sphere by induction.

8. ELECTROSTATIC INDUCTION
There is an uncharged metallic sphere on an insulating stand.
When a negatively charged plastic rod is brought close to the sphere, the free electrons move away due to repulsion and start piling up at the farther end. The near end becomes positively charged due to deficit of electrons. This process of charge distribution stops when the net force on the free electron inside the metal is zero.
When the sphere is grounded, the negative charge flows to the ground. The positive charge remains held due to attractive forces.

9. ELECTROSTATIC INDUCTION
When the sphere is removed from the ground, the positive charge continues to be held at the near end.
When the plastic rod is removed, the positive charge spreads uniformly over the sphere.

10. COULOMB’S LAW
It is an experimental law that deals with the force a point charge exerts on another point charge. Q1 Q2 r
Force F
(outward for like charges, inward for unlike charges)
Coulomb’s law states that the force F between two point charges Q1 and Q2 is
along the line joining them
directly proportional to the product Q1Q2 of the charges
inversely proportional to the square of the distance r between them.

11. Expressed mathematically,
where k is the proportional constant. In SI units, charges Q1 and Q2 are in coulombs (C), the distance r is in meters (m), and the force F is in newtons (N) so that k = 1/4εoεr. Thus, in the SI units
The constant εo (pronounced “epsilon naught”) is known as the permittivity of free space (in farads per metre) and has the value
F/m

12. Relative permittivity of some materials
The constant εr is known as the relative permittivity of the medium in which the charges are in, and εr has a value greater than, or, equal to unity; that is
(dimensionless)
Relative permittivity of some materials
Medium Relative permittivity, εr
Vacuum 1
Air (atmospheric pressure)
Parafin
Rubber 3
Quartz 5
Bakelite 5
Mica 6

13. COULOMB’S LAW IN 3-D RECTANGULAR COORDINATES
If point charges Q1 and Q2 are located at points having position vectors r1 and r2, then the vector force F12 due to Q1, shown in figure below, is given by O x y z Q1 Q2 r1 r2
where F = force, N
Q1 = charge 1, C
Q2 = charge 2, C
= unit vector pointing in direction
of line joining the charges
r =
= distance between point charges, m

14. EXAMPLE
A negative point charge of 1 µC is situated in air at the origin of a rectangular coordinate system. A second negative point charge of 100 µC is situated on the positive x axis at the distance of 500 mm from the origin. What is the force on the second charge? O x y z Q1 Q2 r2

15. SOLUTION Given: Q1 = 1 µC ; Q2 = 100 µC r1 = 0 mm ; r2 = 500 mm
Therefore, O x y z Q1 Q2 r2 N

16. EXAMPLE
Two point charges of 1 C each and each of the same sign are placed 1 mm apart in air. What is the magnitude of the repulsive force?
Q1 = 1 C
Q2 = 1 C
1 mm

17. SOLUTION Given: Q1 = 1 C ; Q2 = 1 C r = 1 mm
From Coulomb’s law the force is N
Q1 = 1 C
Q2 = 1 C
1 mm

18. Example: Hydrogen Atom
In the classical model of the hydrogen atom, the electron revolves around the proton with a radius of 0.53 x m. The magnitude of the charge of the electron and proton is 1.6 x C.
What is the magnitude of the electric force between the proton and the electron?
What is the magnitude of the electric field due to the proton at r?

19. Solution (a) The magnitude of the force is given by
Solutions:
Solution
(a) The magnitude of the force is given by
Now we can substitute our numerical values and find that the magnitude of the force between the proton and the electron in the hydrogen atom is
(b) The magnitude of the electric field due to the proton is given by
(a) The m a
gnitude of the force is given by

20. THE FORCE DUE TO A NUMBER OF POINT CHARGES
If we have more than two point charges, we can use the principle of superposition to determine the force on a particular charge. The principle states that if there are N charges Q1, Q2, …..,QN located respectively at points with position vectors r1, r2, ……, rN, the resultant force F on a charge Q located at point r is the vector sum of the forces exerted on Q by each of the charges Q1, Q2, …..,QN. Hence: Q Q1 Q2 QN z y x r1 r2 rN or
where

21. EXAMPLE
Point charges 1 mC and -2 mC are located at (3,2,-1) and (-1,-1,4) respectively. Calculate the electric force on a 10 nC located at (0,3,1). z y x
Q1 = 1 mC
(3,2,-1)
(-1,-1,4)
(0,3,1)
Q2 = - 2 mC
Q = 10 nC

22. SOLUTION z y x (-1,-1,4) Q2 = -2 mC Q = 10 nC (0,3,1) r r1 R1 Q1 = 1mC
(3,2,1)
(-1,-1,4)
(0,3,1)
Q2 = -2 mC
Q = 10 nC R1 r r1

23. SOLUTION z y x Q1 = 1 mC (3,2,1) (-1,-1,4) (0,3,1) Q2 = - 2 mC
Q = 10 nC r R2 r2

24. SOLUTION
Using
we obtain
Therefore, using mN

25. EXERCISE
Point charges 5 nC and -2 nC are located at (2,0,4) and (-3,0,5) respectively. Determine the force on a 1 nC charge located at (1,-3,7).
(-3,0,5)
Q = 1 nC z
Q2 = - 2 nC
(1,-3,7) y
(2,0,4)
Q1 = 5 nC x
[ nN ]

26. +Q THE ELECTRIC FIELD What’s an electric field?
It is a region around a charged object through which another charge will experience a force.
The behaviour of the fields can be visualised by using field lines. These are lines that follow the direction of the field vectors at convenient points in space, as illustrated in th figure below. +Q

27. FIELD LINES
A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the electric field at any point (electric field lines).
Rules for drawing electric field lines:
Lines begin at positive charges and end at negative charges or
infinity so the lines will show the movement of a “positive test
charge”.
(a) Field lines for a positive charge.
(b) Field lines for a negative charge.

28. Rules for drawing electric field lines:
The number of lines drawn leaving (approaching) a positive (negative) charge is proportional to the magnitude of the charge
No two field lines can cross each other.

29. Rules for drawing electric field lines:
At any point, the direction of the tangent to a curved field line gives the direction of the electric field at that point.

30. Rules for drawing electric field lines:
The field lines are drawn so that the number of lines per unit area, measured in a plane that is perpendicular to the lines, is proportional to the magnitude of E. Thus, E is large where field lines are close together and small where they are far apart.

31. Example
Shown are the electric field lines, the charges that produced the electric
Field are not shown. Rank the magnitude of the electric field for the points labeled A through F.

32. Answer

33. ELECTRIC FIELD INTENSITY
Consider a positive electric point charge Q placed rigidly at the origin of a 2-D rectangular coordinate system. If a unit positive test charge is brought into the vicinity of Q, it is upon acted by a force. This force is directed radially outward and becomes greater as Q2 approaches Q1. It may said that Q1 has a field around it where forces may act. x y F Q
Unit positivetest charge

34. ELECTRIC FIELD INTENSITY
We now define the electric field intensity (or electric field strength) as the force per unit charge when placed in the field. Thus
or simply
The electric field intensity E is obviously in the direction of the force F and is measured in newtons/coulombs or volts/meter.
The electric field intensity at point r due to a point charge located at r is given in vectorial form as

35. EXAMPLE
A negative point charge 10 nC is situated in air at the origin of a rectangular coordinate system. What is the electric field intensity at a point on the positive x axis 3 m from the origin? What is the force acting on a 2 nC positive point charge placed 3 m from the origin? x y F
Q = -10 nC
Unit positivetest charge
3 m

36. SOLUTION y Given Q = -10 nC Therefore F x
Unit positivetest charge
3 m
Given
Q = -10 nC
Therefore
N/C
That is, the electric field is 10 N/C and is in the negative x direction.

37. SOLUTION
The force acting on the - 2 nC charge is calculated from the relation
Putting in Q = - 2 nC and N/C at 3 m from the origin into the above expression we obtain y E
Q = - 2 nC

38. SOLVED EXAMPLE
Find the electric field due to a point charge of 0.5 mC at a distance of 4 cm from it in vacuum. How does the field strength change if the charge is in a medium of relative permittivity 80?
Solution:
q =0.5×10–3 C, d = 4×10–2 m, k = 80 38

39. THE ELECTRIC FIELD OF SEVERAL POINT CHARGES
For N point charges Q1, Q2, ….QN located at r1, r2, ….rN, the electric field intensity at point r is given by the superposition principle as z Q2
where QN r2 Q1 rN r1 y r Q x

40. EXAMPLE
A positive point charge of 1 nC is situated at the origin (x = 0, y = 0), and a negative point charge of -2 nC is situated on the y axis 1 m from the origin (x = 0, y = 1), as shown below. Find the total electric field intensity at the point P on the x axis 2 m from the origin (x = 2, y = 0). y
-2 nC
1 m
2.24 m E E2
120.7o
2 m α x
1 nC P E1

41. SOLUTION E1 y x
r = 2 m
r2 = 1 m
1 nC
-2 nC P
r1 = 0 m r r2 r1

42. SOLUTION y x
1 nC
-2 nC P R1 R2

43. SOLUTION Therefore N/C y x E E2 E1 r = 2 m 2.24 m r2 = 1 m 1 nC -2 nCP α
120.7o
r1 = 0 m

44. Dipoles: the second most important (after a
ELECTRIC DIPOLE
An electric dipole is a system of two charges of the same magnitude q, but opposite sign, separated by a distance d.
Note:
Dipoles are the second most important (after a point charge) configuration of charges.
point charge) configuration of charges.

45. ELECTRIC DIPOLE
The dipole moment describes the strength and orientation of the dipole. It is a vector quantity represented by p . Quantitatively, the dipole moment p is defined as the product of q and d:
The dipole moment vector points from the negative charge to the positive charge.
p = dipole moment = qd
p = qd d

46. Worked Example
Two charges 20 C and – 20 C are 4 mm apart. Determine the electric dipole moment of the system. If the charges are in a medium, will the dipole moment change?

47. Solution
Given: q = 20 x 10-6 C; d = 4 x 10-3 m
p = qd = 20 x 10-6 x 4 x 10-3 = 8 x 10-8 C
Dipole moment is independent of the nature of the medium between the charges.

48. WHY LEARN ELECTRIC DIPOLES ?
Many physical systems from molecules to TV antenna can be described as electric dipoles.

49. WATER MOLECULE AS AN EXAMPLE OF ELECTRIC DIPOLE
A water molecule is an example of an electric dipole. Even though the water molecule as such is neutral, the chemical bonds within the molecule causes displacement of charges. This displacement of charges creates a dipole with a negative charge on the oxygen end and a net positive charge on the hydrogen end.
This gives the water molecule a large dipole moment, allowing hydrogen bonding to form between molecules.
As a result, it’s the most unusual liquid: it is much denser than expected, and as a solid it is much lighter than expected when compared with its liquid form.

50. Anomalies:
1. High freezing point and melting point (due to this our planet is bathed in water,
2. Large heat capacity
3. High thermal conductivity (high water content in organisms contribute to thermal regulation and prevent local temperature fluctuations)
4. High latent heat of evaporation (resistance to dehydration and considerable evaporative cooling
5. Excellent solvent due to its polarity
6. High dielectric constant
7. Etc., etc.

51. TV ANTENNA AS AN EXAMPLE OF ELECTRIC DIPOLE
In radio and telecommunications a dipole antenna is the simplest and most widely used class of antenna. It consists of two identical conductive elements such as metal wires or rods, which are usually bilaterally symmetrical. The driving current from the transmitter is applied between the two halves of the antenna. Each side of the feedline to the transmitter is connected to one of the conductors.
The most common form of dipole is two straight rods or wires oriented end to end on the same axis, with the feedline connected to the two adjacent ends.
From transmitter
Schematic of a (balanced) half-wave dipole antenna connected to a balanced cable.

52. Picture showing electric field emanating from a dipole antenna.
From transmitter
Picture showing electric field emanating from a dipole antenna.

53. THE ELECTRIC FIELD OF A DIPOLE

54. THE ELECTRIC FIELD OF A DIPOLE
Horizontal component of electric field at P is:
And for large r (either large x, large y, or their combination) this becomes:
Vertical component of electric field at P is:

55. DIPOLE IN AN ELECTRIC FIELD
When an electric dipole is placed in a region where there is an external electric field, E, electrostatic forces act on the charged ends of the dipole. If the electric field is uniform, those forces act in opposite directions and with the same magnitude F =qE.
Although the net force on the dipole from the field is zero, and the center of mass of the dipole does not move, the forces on the charged ends do produce a net torque t on the dipole about its center of mass.

56. DIPOLE IN AN ELECTRIC FIELD
The torque is the cross product of the dipole moment and the electric field and the torque will rotate the dipole so that the dipole moment is aligned with the electric field.
Maximum torque occurs when the dipole moment is perpendicular to the electric field.

57. The closer the angle between the dipole moment and the electric field is to 90°, the greater the torque. As the angle  decreases, so does the torque.
When the angle between the dipole moment and the electric field is 0°, the torque is 0 N·m.

58. Maximum torque occurs when the dipole moment is perpendicular to the electric field.
The closer the angle between the dipole moment and the electric field is to 90°, the greater the torque. As the angle  decreases, so does the torque.

59. When the angle between the dipole moment and the electric field is 0°, the torque is 0 N·m.

60. Application Example: Microwave Cooking
In liquid water where molecules are relatively free to move around, the electric field produced by each molecular dipole affects the surrounding dipoles. As a result, the molecules bond together in groups of 2 or 3 because the negative oxygen end of one dipole and a positive hydrogen end of another dipole attract each other. Each time a group is formed, electric potential energy is transferred to the random thermal motion of the group and the surrounding molecules.

61. Application Example: Microwave Cooking
Each time collisions between the molecules breaks up a group, the transfer of energy is reversed. The temperature of the water (which is associated with the average thermal motion) does not change because, on average, the net transfer of energy is zero. When a microwave oven is turned on, the microwaves produce in the oven an electric field that rapidly oscillates back and forth in direction.

62. Application Example: Microwave Cooking
If there is water in the oven, the oscillating field exerts oscillating torques on the water molecules, continually rotating them back and forth to align their dipole moments with the electric field direction. Molecules that are bonded as a pair can twist around their common bond to stay aligned, but molecules that are bonded in a group of 3 must break at least 1 of their 2 bonds.
The energy to break the bonds comes from the electric field produced by the microwaves. Then molecules that have broken away from groups can form new groups, transferring the energy they just gained into thermal energy.

63. Application Example: Microwave Cooking
Thermal energy is added to the water when the groups form but is not removed when the groups break apart, and the temperature of the water increases. Foods that contain water can be cooked in microwave ovens because of the heating of that water. If the water molecule were not an electric dipole, this would not happen and microwave ovens would be useless.

64. END