Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 20 Electric Forces and Fields Topics: • Electric charge • Forces between charged • • objects The field model and the electric field Forces and torques on charged objects in electric fields Sample question: In electrophoresis, what force causes DNA fragments to migrate through the gel? How can an investigator adjust the migration rate? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-1 Late Paper Homework Policy and Solution Posting A. Late papers must be handed in by 5 PM 2 days after assignment is due • Late penalty 10% per day • Solutions will post shortly after 5 PM – no late work accepted after solutions are posted B. Late papers must be handed in by 5 PM 4 days after assignment is due • Late penalty 10% per day • Solutions will post shortly after 5 PM – no late work accepted after solutions are posted Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-13 Discuss Einstein Video and the Nature of Science What make science different from other ways people study and learn about the world around us? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-13 Checking Understanding Two spheres are touching each other. A charged rod is brought near. The spheres are then separated, and the rod is taken away. In the first case, the spheres are aligned with the rod. After the charged rod is removed, which of the spheres (A & B) is: i) Positive ii) Negative iii) Neutral A => Sphere A is + and sphere B is – B => Sphere A is – and Sphere B is + C => Spheres A and B are both + D => Spheres A and B are both – E => Spheres A and B are both neutral Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-13 Checking Understanding Two spheres are touching each other. A charged rod is brought near. The spheres are then separated, and the rod is taken away. In the second case, they are perpendicular. After the charged rod is removed, which of the spheres (C & D) is: i) Positive ii) Negative iii) Neutral A => Sphere C is + and sphere D is – B => Sphere C is – and Sphere D is + C => Spheres C and D are both + D => Spheres C and D are both – E => Spheres C and D are both neutral Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-13 Warm-up 1 Charging 2 spheres You have two conducting spheres. How can you charge them with opposite charges without touching either one with a charged object? (Anything else is fair game) Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-3 Warm-up 2: Charged Spheres & Forces Two identical metal spheres are firmly fastened to and electrically insulated from frictionless plastic air pucks that ride on an air table as shown below. The pucks are held in place as a charge of 2.0 x 10-8 C is placed on sphere A on the left and a charge of 6.0 x 10-6 C is placed on sphere B on the right. The pucks are then released so that the pucks with the spheres attached are now free to move without across the table. A. Draw Free-Body Diagrams for the pucks and spheres (Treat each puck and sphere system as a single object) B. How do the Coulomb forces acting on spheres A & B compare? (Use a ratio) C. Which sphere has the greater acceleration? How would your answer change if the mass of the puck under sphere A was reduced by 50%? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-3 Charged Spheres & Forces Two identical metal spheres are firmly fastened to and electrically insulated from frictionless plastic air pucks that ride on an air table as shown below. The pucks are held in place as a charge of 2.0 x 10-8 C is placed on sphere A on the left and a charge of 6.0 x 10-6 C is placed on sphere B on the right. The pucks are then released so that the pucks with the spheres attached are now free to move without across the table. D. As the two spheres get farther away from one another, how would (if at all) the following quantities change? 1) Force 2) Speed 3) Acceleration Choices: a) Increase b) Decrease c) Stay the same d) Can’t tell Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-3 Coulomb’s Law Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-15 Example Problem 1 Two 0.10 g honeybees each acquire a charge of +23 pC as they fly back to their hive. As they approach the hive entrance, they are 1.0 cm apart. What is the magnitude of the repulsive force between the two bees? How does this force compare with their weight? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-33 Example Problem 1 Two 0.10 g honeybees each acquire a charge of +23 pC as they fly back to their hive. As they approach the hive entrance, they are 1.0 cm apart. What is the magnitude of the repulsive force between the two bees? How does this force compare with their weight? Fe, B1=>B2 = K |q1| |q2| / (r12)2 = (9e9 Nm2/C2)(23e-12 C)2 / (0.010 m)2 Fe, B1=>B2 = 4.8 x 10-8 N Fg, Earth=>B1 = mg = 0.00010 kg x 9.8 m/s2 = 9.8 x 10-4 N The electric force between the bees is much smaller than the weight of the bees on Earth. Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-33 The Force of Gravity Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 6-35 Example Problem 1 – take 2 0.10 g honeybees each acquire a charge of +23 pC as they fly Two back to their hive. As they approach the hive entrance, they are 1.0 cm apart. What is the magnitude of the repulsive force between the two bees? How does this force compare with their weight? Fe, B1=>B2 = K |q1| |q2| / (r12)2 = (9e9 Nm2/C2)(23e-12 C)2 / (0.010 m)2 Fe, B1=>B2 = 4.8 x 10-8 N Fg, B1=>B2 = G m1 m2 / (r12)2 Fg, B1=>B2 = (6.67e-11 Nm2/kg2)(0.0001 kg)2 / (0.010 m)2 Fg, B1=>B2 = 6.7 x 10-15 N The electric force between the bees is much larger than the gravitational forces between them. Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-33 Conceptual Example Problem All charges in the diagrams below are equal magnitude. In each case, a small positive charge is placed at the blank dot. In which cases is the force on this charge: • to the right? • to the left? • zero? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-28 Checking Understanding All charges in the diagrams below are of equal magnitude. In each case, a small, positive charge is placed at the black dot. In which case is the force on the small, positive charge the largest? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-25 Introduction to Electric Field 1. Find the Electric force from a 9.00e-6 C charge on the following charges at 1m, 2m, and 3 m. a) 3.00e-6 C b) -4.00e-6 C c) -10.0e-6 C Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Introduction to Electric Field 1. Find the Electric force from a 9.00e-6 C charge on the following charges at 1m, 2m, and 3 m. a) 3.00e-6 C b) -4.00e-6 C c) -10.0e-6 C 2. Find the force per charge from a 9.00e-6 C charge on the charges above at 1m, 2m, and 3 m. Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Nature of Electric Field E-field Applet 1 http://physics.weber.edu/schroeder/software/EField/ Back-up applet:https://phet.colorado.edu/en/simulation/charges-andfields What observations can we make about E-fields? Source Charges and Test Charges? Superposition of E-fields? Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Nature of Electric Field • Test charge is a small positive charge to sample the E-Field • Charge of test charge is small compared to source charges (source charges are the charges that generate the field) • E-field vectors • E-field is the force per charge • E-field vectors points away from + charges • E-field vectors point towards - charges • E-field for point charges gets weaker as distance from source point charges increases • For a point charge E = Fe / q = [k Q q / r2] / q = k Q / r2 Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The Electric Field Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-34 The Electric Field of a Point Charge Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Slide 20-35 Nature of Electric Field Vectors • Test charge is a small positive charge to sample the E-Field • Charge of test charge is small compared to source charges (source charges are the charges that generate the E-field) • E-field vectors • E-field is the force per charge • E-field vectors points away from + charges • E-field vectors point towards - charges • E-field for point charges gets weaker as distance from source point charges increases • E-fields add as vectors, at a point in space Enet,x = E1x + E2x + … • For a point charge E = Fe / |q| = [k |Q| |qt| / r2] / |qt| = k |Q| / r2 • Electric Force Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley.