1. Static Electricity / Electrostatics
Where have you experienced this?
Shocked a friend? (or yourself)
Clothes in the dryer stick together
Stroke a cat and see the hair standing up on end.
Rub a rubber balloon in somebody’s hair
Bolt of lightning

2. Summary of Subatomic Particles
Proton
Neutron
Electron
In nucleus
Tightly Bound
Positive Charge
Massive
No Charge
Outside nucleus
Weakly Bound
Negative Charge
Not very massive

3. Charged versus Uncharged Particles
The number of electrons which surround the nucleus will determine whether or not an atom is electrically charged or electrically neutral.
Charged versus Uncharged Particles
Positively Charged
Negatively Charged
Uncharged
Possesses more protons than electrons
Possesses more electrons than protons
Equal numbers of protons and electrons

4. Electrons are migrants.
They are attracted to the protons in the nucleus, but they can be persuaded to become attracted to protons in a different nucleus and join another electron shell if energy is added.

5. Charged atoms What condition is true for a charged atom?
They contain unequal amounts of protons and electrons
The same can be said for objects…

6. Positive, Negative or Nuetral?

7. Electric Charge
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.

8. Glass and Silk
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 a excess of electrons.

9. The First Law of Electrostatics
Like charges repel; unlike charges attract.
Pos
Neg
Pos
Neg

11. Here’s the kicker…
Charged objects attract neutral objects!
The electrons in the neutral object are attracted (or repelled) by the electrons (or protons) in the charged object. They move accordingly, and the object becomes polarized.

12. 2. On two occasions, the following charge interactions between balloons A, B and C are observed. In each case, it is known that balloon B is charged negatively. Based on these observations, what can you conclusively confirm about the charge on balloon A and C for each situation.

13. The Quantity of Charge
The quantity of charge (q) can be defined in terms of the number of electrons, but the Coulomb (C) works better
The Coulomb: 1 C = 6.25 x 1018 electrons
Which means that the charge on a single electron is:
1 electron: e- = -1.6 x C

14. Units of Charge
The coulomb (selected for use with electric currents) is actually a very large unit for static electricity. Thus, we often encounter a need 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. = +5.8 x 10^-7 Coulombs (rounded)
Ex: Determine the quantity and type of charge on an object which has x 1012 more protons than electrons.
3.62 x 1012
x C
6.25 x 1018 e
= +5.8 x 10^-7 Coulombs (rounded)

16. Example 1. If 16 million electrons are removed from a neutral sphere, what is the charge on the sphere in coulombs?
1 electron: e- = -1.6 x C
+ +
q = x C
Since electrons are removed, the charge remaining on the sphere will be positive.
Final charge on sphere:
q = pC

17. Coulomb’s Law
The force of attraction or repulsion between two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. F r q q’ - +

18. Coulomb’s constant (k)
Coulomb’s Law r
Charge (Q)
Coulombs (C)
1 C = x 1018 e
e = x C
Force (N)
Distance (m)
Coulomb’s constant (k)
k = x 109 Nm2/C2

19. Electric Charge and Electric Field 16
Coulomb’s Law
The force is along the line connecting the charges, and is attractive if the charges are opposite, and repulsive if they are the same.
Electric Charge and Electric Field 16

20. - + F Draw and label givens on figure: r F = 3.38 x 104 N; Attraction
Example 2. A –5 mC charge is placed 2 mm from a +3 mC charge. Find the force between the two charges.
+3 mC
-5 mC F
Draw and label givens on figure: - q q’ + r
2 mm
F = 3.38 x 104 N; Attraction
Note: Signs are used ONLY to determine force direction.

21. Problem-Solving Strategies
1. Read, draw, and label a sketch showing all given information in appropriate SI units.
2. Do not confuse sign of charge with sign of forces. Attraction/Repulsion determines the direction (or sign) of the force.
3. Resultant force is found by considering force due to each charge independently. Review practice on vectors, if necessary.
4. For forces in equilibrium: SFx = 0 = SFy = 0.

22. - + 1 nC = 1 x 10-9 C 1. Draw and label. r2 r1 F1 F2 2. Draw forces.
Example 3. A –6 mC charge is placed 4 cm from a +9 mC charge. What is the resultant force on a –5 mC charge located midway between the first charges?
1 nC = 1 x 10-9 C
1. Draw and label. - +
2 cm
+9 mC
-6 mC q1 q2 r2 r1 F1 F2
2. Draw forces. q3
3. Find resultant; right is positive.
F1 = 675 N
F2 = 1013 N

23. Example 3. (Cont.) Note that direction (sign) of forces are found from attraction- repulsion, not from + or – of charge. + - +
2 cm
+9 mC
-6 mC q1 q2 r2 r1 q3 F2 F1
F1 = 675 N
F2 = 1013 N
The resultant force is sum of each independent force:
FR = N
FR = F1 + F2 = 675 N N;

24. + - - Draw free-body diagram. F2 F1
Example 4. Three charges, q1 = +8 mC, q2 = +6 mC and q3 = -4 mC are arranged as shown below. Find the resultant force on the –4 mC charge due to the others.
Draw free-body diagram. + -
4 cm
3 cm
5 cm
53.1o
+6 mC
-4 mC
+8 mC q1 q2 q3 -
53.1o
-4 mC q3 F1 F2
Note the directions of forces F1and F2 on q3 based on attraction/repulsion from q1 and q2.

25. + - F2 F1 Thus, we need to find resultant of two forces:
Example 4 (Cont.) Next we find the forces F1 and F2 from Coulomb’s law. Take data from the figure and use SI units. + -
4 cm
3 cm
5 cm
53.1o
+6 mC
-4 mC
+8 mC q1 q2 q3 F2 F1
Thus, we need to find resultant of two forces:
F1 = 115 N, 53.1o S of W
F2 = 240 N, West

26. Example 4 (Cont.) We find components of each force F1 and F2 (review vectors).-
-4 mC q3
F1= 115 N
F1y
F1x
F1x = -(115 N) Cos 53.1o = N
F1y = -(115 N) Sin 53.1o = N
Now look at force F2:
F2x = -240 N; F2y = 0
Rx = SFx ; Ry = SFy
Rx= -309 N
Rx = – 69.2 N – 240 N = -309 N
Ry= N
Ry = N – 0 = N

27. Example 4 (Cont. ) Next find resultant R from components Fx and Fy
Example 4 (Cont.) Next find resultant R from components Fx and Fy. (review vectors).
Rx= -309 N
Ry= N -
-4 mC q3
Ry = N
Rx = -309 N
We now find resultant R,q: f R
R = ( (309)2 + (92.1)2 )1/2 = 322.4
Thus, the magnitude of the electric force is:
R = N

28. Resultant Force: R = 322 N, q = 253.40
Example 4 (Cont.) The resultant force is 317 N. We now need to determine the angle or direction of this force.
-69.2 N -
-309 N f R q
-92.1 N
-92.1 N
The reference angle is: f = 73.40S of W
Or, the polar angle q is: q = =
Resultant Force: R = 322 N, q =

29. Summary of Formulas: Like Charges Repel; Unlike Charges Attract.
1 mC = 1 x 10-6 C
1 nC = 1 x 10-9 C
1 pC = 1 x C
1 electron: e- = -1.6 x C

30. Electric Charge and Electric Field 16
The Electric Field
Electric field of a point charge +Q r
Electric Charge and Electric Field 16

31. Electric Charge and Electric Field 16
A proton is released in a uniform electric field, and it experiences an electric force of 3.75 × N toward the south. What are the magnitude and direction of the electric field? E + F
south
Electric Charge and Electric Field 16

32. Electric Charge and Electric Field 16
What are the magnitude and direction of the electric field at a point midway between a +7.0 mC and a -8.0 mC charge 8.0 cm apart? Assume no other charges are nearby. d E1 + - q1 q2 E2
Electric Charge and Electric Field 16

33. Electric Charge and Electric Field 16
A proton (m = 1.67 x kg) is suspended at rest in a uniform electric field E. Take into account gravity at the Earth's surface, and determine E. E + q
Electric Charge and Electric Field 16

34. Electric Charge and Electric Field 16
The Electric Field
The electric field can be represented by field lines. These lines start on a positive charge and end on a negative charge.
Electric Charge and Electric Field 16

35. Electric Charge and Electric Field 16
The Electric Field
The number of field lines starting (ending) on a positive (negative) charge is proportional to the magnitude of the charge. E
The electric field is stronger where the field lines are closer together. E A B C
Electric Charge and Electric Field 16

36. Electric Charge and Electric Field 16
The Electric Field
Electric field of two charges:
Electric Charge and Electric Field 16

37. Electric Charge and Electric Field 16
The Electric Field
The electric field between two closely spaced, oppositely charged parallel plates is constant.
Electric Charge and Electric Field 16

38. First, we note that the direction of E is toward –Q (down).
A +2 nC charge is placed at a distance r from a –8 mC charge. If the charge experiences a force of 4000 N, what is the electric field intensity E at point P?
Electric Field . - -Q P
+2 nC + +q
4000 N E E r
–8 mC
First, we note that the direction of E is toward –Q (down).
E = 2 x 1012 N/C Downward
Note: The field E would be the same for any charge placed at point P. It is a property of that space.

39. The E-field is downward, and the force on e- is up. E . F
A constant E field of 40,000 N/C is maintained between the two parallel plates. What are the magnitude and direction of the force on an electron that passes horizontally between the plates.c - e-
The E-field is downward, and the force on e- is up. E . F
F = 6.40 x N, Upward

40. The E-Field at a distance r from a single charge Q
Consider a test charge +q placed at P a distance r from Q. + Q . r P E + Q . r P F + +q
The outward force on +q is:
The electric field E is therefore:

41. First, find the magnitude: . r 3 m
What is the electric field intensity E at point P, a distance of 3 m from a negative charge of –8 nC?
First, find the magnitude: . r P -Q
3 m
-8 nC
E = ?
E = 8.00 N/C
The direction is the same as the force on a positive charge if it were placed at the point P: toward –Q.
E = 8.00 N, toward -Q

42. The Resultant Electric Field.
The resultant field E in the vicinity of a number of point charges is equal to the vector sum of the fields due to each charge taken individually.
Consider E for each charge. E2 + - q1 q2 q3 A E1
Vector Sum:
E = E1 + E2 + E3 ER E3
Magnitudes are from:
Directions are based on positive test charge.

43. - + · q1 q2 E for each q is shown with direction given. E1 A E2
Example. Find the resultant field at point A due to the –3 nC charge and the +6 nC charge arranged as shown. + - q1 q2
4 m
3 m
5 m
-3 nC
+6 nC
E for each q is shown with direction given. E1 A E2
Signs of the charges are used only to find direction of E

44. - + · q1 q2 E2 E1 A E1 = 3 N/C , North E2 = 3.38 N, West
Example . (Cont.)Find the resultant field at point A. The magnitudes are: + - q1 q2
4 cm
3 cm
5 cm
-3 nC
+6 nC E2 E1 A
E1 = 3 N/C , North
E2 = 3.38 N, West E2 E1 ER
Next, we find vector resultant ER f

45. Resultant Field: ER = 4.25 N; 138.40
Example . (Cont.)Find the resultant field at point A using vector mathematics. E2 E1 ER
E1 = 3 N, West
E2 = 3.38 N, North f
Find vector resultant ER
3.00N
3.38N
f = N of W; or q =
Resultant Field: ER = 4.25 N;