Applications of Coulomb’s Law

1. Applications of Coulomb’s Law Physics 12

2. Joke of the day/clip of the day: • Minute physics again! • http://www.youtube.com/watch?v=eTfm03T3VAE&list=PLED25F943F8D6081C&index=17

3. Review: Coulomb’s Law • Fe is electrical force (N) • k is the Coulomb constant • q1 is the electrical charge on object 1 (C) • q2 is the electrical charge on object 2 (C) • r is distance between the objects (m)

4. Part 1 3 Charges Straight line

5. Sample Problem 1: • Three charges are arranged in a line; if the three charges are 15μC, -12μC and 18μC respectively. • The distance between the first two charges is 0.20m and the second and third charges is 0.30m. • Draw a diagram • Draw FB diagrams for each charge • What is the force (Fnet) experienced by the charge A? • What is the force (Fnet) experienced by the charge B? • What is the force (Fnet) experienced by the charge C?

6. Practise in a straight line: • Coulomb’s Law WS#2 • Questions 1-3

7. Part 2 3 Charges with angles!

8. What happens if the charges are NOT in a straight line? • We have to find the components (x and y) for each force and do some vector addition!

9. Force (Vector) Addition • To add forces, resolve each force into its components and treat the forces in the x-direction and y-direction independently • Once you sum the x and y components, use Pythagorean Theorem and Trigonometry to resolve into a resultant force

10. Example: Force Addition • A point P has forces of 12.0N at 24.3°, 17.6N at 112°, 6.78N at 241° and 10.2N at 74.4°. • A diagram may be helpful! • Determine the resultant vector.

11. Example cont`d: • X Ax = 12.0cos24.3= 10.9 Bx= 17.6cos112= -6.59 Cx= 6.78cos241= -3.29 Dx= 10.2cos74.4= 2.74 +3.76 • Y Ay = 12.0sin24.3= 4.94 By = 17.6sin112= 16.3 Cy = 6.78sin241= -5.93 Dy= 10.2sin74.4= 9.82 +25.1

12. Example cont`d θ

13. Practise vector addition: • Add the following vectors head to tail, mark angle! • F(x) = -2 , F(y) = 4 • F(x) = -0.2 , F(y) = -0.6 • F(x) = 100 , F(y) = 50 • F(x) = 3 , F(y) = -2 • F(x) = -0.25 , F(y) = 0.11

14. Coulomb’s Law and Vector Addition • When we consider an electrostatic system, we need to use Coulomb’s Law to determine the magnitude and direction of each force • Once the magnitude and direction of each force has been determined, then the vector sum can be completed

15. Sample problem 2: • Three charges are arranged as follows; (A) +2.2μC is placed 2.3m due north of (B) -3.7μC charge and (C) +1.9μC charge is 3.1m due east of (B). • Draw a diagram • What is the force experienced by the (B)?

16. Coulomb’s WS #2 • Try questions 4, 5

17. Sample problem 3: • Three charges are arranged as follows; a -2.0μC is placed 4.0m due north of a 3.0μC charge and 3.0m due west of a 5.0μC charge. • Draw a diagram • What is the force experienced by the 3.0μC charge?

18. Concerned with forces on 3.0μC charge: -2.0μC 5.0 μ C 1 3.0m 3 4.0m 5.0m 3.0 μ C 2

19. F1on2is only in y anyway, need to find angle for F3on2: -2.0μC 5.0 μ C 1 3.0m 3 θ tan θ = 4.0/3.0 θ = 53° 4.0m 3.0 μ C Total angle from x-axis: 2

20. Summary of components: 3.0 μ C 2

21. Use the x and y component data to determine the resultant force vector: θ tan θ = (-9.0x10-4) (-3.2x10-3) θ = 16° Use Pythagorean to find resultant force: (-3.2x10-3)2 + (-9.0x10-4)2 = c2 3.3 x10-3N= c

22. Finish Coulomb’s WS #2 • Question 6 • Page 640 • Questions 6, 7, 8

23. Part 3 Hanging angles!

24. Pith Ball Demo:

25. Sample problem 4: A negatively charged pith ball, with a mass of 1.25g is suspended from a thread from above. It makes a angle of 22° with the vertical when another negatively charged pith ball is brought near. The distance between the two pith balls is 3.35cm. What is the electrostatic force on the first pith ball?

26. Pith Ball

27. Coulomb’s WS #2 • Question 7 • Page 641 • Questions 9, 10