Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Electric charge Forces between charged objects The field model and the electric field Forces and torques on charged objects in electric fields Chapter 20 Electric Forces and Fields Topics: Sample question: In electrophoresis, what force causes DNA fragments to migrate through the gel? How can an investigator adjust the migration rate? Slide 20-1

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Late Paper Homework Policy and Solution Posting Slide 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. Discuss Einstein Video and the Nature of Science Slide 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. 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 Slide 20-13

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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 Slide 20-13

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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) Slide 20-3

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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 C is placed on sphere A on the left and a charge of 6.0 x 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%? Slide 20-3

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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 C is placed on sphere A on the left and a charge of 6.0 x 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) Force2) Speed3) Acceleration Choices: a) Increase b) Decrease c) Stay the same d) Can’t tell Slide 20-3

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. Coulomb’s Law Slide 20-15

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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? Example Problem 1 Slide 20-33

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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? F e, B1=>B2 = K |q 1 | |q 2 | / (r 12 ) 2 = (9e9 Nm 2 /C 2 )(23e-12 C) 2 / (0.010 m) 2 F e, B1=>B2 = 4.8 x N F g, Earth=>B1 = mg = kg x 9.8 m/s 2 = 9.8 x N The electric force between the bees is much smaller than the weight of the bees on Earth. Example Problem 1 Slide 20-33

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The Force of Gravity Slide 6-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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? F e, B1=>B2 = K |q 1 | |q 2 | / (r 12 ) 2 = (9e9 Nm 2 /C 2 )(23e-12 C) 2 / (0.010 m) 2 F e, B1=>B2 = 4.8 x N F g, B1=>B2 = G m 1 m 2 / (r 12 ) 2 F g, B1=>B2 = (6.67e-11 Nm 2 /kg 2 )( kg) 2 / (0.010 m) 2 F g, B1=>B2 = 6.7 x N The electric force between the bees is much larger than the gravitational forces between them. Example Problem 1 – take 2 Slide 20-33

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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? Conceptual Example Problem Slide 20-28

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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? Slide Checking Understanding

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

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 Back-up applet: fieldshttps://phet.colorado.edu/en/simulation/charges-and- fields 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 = F e / q = [k Q q / r 2 ] / q = k Q / r 2

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The Electric Field Slide 20-34

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. The Electric Field of a Point Charge Slide 20-35

Copyright © 2007, Pearson Education, Inc., Publishing as Pearson Addison-Wesley. 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 E net,x = E 1x + E 2x + … For a point charge E = F e / |q| = [k |Q| |q t | / r 2 ] / |q t | = k |Q| / r 2 Electric Force