Unit 3B. Gravitational Fields Electric Fields

Transcription

1 Unit 3B Gravitational Fields Electric Fields 1

2 Force of gravity can be calculated using Newton s Universal Law of Gravity FG F G m m 1 r Force of gravity is directly proportional to the masses involved Force of gravity is inversely proportional to the square of the distance (r) between the centre of the masses of the objects. F G mm r

3 can be replaced by = if we have the appropriate proportionality constant. For our universe the constant is G which is called the universal gravitational constant G Nm 2 kg F G Gm m r

4 Calculate your weight on the following planets and moons Mass Earth Our Moon Jupiter Mars 5.97x10 24 kg 7.346x kg x kg x10 kg Radius 6.366x 10 3 km 1.74x10 3 km x10 km x 10 km Fg g 4

5 Problems 1. What happens to the gravitational force when: a)the distance between the two masses doubles? b)double one mass and triple the other? c)double one mass and half the radius? 5

6 2. What is the gravitational force of attraction between a 450 kg rock and a 1100 kg car that are separated by a distance of 3.5 m? 6

7 3. Calculate the mass of two identical objects that experience a gravitational force of attraction of 2.16 x N when there is 2.5 m of separation between them. 7

8 Orbits and Orbital Speed What is an orbit? An orbit is a regular, repeating path that an object in space takes around another one. An object in an orbit is called a satellite. A satellite can be natural, like the moon, or human made. 8

9 Orbits and Orbital Speed In our solar system, the Earth orbits the Sun, as do the other planets. They all travel on or near the orbital plane, ecliptic plane an imaginary disk-shaped surface in space. All of the orbits are either circular or elliptical in their shape. 9

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11 Alice and Bob: Why doesn t the moon fall down? 11

12 What is shown here? Space Debris 12

13 Orbital Speed It is the speed of a satellite to maintain orbit at a certain altitude. What type of motion is a satellite experiencing? Circular We are going to assume it is uniform circular motion. 2 mv F c r 13

14 What provides the force for this motion? Gravity Gm m 1 2 F G 2 r Lets derive a formula for orbital speed of an object with a mass, m 1 at a certain distance, r, from the centre of a planet of mass m 2 14

15 1. What is the speed of the International Space Station (ISS) in its typical orbit of 340 km above Earth s surface? 15

16 2. What is the speed of a satellite orbiting Earth at an altitude of 22,500 km (above Earth s surface)? 16

17 Geosynchronous orbit One application of the law of gravity is to figure out how to put a satellite into orbit so that it remains in the same position relative to ground all the time. A satellite in such a position can be used to relay microwave communication signals from one part of the globe to another. Before communications satellites, such signals had to be relayed via earth-based relay towers which had to be placed in sight of one another. 17

18 Geosynchronous orbit Such satellites are of major importance in our daily lives: Satellite TV GOES weather satellites Communication satellites 18

19 Questions: 1. What is the period of a geosynchronous satellite? 19

20 2. A)What is the orbital radius of a geosynchronous satellite situated above the equator? B) What is its altitude above the equator? Radius 6378 Km 20

21 Other random questions: 3. What speed is required for an object to stay in an orbit two Earth radii above Earth s surface? 21

22 4. The planet Mars has a satellite, Phobos, that travels in a circular orbit of radius 9.4 x 10 6 m with a period of 7 h and 39 min. What is the mass of Mars? 22

23 5. What is the period of a satellite orbiting Earth at an altitude equal to 1 earth radius? 23

24 Fun One The GPS satellites and complete two full orbits every day. The GPS satellites are not in a geostationary orbit, but rise and set two times per day. What is their altitudes? Ans: They circle the Earth at an altitude of about 20,000 km 24

25 ELECTRIC FIELDS & ELECTRIC FORCE 25

26 Text Page 543 Shielding of Cell Phones 26

27 Introduction to Electric Field Strength You can exert a force on things because you have strength. Since an electric field exerts a force on a charge that is in the field, the field must also have strength. Say that the test charge, q t in the picture is being repelled with an electric force of 10 N. Further, assume that the test charge is 2.0 C.

28 Question What is the force per coulomb on the test charge? Answer 10 N on 2.0 C means 5 N per C, or 5 N/C. 5 N/C (or 10 N 2 C) is the Electric Field Strength represented by the symbol in the previous picture. Therefore, electric field strength at a particular point in the field is the force per unit charge at that point, or in mathematical terms: F q e t

29 The units of electric field strength are N/C. If it becomes critical that the direction be maintained in the computations, we will compute the magnitude E from: Fe q t And then add the appropriate direction to the final answer if required. The direction of the electric field is ALWAYS determined by the direction that a POSITIVE TEST CHARGE would move when placed at that point. In many textbooks the term electric field strength (or intensity) is shortened to electric field.

30 Example: 1.A charged rubber balloon has an electric field intensity around it. At a particular point in the field a test charge of 2.5 nc experiences an 4 electric force of N A) What is the electric field strength at that point?

31 B) If a test charge of 4.5 mc was placed at that same point, what force would it experience?

32 C) What is the direction of the force? That depends of the type of charge on the balloon. If the charge on the balloon is negative, then the electric field will be directed from towards the balloon.

33 Coulombs (C) Note: In the definition of electric field, the charges are measured in. This is the SI unit of charge. It is based upon the charge on one electron (or proton) called the. elementary charge e = x C

34 Examples 1.How many electrons are in 1 C of charge? Q n ne Q e Not on Formulae Sheet 1 C C / e e

35 2. How many electrons are in a bolt of lightning with 20 C of charge?

36 3. Calculate the number of excess electrons or protons on a body, given its net charge is mc. n Q e C C / e excess protons

37 You should know these prefixes and symbols from to QUIZ NEXT CLASS

38 ELECTRIC FORCE electric Describes the force between charges. like It is a repulsive force between charges (i.e. two positives or two negatives) It is an attractive force between different charges. (i.e. a positive and negative charge) It increases with larger charges and closer distances.

39 It is the force which keeps atoms and molecules together. It is responsible for tensing our muscles, and for cooking our food. It is the cause of most common forces that we have discussed so far. (other than gravity) Magnetic force is a form of electric force.

40 Sample Questions on direction of electric fields and forces and electric charges 1. An object with a charge of 4.80 mc experiences an electrostatic force having a magnitude of 6.00 x 10-2 N when placed near a negatively charged metal sphere. What is the electric field strength at this location? (A) 1.25 x 10 4 N/C directed away from the sphere (B) 1.25 x 10 4 N/C directed toward the sphere (C) 2.88 x 10-8 N/C directed away from the sphere (D) 2.88 x 10-8 N/C directed toward the sphere 40

41 2. Three rods, A, B and C, are brought near each other. Rod A attracts rod B but repels rod C. What type of charge is on each rod? rod A rod B rod C (A) negative negative negative (B) negative negative positive (C) negative positive negative (D) negative positive positive 41

42 Charles Augustin de Coulomb ( ) was the first to experiment with the forces that exist between any two electric point charges. Point charges are extremely small particles that have no measurable dimensions but carry an electric charge. By using Cavendish s torsion balance, Coulomb was able to measure the torque and thus the force applied between the charges placed at specific distances apart

43 F Coulomb s Law F kq q 1 2 r 2 Electrical Force (Newton) q 1 charge on one object (C - coulombs) q 2 charge on other object (C ) r distance between the centers of the objects (metres) k Coulombs Constant k = 9.0 x 10 9 N m 2 / C 2

44 The electric force between two charged objects is directly proportional to the product of the charges If q 1 is doubled and q 2 is tripled, then the force increases by a factor of (because is directly proportional to q 1 x q 2 ). The signs of the charges are not included in calculating the magnitudes of the forces (magnitude is always positive). The direction of the force is found from a freebody diagram.

45 What is the direction of the electric force in the following situations? A) B) 45

46 The F between two charged particles varies inversely as the square of the distance between their centres. If distance doubles, then decreases by a factor of 1/4 ; If the particles are moved 3 times closer (i.e., distance decreases by a factor of 1/3), then increases by a 9 factor of. F F

47 Examples of Coulomb s LAW 1. If one charge increases by a factor of 3 and the other by a factor of 2 while the distance between their centers increases by a factor of 2, what is the overall affect on the force? F 2 kq q r

48 2. Calculate the repelling force between two charged spheres, each having a charge of +2.0 micro coulombs, whose centres are 10 cm apart. F kq q r 1 2 2

49 3. Determine the electric force between two point charges of +4.0 x 10-8 C and -8.0 x 10-7 C, separated by a distance of 3.0 mm. Is the force one of attraction or repulsion.

50 NOTES: More than two charges might be involved in a level three question. Read the F forces like this: BA as the force of B on A. C A What is the net force on A due to the presence of B and C? B Q A = -2.0 x 10-6 C, Q B = +4.0 x 10-6 C, Q C = -3.0 x 10-6 C r ba = 15 cm r ca = 10. cm

51 F BA F BA kq q r 2 BA Given: Q A = -2.0 x 10-6 C, Q B = +4.0 x 10-6 C, r BA = 15 cm B A C A B

52 F CA F CA kq q C A r 2 CA Given: Q A = -2.0 x 10-6 C, Q C = -3.0 x 10-6 C, r CA = 10. cm C A B

53 Fnet on A

54 Coulomb Worksheet

55 1. Calculate the electric force between two -3.0 x 10-7 C charges placed 3.5 mm apart. 55

56 2. Find the electric force between 2 electrons that are 30 micrometres apart. 56

57 3. Find the distance between 2 protons in a helium nucleus given the electrostatic force that each proton exerts on each other is N. 57

58 4. Determine the Electric force acting on B for each of the following: A) Q Q Q A B C 3.6C 4.2C 5.6C 58

59 4.B) Q A C 8.6mC Q 4.2mC B Q 7.9mC 59

60 5.Two small styrofoam balls are covered in a metallic paint and given identical charges. The balls each have a mass of 5.0 g and hang at the ends of 45 cm non-conducting string as shown. Due to the repelling force between the balls, the strings form an angle of 44 o. What is the charge on each ball? 60

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62 Text page 538 #5,7A, 8 62

63 Different Form of Formula The electric field strength can also be determined from the charge on the object, and the distance from the object. To do this we derive a new form of using Coulomb s Law.

64 Charged Object Q r Consider Distance from centre of charged object to test charge Test charge q Q and q are the respective charges on the charged object and the test charge F e q kqq 2 r q kqq 2 r 1 q kq r 2

65 Examples for kq r 2 1. The ball of a Van der Graff generator has a charge of 7.00 nc on it. What is the electric field at a distance of cm from the centre of the ball?

66 2. The electric field is 2.67 x 10 3 N/C at a point 68.4 mm from a charged object. What is the charge on the object? (Express the charge in micro coulombs)

67 3. At what distance from a point charge of 6.2 mc is the electric field 4.2 x 10 2 N/C?

68 4.The ball of a Van der Graff generator has an excess of 3.0 X electrons. A) What is the electric field acting on a 1.0 nc test charge located 6.0 m from the generator? B) What is the force acting on the test charge?

69 TEXT Page 553 # 1-9

70 Electric Fields around or between Multiple Charges

71 Linear charges 1. Find the electric field at the point X 0.50 m A mc X 2.0 m B mc

72 2. Two point charges are 0.35 m apart. The charge on q 1 is 2.4 x 10-9 C and the charge on q 2 is -1.6x10-8 C. A) Calculate the net electric field at point P. Include direction 72

73 B) A new charge of 3.1x10-12 C is placed at P. Calculate the electric force on the new charge. 73

74 Public questions from June 2007 exam A negatively charged particle with a mass of kg is at rest between two horizontal parallel charged plates as shown. If there is an excess of 5.0 x 10 2 electrons on the particle, calculate the electric field strength between the parallel plates.

75 Text Questions on electric fields and force Page # 6-8, 17, 18, 27

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