Lesson 17 Electric Fields and Potential

1. Lesson 17Electric Fields and Potential Eleanor Roosevelt High School Chin-Sung Lin

2. Electric Fields

3. Gravitational & Electric Forces • What are the formulas for the following physics laws? • Law of Universal Gravitation • Coulomb’s Law

4. Gravitational & Electric Forces m1m2 q1 q2 Fg = G Fe = k • Law of Universal Gravitation • Coulomb’s Law r2 r2

5. Gravitational Field • What’s the definition of gravitational field?

6. Gravitational Field • Gravitational Field: Force per unit mass Fg • Fg: gravitational force (N) • m: mass (kg) • g: gravitational field strength (N/kg, or m/s2) g = m

7. Electric Field • Electric Field: Force per unit charge Fe • Fe: electric force (N) • q: charge (C) • E: electric field strength (N/C) E = q

8. Gravitational & Electric Fields • Gravitational Field Fe Fg • Electric Field E = g = q m

9. Electric Field Source Charge Test Charge +q +Q r Fe Fe E = q

10. Electric Field Source Charge Test Charge +q • –Q Fe r Fe E = q

11. Electric Field • Electric field is a vector • A vector includes ___________ and ____________ +q • –Q Fe Source Charge Test Charge r Fe E = +q +Q q r Fe

12. Electric Field • Electric field is a vector • A vector includes direction and magnitude +q • –Q Fe Source Charge Test Charge r Fe E = +q +Q q r Fe

13. Electric Field • Can you apply Coulomb’s law to this formula and then simplify it? Fe E = = ??? q Source Charge Test Charge +q +Q r Fe

14. Electric Field • Electric Field: Force per unit charge • Fe: electric force (N) • q: test charge (C) • Q: source charge (C) • E: electric field strength (N/C) • r: distance between charges (m) • k: electrostatic constant (N m2/C2) Fe q Q Q E = = k = k q r2q r2

15. Electric Field Example • What is the magnitude of the electric field strength when an electron experiences a 5.0N force?

16. Electric Field Example • What is the magnitude of the electric field strength when an electron experiences a 5.0N force? E = Fe / q E = 5 N / (1.6 x 10-19 C) = 3.13 x 1019 N/C

17. Electric Field Example • What are the magnitude and direction of the electric field 1.5 m away from a positive charge of 2.1*10-9 C?

18. Electric Field Example • What are the magnitude and direction of the electric field 1.5 m away from a positive charge of 2.1*10-9 C? E = kQ / r2 E = (8.99 x 109 N m2/C2) (2.1 x10-9 C) / (1.5 m)2 = 8.4 N/C Direction: away from the positive charge

19. Electric Field Exercise • There is a negative charged particle of 0.32 C in the free space. (a) What are the magnitude and direction of the electric field 2.0 m away from the particle? (b) What are the magnitude and direction of the electric force when an electron is placed 2.0 m away from this particle? [3 minutes] e – – 0.32 C 2.0 m

20. Electric Field Exercise • There is a negative charged particle of 0.32 C in the free space. (a) What are the magnitude and direction of the electric field 2.0 m away from the particle? E = kQ / r2 E = (8.99 x 109 N m2/C2) (0.32 C) / (2.0 m)2 = 7.2 x 108 N/C Direction: toward the negative charge

21. Electric Field Exercise • There is a negative charged particle of 0.32 C in the free space. (b) What are the magnitude and direction of the electric force when an electron is placed 2.0 m away from this particle? E = Fe / q Fe = q E Fe = (1.6 x 10-19 C) (7.2 x 108 N/C) = 1.15 x 10-10 N

22. Aim: ElectricField • DoNow: (4 minutes) • Write down the definition of Electric Fieldin words • Write down theformulasof Electric Field in two different forms • Define every symbol in the formula and identify their units • Identify the relationships between Electric Field and other variables

23. Aim: Electric Field

24. Electric Field • Electric Field: Force per unit charge • Fe: electric force (N) • q: test charge (C) • Q: source charge (C) • E: electric field strength (N/C) • r: distance between charges (m) • k: electrostatic constant (N m2/C2) Fe Q E = = k q r2

25. Electric Field • Electric Field: Force per unit charge • Fe: electric force (N) • q: test charge (C) • Q: source charge (C) • E: electric field strength (N/C) • r: distance between charges (m) • k: electrostatic constant (N m2/C2) 1 1 E~ Q E~ Fe E ~ E ~ q r2

26. Electric Field • If you shift the test charge around, where can you find the electric field with the same magnitude? Source Charge Test Charge +q +Q Fe Fe E = q

27. Electric Field Source Charge Test Charge +q +Q E Fe E Fe E Fe E Fe

28. Electric Field • What will happen if you move the test charges away from the source charge? Source Charge Test Charge +q +Q E Fe E Fe E Fe E Fe

29. Electric Field Source Charge Test Charge +q +Q E Fe E Fe E Fe E Fe

30. Electric Field Source Charge Test Charge +q +Q E E Fe Fe E Fe E E Fe Fe E E Fe Fe E Fe

31. Electric Field Test Charge Source Charge Test Charge +q +Q +q E E Fe Fe E Fe E E Fe Fe E E Fe Fe E Fe

32. Electric Field Representation Vector representation - +

33. Electric Field Representation Line-of-Force representation - +

34. Electric Field Representation • How do you decide the strength of electric field? - +

35. Electric Field Representation • When the field lines are denser, the field is stronger - +

36. Electric Field Representation • Where can you find the the strongest electric field? E C A D B

37. Electric Field: Point Charge Line-of-Force representation - +

38. Electric Field: Pair of Charges Line-of-Force representation

39. Electric Field: Pair of Charges • Sketch the electric field for like charges? • + +

40. Electric Field: Pair of Charges Line-of-Force representation + • +

41. Electric Field: Pair of Charges Line-of-Force representation

42. Electric Field: Parallel Plates Line-of-Force representation

43. Electric Field: Parallel Plates • Anything special for the electric field between the parallel plates charged with opposite charges?

44. Electric Field: Parallel Plates • The electric field between the parallel plates is uniform except at both ends

45. Electric Field Example • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (a) What is the charge on the particle? (b) By how many electrons is the particle deficient?

46. Electric Field Example • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (a) What is the charge on the particle? E = Fe / q Fe = E qFg = mg Fe = Fg E q = mg (1.2 x 107 N/C) q = (5.87 x 10-10 kg) (9.81 m/s2) q = 4.80 x 10-16 C

47. Electric Field Example • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (b) By how many electrons is the particle deficient? e- = 1.6 x 10-19 C number of e- = 4.80 x 10-16 C / 1.6 x 10-19 C = 3000 e –

48. Electric Field Exercise • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 9.6 x 106 N/C and are 2.00 mm apart. What is the charge on the particle?

49. Electric Field Exercise • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 9.6 x 106 N/C and are 2.00 mm apart. What is the charge on the particle? E = Fe / q Fe = E qFg = mg Fe = Fg E q = mg (9.6 x 106 N/C) q = (5.87 x 10-10 kg) (9.81 m/s2) q = 6.0 x 10-16 C

50. Electric Field Exercise • A positively charged ball with mass 20 g is hanging between two charged parallel plates from the ceiling through an insulating wire with length 0.1 m. The electric field strength of the charged parallel plates is 4.2 x 109 N/C. When the ball is in balance, the wire and the vertical line form an angle of 60o. What is the charge of the ball?