1. Opposites Attract

3. Electric Fields

4. In 1975, a French scientist Coulomb investigated the magnitude of the force acting between charges and found it to follow an inverse square law
Charles-Augustin de Coulomb
(June 14, 1736 – August 23, 1806)

5. Electric Force : Coulomb’s Law-Q _ F F +q +

6. FE acts along the line joining the two charges.
The electric force between the two charges form an action-reaction pair,
Coulomb’s law is applicable only for point charges. Uniformly charged conducting spheres separated by large distances can be treated as point charges.

7. FIELD

8. attractive repulsive? Quick Check 1
The masses of an electron and a proton are 9.11 x kg and 1.67 x kg respectively. Given that the electron and proton in a hydrogen atom are separated by a distance of about 5.3 x m and each carries a charge of magnitude 1.6 x C.
(i) Is the force between them
attractive
repulsive?

9. (b) (ii) Is the magnitude of force the proton exert on the electron,
(a) greater,
(b) the same ,
(c) or less force, than the electron exert on the proton?
(b)

10. To find magnitude of electric force:
Using

11. FG is negligible compared to the corresponding FE
FG = GMm/r2
= 3.6x10-47N
Ratio of electric force to gravitational force =
FE : FG = 3.6x1029
FG is negligible compared to the corresponding FE

12. Electric Field
A charge causes a field around it such that other charges in that field will experience an electric force. This field affects any other charge placed within it, but not the charge producing it.
i.e. electric field thus refers to the region where an electric force is experienced by a test charge placed at any point in this region.

13. Representation of an Electric FieldQ _
Negative charge

14. positive charge Q +

16. Examples of Electric Field lines- +

18. Examples of Electric Field lines
more

20. Note:
Electric field lines begin from the positive charges and end on the negative charges.
Electric field lines do not cross each other.
The number of field lines per unit cross sectional area is proportional to the magnitude of the electric field strength.

21. The electric field strength at a point is defined as the electric force acting per unit positive charge placed at that point.

22. V m-1 or N C-1
vector
direction of E and F are the same for positive charge at that point
direction of E and F are opposite for negative charge at that point

23. The lines are drawn such that the tangent to a field line at a point gives the direction of E at that point. EA EB A B

24. Quick Check 2 Decreases as we move from inner to outer.
Notice that the field lines are perpendicular to the surface and points from +ve to –ve.
Field lines are closer near inner cylinder implying stronger field P

25. Quick Check 3
A test charge of + 3 μC is placed at a point X where an external field is directed to the right and has a magnitude of 4 μ 106 N C-1. If the test charge is replaced with another test charge of – 3 μC, the external electric field at X
a) Is unaffected
b) Reverses direction.
c) Changes in a way that cannot be determined.
(A)

26. - Graphical representation
For positive point charge
- Graphical representation r E R + -R
For r >R
For positive spherical conductor

27. Worked Example 1: Deduce the direction: using the type of charges
Find the magnitude of each force using Coulomb’s Law
Find the resultant of the two forces as a vector.

28. Fq sin Fq Fq cos   Fq cos Fq Fq sin Q Horizontal:
Fhort = Fqcos + Fqcos
= 0.46N Fq
Fq cos 
Vertical:
Fvert = Fqsin - Fqsin
= 0N Q 
Fq sin
Fq cos Fq
Resultant:
Fresult = Fhort
= 0.46N

29. Worked Example 1:
If we decide to work with Electric field, similar steps can be taken:
Deduce the direction of E field from the type of charges.
Find the magnitude of each E field
Find the resultant of the two E field as a vector (= ET )
Use F = QET to find the force.

30. Eq sin Eq Eq cos   Eq cos Eq Eq sin Q Horizontal:
Ehort = Eqcos + Eqcos
= 1.2 x 105 N C-1 Eq
Vertical:
Fvert = Eqsin - Eqsin
= 0 N C-1
Eq cos  Q 
Eq sin
Eq cos
Resultant:
Eresult = Ehort
= 1.2 x 105 N C-1
F = Eresult = 0.46N Eq

31. For some fun and laughter…

34. Electric Potential Energy+ A Q q FE
Fext r B C
Electric potential energy of a charge at any point in an electric field is defined as the work done by an external agent to bring the charge from infinity to that point.

35. It is an entity of the system.
unit : J
scalar
It is an entity of the system.
Electric Potential Energy

36. Quick Check 4
Find the work done in bringing 2 point charges, -12 C and –8.0 C, from a separation of 10 cm to 6.0 cm in vacuum.

37. Solution
This question cannot be done using W.D = F  d
since F is NOT constant.
Instead, W.D = ΔUE
= Ufinal - Uinitial
= (q1q2/40)[ 1/rf – 1/ri ]
= [ (-12  10-6)(-8.0  10-6) /40 ] [1/0.06 – 1/0.1]
= J

38. Q1: What if the charges are of opposite sign but the same magnitude?
- Same with positive works done by external agent meaning that external agent MUST do work to bring them closer.
Q2: What if one of the charges are of opposite sign but the same magnitude?
- Workdone is negative by the external agent.

39. Electric Potential
The electric potential at a point is defined as the work done per unit positive charge by an external agent in bringing a charged body from infinity to that point.
unit V or J C-1
scalar

40. Considering points A and B in the previous diagram, potential difference between A and B is:
VBA = VA – VB (VA > VB)
where UA and UB are the electric potential energies at points A and B respectively, and WBA is the work done by Fext to bring the test charge from point B to point A. + A Q q FE
Fext B C

41. - Graphical representationV
For positive point charge
For negative point charge

42. Equipotentials
Equipotentials are lines or surfaces that are of the same potential V.
Similar to contour lines that you learn in geography

43. E-field lines
Equipotentials
FIELD

44. Quick Check 5
Example 8:
Soln:
K Q r V
-14V
, r V
-9V
- 6V

45. Try the following link to see how the field lines, equipotential lines look like around charges!

46. E V Relationship between E and V k Q q r 2 F k Q q r U F +q U +q k Q

47. E F Relationship between E and V - V r d - U r d
Numerically, E = -ive potential gradient
- V r E d
and F = -ive potential energy gradient
- U r F d
The negative sign just means that E-field points towards the direction of decreasing potential.

48. A short break:

51. Special Case: Uniform Electric Field
A uniform field means that a charge placed at any point in the field will experience the same force.
implies that the field lines must be parallel and equidistant from one another in the field.

52. Uniform Electric Field_ + A B
Equipotential lines V
- Field lines must be parallel and equidistant from one another in the field. r A B E A B r

53. Uniform Electric Field
can now be re-written as:
The negative sign indicates that E points in the direction of decreasing V and is often ignored during calculations.

54. Quick Check 6
Solution
E = V/d
= 12 / 0.3 x 10-2
= 4000 V m-1

55. Quick Check 7
The figure below shows two equipotential lines X and Y. The electric field strength between X and Y is 5.0 V m-1 and points in the direction shown. The potential of X is V. The distance between X and Y is 4.0 m. E a b c
4.0 m X Y

56. E a b c
4.0 m X Y
Uniform Electric Field
(a) Find the electric potential of Y
E = V /d = 5.0  (Vx - Vy )= 5.0 x 4.0
Vy = Vx x 4.0
= (-20.0) – (5.0 x 4.0) = V
(X is at higher potential)

57. E a b c
4.0 m X Y
Uniform Electric Field
(b) Calculate the work done to bring a
-5.0 C charge from
a to b,
from a to c
W = qV = -5.0 (0) = 0 J
W = qV = q (Vf - Vi )
= -5.0 ( – (-20.0))
= 100 J

58. (c) If an electron is released from rest midway between X and Y, state whether it will accelerate towards X or Y. Calculate the speed of the electron when it reaches either X or Y.
towards X because it is of higher potential
Using conservation of energy:
½ mv2 = qV
½ (9.11 x 10-31)v2 = 1.6 x x 10.0
v = 1.87 x 106 m s-1

59. Motion of charged particle in a Uniform E-field
1. Positive charge placed in E-field: E + F q
Motion: charge accelerates in the same direction as E.
FE = qE = ma
a = qE / m

60. Motion of charged particle in a Uniform E-field
2. Negative charge placed in E-field: E F -
Motion: charge accelerates in the opposite direction to E.

61. Motion of charged particle in a Uniform E-field
3. Positive charge projected with constant speed perpendicular into E-field: E
Negative
Positive + v F
Motion: Uniform speed upwards, accelerates to the right. Path of motion is parabola (recall Projectile Motion in Kinematics)

62. Look at Worked example 6 for a “replay” of projectile motion!
Another indication that topics are related
and you need to master earlier concepts!

63. Summary:
Concept Map
- Look at similarity between E and g field and if you master one, you can apply almost everything to the other topic!

64. May the “force” be with you.. .
“study hard. you must….”
“Or regret, you will….”
… and finally,

65. The End

67. A Butterfly’s Lesson
”One day, a small opening appeared in a cocoon; a man sat and watched for the butterfly for several hours as it struggled to force its body through that little hole.

68. Then, it seems to stop making any progress.
It appeared as if it had gotten as far as it could and it could not go any further.

69. So the man decided to help the butterfly: he took a pair of scissors and opened the cocoon.
The butterfly then emerged easily.
But it had a withered body, it was tiny and shriveled wings.

70. The man continued to watch because he expected that, at any moment, the wings would open, enlarge and expand, to be able to support the butterfly’s body, and become firm.

71. Neither happened!
In fact, the butterfly spent the rest of its life crawling around with a withered body and shriveled wings. It never was able to fly.

72. What the man, in his kindness and his goodwill did not understand was that the restricting cocoon and the struggle required for the butterfly to get through the tiny opening, were God’s way of forcing fluid from the body of the butterfly into its wings, so that it would be ready for flight once it achieved its freedom from the cocoon.

73. Sometimes, struggles are exactly what we need in our life.
If God allowed us to go through our life without any obstacles, it would cripple us. We would not be as strong as we could have been. Never been able to fly.

74. I asked for Strength...
and God gave me difficulties to make me strong.
I asked for Wisdom...
and God gave me problems to solve.
I asked for prosperity...
and God gave me a brain and brawn to work.

75. I asked for Courage…..
and God gave me obstacles to overcome.
I asked for Love...
and God gave me troubled people to help.

76. I asked for Favors...
And God gave me Opportunities.
“I received nothing I wanted...
But I received everything I needed."

77. Live life without fear, confront all obstacles and know that you can overcome them.
Send this message to your friends and show them how much you care. Send it to anybody that you consider a FRIEND, even if this means to send it to the same person that conveyed it to you. If this message returns to you, you can be sure that your circle of friendship is made out of true friends.