1. Physics 2112 Unit 1: Coulomb’s Law
Today’s Concepts:
A) Coulomb’s Law
B) Superposition

2. What we thought…. Why are electrons fixed and stationary in plastics?
Nothing was difficult to understand. Videos were clear and concise.
Would like to see a variety of worked-out examples to strengthen understanding.
The most difficult is listening to his voice.
donuts?
I think I understood this pre lecture for the most part.
HA AP Physics
i am confused about the forces relating to magnetism.
Feel ok about the check point. However was a bit confused in one of the calculations, they didn't include one of charges and im not sure why. Would be nice to do a few calculations in class. The lecture seemed fine.
You're going to be so proud of me! Like OMG. For the first time ever I actually watched the whole pre-lecture and I actually thought about the Checkpoint questions. Assuming I actually got these questions right, I didn't have trouble with anything. :)
This is a recommended way to incorporate some of the free response answers from the CheckPoints into your lecture.

3. Electric Charges
Electric charges come in two sign: + and - Like sign charges repel, Opposite sign charges attract. Electric charge is “quantized”. qe=1.602 X 10-19C Charge of electron is exactly same as charge of proton
Why?
Why?
Why?
How?!

4. Electric charges is conserved
At Fermilab -1 +1
anti-proton
proton

5. Tribo-electric Series
Electric Charges
Tend to lose electrons when rubbed
Tribo-electric Series
Tend to gain electrons when rubbed

6. QUESTION
When we place near the positively charged glass rod, the ruler was attracted to it. What will happen when it is placed near the charged PVC rod?
A. It will be attracted to it
B. It will be repelled by it
C. There will be no effect

7. QUESTION
In our simulation, we have now rubbed the sweater and the balloon, transferring charge from one to the other. Both are pretty good insulators. What can we say about the force between the sweater and the balloon?
They will attract each other.
They will repel each other.
There will be no force between them.
This series of questions is meant to be used in conjunction with the PhET simulation involving the sweater and the balloon on static electricity.
Key point is that the attraction to the sweater and the attraction too the wall are caused by two different situations.

8. QUESTION
In our simulation, we have now rubbed the sweater and the balloon, transferring charge from one to the other, but the wall remained neutral. All three are pretty good insulators. What can we say about the force between the wall and the balloon?
They will attract each other.
They will repel each other.
There will be no force between them.

9. Conductors / Insulators
Conductors – charges are free to move anywhere on the conductor Insulators - charges remain where they are place except in cases of “extreme” force

10. Opposite signs attract
Electro-static Force
The force on a charge due to another charge is proportional to the product of the charges and inversely proportional to the separation squared. q2 q1 r
The force is always parallel to a line connecting the charges, but the direction depends on the signs of the charges:
Opposite signs attract q2 q1 q2 q1
Like signs repel

11. Unit of Charge is Columb (C)
Coulomb’s Law
k = 9 X 109 N m2/C2
Unit of Charge is Columb (C)
(big!) qe=1.602 X 10-19C
Charles-Augustin de Coulomb – French engineer who was a pioneer in torsion and soil mechanics.

12. QUESTION
A hydrogen atom is composed of a nucleus containing a single proton, about which a single electron orbits. When the electron is in the lowest orbital position, the electric force between the two particles is 2.3 X 1039 greater than the gravitational force! How would we adjust the separation between the two particles so the electric and gravitational forces are equal? A) We must move the particles farther apart. B) We must move the particles closer together. C) It can’t be done at any distance.

13. Coulomb’s Law Our notation: is the force by 1 on 2 (think “by-on ”)
is the unit vector that points from 1 to 2.
Examples:
If the charges have the same sign, the force by charge 1 on charge 2 would be in the direction of r12 (to the right). q2 q1
If the charges have opposite sign, the force by charge 1 on charge 2 would be opposite the direction of r12 (left). q1 q2

14. Coulomb Force
Two paperclips are separated by 10 meters. Then you remove 1 electron from each atom on the first paperclip and place it on the second one.
k = 9 x 109 N m2 / C2
electron charge = 1.6 x Coulombs
NA = 6.02 x 1023
What will the direction of the force be?
A) Attractive B) Repulsive
F = 9e9 x (1.6e-19 * 3e22)^2 / 100 = 2 e 15 Newtons
Equivalent to weight of 2e14 kg object near surface of the earth
paperclip 1e-3 kg
textbook 1 kg
person kg
car kg
Aircraft carrier tons = 1e5 x 1e3 = 1e 8 kg
Mt Everest m height,. Estimate volume as 1/3 h^3 = 1/3 * (9e3)^3 = 243 e 9 m^3. (if density = 1000 kg/m^3) = 2.43e14 kg.

15. Question
Two paperclips are separated by 10 meters. Then you remove 1 electron from each atom on the first paperclip and place it on the second one.
k = 9 x 109 N m2 / C2
electron charge = 1.6 x Coulombs
NA = 6.02 x 1023
Which weight is closest to the approximate force between those paperclips (recall that weight = mg, g = 9.8 m/s2)?
A) Paperclip (1 g x g)
B) Text book (1 kg x g)
C) Truck (104 kg x g)
D) Aircraft carrier (108 kg x g)
E) Mt. Everest (1013 kg x g)
F = 9e9 x (1.6e-19 * 3e22)^2 / 100 = 2 e 15 Newtons
Equivalent to weight of 2e14 kg object near surface of the earth
paperclip 1e-3 kg
textbook 1 kg
person kg
car kg
Aircraft carrier tons = 1e5 x 1e3 = 1e 8 kg
Mt Everest m height,. Estimate volume as 1/3 h^3 = 1/3 * (9e3)^3 = 243 e 9 m^3. (if density = 1000 kg/m^3) = 2.43e14 kg.

16. Example 1.1 (forces between paper clips)
Two 1 gram paperclips are separated by 10 meters. Then you remove 1 electron from each atom on the first paperclip and place it on the second one.
What is the force between the two clips?
F = 9e9 x (1.6e-19 * 3e22)^2 / 100 = 2 e 15 Newtons
Equivalent to weight of 2e14 kg object near surface of the earth
paperclip 1e-3 kg
textbook 1 kg
person kg
car kg
Aircraft carrier tons = 1e5 x 1e3 = 1e 8 kg
Mt Everest m height,. Estimate volume as 1/3 h^3 = 1/3 * (9e3)^3 = 243 e 9 m^3. (if density = 1000 kg/m^3) = 2.43e14 kg.

17. Example 1.2 (forces between students)
What would the force be between two 80kg students sitting two meters apart if the charge of the proton were 10-10% greater than the charge of an electron?
F = 9e9 x (1.6e-19 * 3e22)^2 / 100 = 2 e 15 Newtons
Equivalent to weight of 2e14 kg object near surface of the earth
paperclip 1e-3 kg
textbook 1 kg
person kg
car kg
Aircraft carrier tons = 1e5 x 1e3 = 1e 8 kg
Mt Everest m height,. Estimate volume as 1/3 h^3 = 1/3 * (9e3)^3 = 243 e 9 m^3. (if density = 1000 kg/m^3) = 2.43e14 kg.

18. CheckPoint: Forces on Two Charges
Two charges q = + 1 μC and Q = +10 μC are placed near each other as shown in the figure below.
Which of the following diagrams best depicts the forces acting on the charges:

19. CheckPoint Results: Forces on Two Charges
Since they are both positive they have to repel each other, and I would think the one with the stronger charge would repel more.
Just like a fly having the same amount of force impact on a bus that the bus has on the fly, two charges will have equal and opposite effects on each other in accordance with Newton's Third Law.
The force of 2 (+10 uC) on 1 (+1 uC) is 10 times greater than the force of 1 on 2 consequently causing charge 1 to become displaced at a much larger magnitude than charge 2.

20. Question
Two charges, -Q and +2Q are placed a distance R apart. The magnitude and direction of the force F on the -Q charge is shown by the arrow. Which of the following arrows represents the approximate magnitude and direction of the force on the particle with the larger (+2Q) charge? A. C. D. E. B.
+2Q -Q

21. Question
Two charges, -Q and +2Q are placed a distance R apart. The magnitude and direction of the force F on the -Q charge is shown by the arrow. Next, the –Q charge is changed to +2Q. Which of the following arrows represents the approximate magnitude and direction of the new force on the bottom +2Q?

22. Question
Two charges, -Q and +2Q are placed a distance R apart. The magnitude and direction of the force F on the -Q charge is shown by the arrow. The charges are then moved to a separation distance of R/2. Which arrow’s direction and length represents the force on the +2Q charge in this new position?e new force on the +2Q charge?

23. Superposition
If there are more than two charges present, the total force on any given charge is just the vector sum of the forces due to each of the other charges: q2
F2,1
F3,1
F4,1 F1
F4,1
F2,1 q1 F1
F3,1 q4 q3

24. Question: Superposition
What happens to the magnitude of Force on q1 if its sign is changed?
A) |F1| increases
B) |F1| remains the same
C) |F1| decreases
D) Need more information to determine q2 q1 q4 q3

25. The direction of all forces changes by 180o – the magnitudes stay the same:q1 q2 q3 q4
F2,1
F3,1
F4,1 F1 q1 q2 q3 q4
F4,1
F2,1 F1
F3,1 F1 F1
F2,1
F2,1
F3,1
F3,1
F4,1
F4,1

26. CheckPoint: Compare Forces
Compare the magnitude of the net force on q in the two cases.
A) |F1 | > |F2|
B) |F1 | = |F2|
C) |F1 |< |F2|
D) Depends on sign of q q -Q +Q
Case 1
Case 2
If the CheckPoint was done before the first class, you could use this opportunity to show results. If not, the next slide (hidden) present the CheckPoint as a Clicker Question
Case 1
Case 2

27. CheckPoint: Compare Forcesq -Q +Q
Case 1
Case 2
Compare the magnitude of the net force on q in the two cases.
A) |F1 | > |F2|
B) |F1 | = |F2|
C) |F1 |< |F2|
D) Depends on sign of q F1
F2 = 0
No matter the sign of charge q, there will be an attractive and repulsive force in case 1 (same direction), and two forces (attractive or repulsive depending on the sign of the charge on q) that cancel so that the net force on q is zero.
the distance between the two particles from charge q is the same and the quantity of charge on +Q and -Q is the same. so the magnitude of the net force will be the same in both cases.
I believe that if the sign of q is negative, the force on q will be less than if the sign of q is positive, as there will be two repelling forces acting on it rather than one repelling and one attracting force.
If the CheckPoint was done before the first class, you could use this opportunity to show results. If not, the next slide (hidden) present the CheckPoint as a Clicker Question

28. CheckPoint: Compare Forcesq -Q +Q
Case 1
Case 2
Compare the magnitude of the net force on q in the two cases.
A) |F1 | > |F2|
B) |F1 | = |F2|
C) |F1 |< |F2|
D) Depends on sign of q
F2 = 0 F1
D) because if the central charge is neutral there is no net force on either. If the central charge isnt neutral, the net force would be greater in case 1.
Good point. From now on, if I refer to a q, you can assume |q| = 0
If the CheckPoint was done before the first class, you could use this opportunity to show results. If not, the next slide (hidden) present the CheckPoint as a Clicker Question

29. CheckPoint: Force from Four Charges
A positive charge Q is placed at the origin. Four other positive charges are placed on a circle in the x-y plane centered on Q as shown. The top charge is located on the y axis and has a magnitude of 3q. The three bottom charges are symmetric about the y axis as shown and each has magnitude q. y 3q x Q q q
What is the direction of horizontal force on Q?
A) Fx > B) Fx = C) Fx < 0 q

30. Question y
A positive charge Q is placed at the origin. Four other positive charges are placed on a circle in the x-y plane centered on Q as shown. The top charge is located on the y axis and has a magnitude of 3q. The three bottom charges are symmetric about the y axis as shown and each has magnitude q. 3q x Q q q
What is the vertical force on Q?
A) Fy > B) Fy = C) Fy < 0 q
Next do calculation

31. Question What is the vertical force on Q?y
A positive charge Q is placed at the origin. Four other positive charges are placed on a circle in the x-y plane centered on Q as shown. The top charge is located on the y axis and has a magnitude of 3q. The three bottom charges are symmetric about the y axis as shown and each has magnitude q. 3q x Q
What is the vertical force on Q?
A) Fy > B) Fy = C) Fy < 0 q q q
The y distance between the bottom left and bottom right charges is less than the y distance of the 3q on top, so the net force in the y direction would be upwards
because the magnitude of the 3q charge will equal the magnitude of the three q charges.
all force of the 3q is directed along the negative y axis. some of the force exerted on Q by two of the q charges is directed along the x axis. therefore the net force is along the negative y axi
Next do calculation

32. Example 1.3 (Force on charge)q2 2m
+10uC q1 q3
-30uC 2m
+25uC
What is the force on q2?

33. + + Charge on a Conductor Charges are free to move on conductor
Charges always on very edge of conductor