1. ELECTROSTATICS:
The study of the behavior of stationary charges
ELECTRIC CHARGE
There are two types of electric charge, arbitrarily called
positive and negative. Rubbing certain electrically neutral objects together (e.g., a glass rod and a silk cloth) tends to cause the electric charges to separate. In the case of the glass and silk, the glass rod loses negative charge and becomes positively charged while the silk cloth gains negative charge and therefore becomes negatively charged. After separation, the negative charges and positive charges are found to attract one another.

2. When a rubber rod is rubbed against fur, electrons are removed from the fur and deposited on the rod.
Electrons move from fur to the rubber rod.
positive
negative --
The rod is said to be negatively charged because of an excess of electrons. The fur is said to be positively charged because of a deficiency of electrons.

3. When a glass rod is rubbed against silk, electrons are removed from the glass and deposited on the silk.
silk
glass
Electrons move from glass to the silk cloth.
positive
negative
+ +
The glass is said to be positively charged because of a deficiency of electrons. The silk is said to be negatively charged because of an excess of electrons.

4. Gold-leaf Electroscope Pith-ball Electroscope
Laboratory devices used to study the existence of two kinds of electric charge.
Gold-leaf Electroscope
Pith-ball Electroscope

5. The two negative charges repel each other.
1. Charge the rubber rod by rubbing against fur.
2. Transfer electrons from rod to each pith ball.
The two negative charges repel each other.

6. 1. Charge the glass rod by rubbing against silk.
2. Touch balls with rod. Free electrons on the balls move to fill vacancies on the cloth, leaving each of the balls with a deficiency. (Positively charged.)
The two positive charges repel each other.

7. Opposite Charges Attract!
fur
Rubber
silk
glass
Attraction
Note that the negatively charged (green) ball is attracted to the positively charged (red) ball.
Opposite Charges Attract!

8. Like charges repel; unlike charges attract.
Pos
Neg
Pos
Neg

9. Charging by Contact  
Some electrons leave rod and spread over sphere.

10. Charging by Induction    
Rod does not touch sphere.  It pushes electrons out of the back side of the sphere and down the wire to ground.  The ground wire is disconnected to prevent the return of the electrons from ground, then the rod is removed.

11. The law of conservation of electric charge: "The net amount of electric charge produced in any process is zero." Another way of saying this is that in any process electric charge cannot be created or destroyed, however, it can be transferred from one object to another.
Charged comb attracts neutral water molecules.
Charged comb attracts neutral bits of paper.

12. Applications of Electrostatic Charging
                               
Fine mist of negatively charged gold particles adhere to positively charged protein on fingerprint.
                                                     
Negatively charged paint adheres to positively charged metal.

13. The quantity of charge (q) can be defined in terms of the number of electrons, but the Coulomb (C) is a better unit for later work.
The Coulomb: 1 C = 6.25 x 1018 electrons
The charge on a single electron is:
1 electron: e- = -1.6 x C

14. The coulomb (selected for use with electric currents) is actually a very large unit for static electricity. It is common to use the metric prefixes.
1 mC = 1 x 10-6 C
1 nC = 1 x 10-9 C
1 pC = 1 x C

15. COULOMB’S LAW
Coulomb’s Law states that two point charges exert a force (F) on one another that is directly proportional to the product of the magnitudes of the charges (q) and inversely proportional to the square of the distance (r) between their centers. The equation is:
F = electrostatic force (N)
q = charge (C)
k = 9x109 N•m2/C2
r = separation between charges (m)

16. The value of k can also be expressed in terms of the permittivity of free space (εo):
9x109 N. m2/C2
The proportionality constant (k) can only be used if the medium that separates the charges is a vacuum. If the region between the point charges is not a vacuum then the value of the proportionality constant to be used is determined by dividing k by the dielectric constant (K).
For a vacuum K = 1, for distilled water K = 80, and for
wax paper K = 2.25

17. Problem-Solving Strategies
1. Draw and label a figure indicating positive and negative charges along with the given distances.
2. Draw the force of attraction or repulsion on the given charge on a neat, labeled FBD.
3. Find the resultant force.
Important: Do not use the signs of the charges when applying Coulomb's law!

18. = 80 N = 120 N FR = 80 + 120 = 200 N, to the right
11.1 Two charges q1 = - 8 μC and q2= +12 μC are placed 120 mm apart in the air. What is the resultant force on a third charge
q3 = - 4 μC placed midway between the other charges? FR F2
q1 = - 8x10-6 C
q2= +12x10-6 C
q3 = - 4x10-6 C
r = m F1 - q1 - q3 + q2
0.06 m
0.06 m
= 80 N
= 120 N
FR =
= 200 N, to the right

19. 11.2 Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are arranged as shown. Find the resultant force on q3 due to the other two charges. F1 FR
q1 = +4x10-9 C
q2= -6x10-9 C
q3 = -8x10-9 C F2
37˚ θ
= 2.88x10-5 N
= 6.75x10-5 N

20. From the FBD: Σ Fy = F1 sin 37˚ Σ Fx = F2 - F1 cos 37˚θ
Σ Fy = F1 sin 37˚
= (2.88x10-5)(sin 37˚)
= 1.73x10-5 N
Σ Fx = F2 - F1 cos 37˚
= (6.75x10-5) - (2.88x10-5)(cos 37˚)
= 4.45x10-5 N
= 4.8x10-5 N
FR (4.8x10-5 N, 21˚)
= 21˚

21. ELECTRIC FIELD
An electric field is said to exit in a region of space in which an electric charge will experience an electric force. The magnitude of the electric field intensity is given by:
Units: N/C

22. The direction of the electric field intensity at a point in space is the same as the direction in which a positive charge would move if it were placed at that point. The electric field lines or lines of force indicate the direction. The electric field is strongest in regions where the lines are close together and weak when the lines are further apart.

24. F = qE = 1.6x10-19 (6x104) = 9.6x10-15 N, upward
11.3 The electric field intensity between two plates is constant and directed downward. The magnitude of the electric field intensity is 6x104 N/C. What are the magnitude and direction of the electric force exerted on an electron projected horizontally between the two plates?
E = 6x104 N/C
qe = 1.6x10-19 C
F = qE
= 1.6x10-19 (6x104)
= 9.6x10-15 N, upward

25. 11.4 Show that the gravitational force on the electron of example 11-3 may be neglected.
me = 9.11x10-31 kg
FG = mg
= 9.11x10-31 (9.8)
= 8.92x10-30 N
The electric force is larger than the gravitational force by a factor of 1.08x1015!

26. The electric field intensity E at a distance r from a single charge q can be found as follows:
Units: N/C

27. 11.5 What is the electric field intensity at a distance of 2 m from a charge of -12 μC?
r = 2 m
q = -12 μC
= 27x103 N/C, towards q

28. When more than one charge contributes to the field, the resultant field is the vector sum of the contributions from each charge.
Units: N/C

29. 11.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field a. At point A
q1 = -6 x10-9 C
q2 = +6 x10-9 C E1 ER E2
= 3.38x104 N/C, left
= 8.44x103 N/C, left

30. E1 ER E2
ER = E1 + E2
= 3.38x x103
= 4.22x104 N/C, left

31. 11.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart, as shown in the figure. Determine the electric field b. At point B E2
q1 = -6x10-9 C
q2 = +6x10-9 C
37º θ ER E1
= 6.67x103 N/C
= 2.4x103 N/C

32. From vector diagram: Σ Ex = - E2cos 37˚ = - (2.4x103)(cos 37˚)
37º
Σ Ex = - E2cos 37˚
= - (2.4x103)(cos 37˚)
= N/C θ ER E1
Σ Ey = E2 sin 37˚- E1
= (2.4x103)(sin 37˚) - (6.67x103)
= N/C
= N/C

33. = 70˚ E2
37º θ
180˚ + 70˚ = 250˚
ER (5566 N/C, 250˚) ER E1

34. GRAVITATIONAL POTENTIAL ENERGY
Consider that a mass m is moved from level A to level B. An external force F equal to the weight mg must be applied to move the mass against gravity. The work done by this force is W = mgh

35. When the mass m reaches level B, it has a potential for doing work relative to level A. The system has a potential energy (U) that is equal to the work done against gravity. The potential energy coverts into kinetic energy as the mass falls from level B.

36. ELECTRIC POTENTIAL ENERGY
When a charge q is moved against a constant electric force for a distance d, the potential energy of the system is equal to the work done to move the charge:
PE = W = F d = qE d

37. If the charge is released, it will acquire a kinetic energy as it returns the same distance.

38.  An important difference between gravitational potential energy and electric potential energy is that in the case of gravity, there is only one kind of mass, and the forces involved are always forces of attraction.
Therefore, a mass at higher elevations always has the greater potential energy relative to the earth.

39. This is not true in the electrical case because of the existence of the negative charge. In the figure below, a positive charge has a greater potential energy at point B than at point A. Work has been done against the electric field.

40. On the other hand, if a negative charge were moved from point A to B, work will be done by the field.
A negative charge would have a lower potential energy at B, exactly the opposite situation of a positive charge.
Whenever a positive charge is moved against an electric field, the potential energy increases.
Whenever a negative charge is moved against an electric field, the potential energy decreases.

41. +
When a positive charge is placed in an electric field:
- It moves in the direction of the field
- It moves from a point of higher potential to a point of lower potential
- Its electrical potential energy decreases and its kinetic energy increases -

42. +
When a negative charge is
placed in an electric field
- It moves opposite to the
direction of the field
- It moves from a point of
lower potential to a point of
higher potential
- Its electrical potential energy
increases and its kinetic
energy increases
- Work has to be done on the
charge for it to move from
point A to point B -

44. ELECTRIC POTENTIAL ENERGY
The electric potential energy of a system composed of a charge q and another charge Q separated by a distance r is equal to the work done against the electric forces in moving a charge +q from infinity to that point.
Joules (J)

45. = 3.6x10-4 J = 9.0x10-4 J ΔPE = 9.0x10-4 - 3.6x10-4 = 5.4x10-4 J
11.7 A charge of + 2 nC is 20 cm away from another charge of μC. a. What is the potential energy of the system?
q = +2x10-9C
Q = + 4x10-6 C
r = 0.20 m
= 3.6x10-4 J
b. What is the change in potential energy if the 2 nC charge is moved to a distance of 8 cm from the 4 μC charge?
r = 0.08 m
= 9.0x10-4 J
ΔPE = 9.0x x10-4
= 5.4x10-4 J

46. If the charges have the same sign, PE is positive
Positive work must be done to force the two charges near one another
The like charges would repel
If the charges have opposite signs, PE is negative
The force would be attractive
Work must be done to hold back the unlike charges from accelerating as they are brought close together

47. ELECTRIC POTENTIAL
The advantage of the concept of an electric field as force per unit charge was to assign an electrical property to space. If the electric field intensity is known at some point, the force on a charge placed at that point can be predicted.
It is equally convenient to assign another property to the space surrounding a charge. This will allow us to predict the potential energy due to another charged placed at any point. This property of space is called electric potential.

48. The electric potential V is defined in terms of the work to be done on a charge to move it against an electric field.
The electric potential V is a scalar quantity defined as the potential energy per unit charge.
Units: J/C = Volt (V)
The potential at a point is defined in terms of a positive charge. The potential due to a positive charge is positive, and the potential due to a negative charge is negative.

49. Substitution of the equation for potential energy yields:
Units: Volt (V)

50. ELECTRIC POTENTIAL due to POINT CHARGES
The potential V at a point a distance r from a charge Q is equal to the work per unit charge done against electric forces in bringing a positive charge +q from infinity to that point.
The point of zero electric potential is taken to be at an infinite distance from the charge
In other words, the potential at some point A as shown in the figure, is equal to the potential energy per unit charge. A

51. The potential is the same at equal distances from a spherical charge
The potential is the same at equal distances from a spherical charge. The dashed lines in the figure below are called equipotential lines. Note that the lines of equal potential are always perpendicular to the electric field lines.

52. Equipotential Lines

53. An equipotential surface is a surface on which all points are at the same potential
No work is required to move a charge at a constant speed on an equipotential surface

55. Electric Potential Due to Point Charges
These plots show the potential due to
(a) positive and
b) negative charge.

56. = - 6x104 V PE = qVA = 4x10-9 (-6x104) = -24x10-5 J
11.8 a. Calculate the potential at a point A that is 30 cm distant from a charge of- 2 μC.
Q = - 2x10-6 C
r = 0.3 m
= - 6x104 V
b. Find the potential energy if a + 4 nC is placed at point A.
q = +4x10-9 C
PE = qVA
= 4x10-9 (-6x104)
= -24x10-5 J

57. The potential in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge.

58. positive = the field does the work
11.9 Two charges + 6 μC and - 6 μC are separated by 12 cm, as shown. Calculate the potential a) at point A
Q1 = + 6x10-6 C
Q2 = - 6x10-6 C
r1 = 4x10-2 m
r2 = 8x10-2 m
= 6.75x105 V
positive = the field does the work

59. negative = an external source of energy must perform the work
b) at point B.
= x106 V
negative = an external source of energy must perform
the work

60. 11.10 What is the electric potential at the center of a square of sides equal 1 m if the charges placed at the corners are +1 nC, -2 nC, nC and -4 nC (read in a CW direction)
The center of the square is equidistant from all four charges, a distance r of +1 -2
= 0.707
- 4 +3
V = V1 + V2 + V3 + V4
= - 25 V

61. The potential difference between two points A and B to be equal to:
In other words the potential difference between two points is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential.

62. ΔV = VA-VB = 6.75x105 - (-1.01x106) = 1.69x106 V W = q ΔV
11.11 Consider the two charges on problem 11.8
a. What is the potential difference between points A and B?
VA = 6.75x105 V
VB = -10.1x105 V
ΔV = VA-VB
= 6.75x105 - (-1.01x106)
= 1.69x106 V
b. How much work is done by the electric field in moving a -2 nC charge from A to B?
q = -2x10-9 C
W = q ΔV
= (-2x10-9)(1.69x106)
= x10-3 J

63. POTENTIAL AND ELECTRIC FIELD
The potential difference between two oppositely charged plates is equal to the product of the field intensity and the plate separation.
W = q ΔV
= F d
= q E d
ΔV = E d
Units:
ΔV = Volts (V)
E = V/m or N/C

64. 11.12 The potential difference between two plates 5 mm apart is
10 kV. Determine the electric field intensity between the two plates.
d = 5x10-3 m
V = 10x103 V
= 2x106 V/m

65. CAPACITORS
A capacitor is a device that can store electric charge and consists of two conducting objects placed near one another but not touching. A typical capacitor consists of a pair of parallel plates of area A separated by a distance d. Often the two plates are rolled into the form of a cylinder with paper or other insulator.

67. If a potential difference or voltage is applied to a capacitor it becomes charged. The amount of charge acquired by each plate is proportional to the voltage.
C is the capacitance in farads (F),
Q is the charge in (C)
V is the voltage (V)

69. The capacitance of a parallel-plate capacitor depends on the geometry of the capacitor according to this equation:
εo = 8.85x10-12 C2/Nm2
Where C is the capacitance (F), A is the area of the plates (m2) and d is the separation between the plates (m)

70. Keyboards and Defibrillators
Applications of Capacitors:  
            
 Keyboards and Defibrillators                                              

71. Computers use capacitors in the keyboard:
The keyboards use capacitors at the bases of the keys
When the key is pressed, the capacitor spacing decreases and the capacitance increases
The key is recognized by the change in capacitance

72. 11.13 a. Find the capacitance of a capacitor whose plates are cm x 12 cm separated 1 mm by an air gap.
A = 0.12 x 0.12 m2
d = m
V = 12 V
= 1.27x10-10 F
b. Find the charge on each plate if the capacitor is connected to a 12 V battery.
Q = CV
= 1.27x10-10 (12)
= 1.53x10-9 C

73. ENERGY OF A CHARGED CAPACITOR
Consider a capacitor initially uncharged. When a source of voltage is connected to it, the voltage between the plates increases as charge is transferred.
The energy of a charged capacitor can be found with these equations:
Units: Joules (J)

74. CAPACITORS IN SERIES AND PARALLEL
These are the symbols used in different arrangements of capacitors:

75. CAPACITORS IN SERIES
The concept of equivalent capacitance refers to the capacitance of a single capacitor that could replace all the capacitors in a circuit.

76. CAPACITORS IN SERIES Series capacitors always have the same charge.
The voltage across the equivalent capacitor Ceq is the sum of the voltage across both capacitors.

77. Charge is same: series connection of capacitors.
Q = Q1 = Q2 =Q3
Battery C1 C2 C3 + - Q1 Q2 Q3

78. series connection = Sum of voltages
Total voltage (V)
series connection = Sum of voltages
V = V1 + V2 + V3
Battery C1 C2 C3 + - V1 V2 V3 • A B

79. Equivalent Capacitance in Series: Where does the equation come from?
Q1= Q2 = Q3 + - C1 C2 C3 V1 V2 V3
V = V1 + V2 + V3
Equivalent Ce for capacitors in series:

80. CAPACITORS IN PARALLEL

81. CAPACITORS IN PARALLEL
Parallel Capacitors always have the same voltage drop across each of them.
The charge on the equivalent capacitor CEQ is the sum of the charges on both capacitors.

82. Equivalent Capacitance in Parallel: Where does the equation come from?
Parallel capacitors in Parallel: C2 C3 C1 + -
Q = Q1 + Q2 + Q3
Equal Voltages: CV = C1V1 + C2V2 + C3V3
Equivalent Ce for capacitors in parallel:
Ce = C1 + C2 + C3

83. C3 is now in parallel with C2,4 Ceq= C3 + C2,4 = 3 +1.33 = 4.33 μF
11.14 a. Find the equivalent capacitance of the circuit.
C2 = 2 μF, C3 = 3 μF, C4 = 4 μF
V = 120 V
C2 and C4 are in series
= 1.33 μF
C3 is now in parallel with C2,4
Ceq= C3 + C2,4 =
= 4.33 μF

84. First, we need to find the total charge of the system QT = CeqV
b. Determine the charge on each capacitor.
First, we need to find the total charge of the system
QT = CeqV
= 4.33 (120)
= 520 μC
Q3 = C3V
= 3(120)
= 360 μC
Q2 and Q4 have the same charge since they are in series:
Q2 = Q4 = QT - Q3 =
= 160 μC
Q3 = 360 μC, Q2 = Q4 =160 μC

85. The remaining voltage (120 - 40 = 80 V) goes through the C2 capacitor.
c. What is the voltage across the 4 μF capacitor?
= 40 V
The remaining voltage ( = 80 V) goes through the C2 capacitor.