Coulomb's Law and Electric Fields in Physics Lecture

1. Physics 102:Lecture 02 Coulomb’s Lawand Electric Fields • Today we will … • get some practice using Coulomb’s Law • learn the concept of an Electric Field Note the change in the Daily Planner

2. But first, a leftover from last lectureACT: Induced Dipole • An uncharged conducting sphere is hung next to a charged sphere. What happens when the uncharged sphere is released? 1) Nothing 2) Attracted to charged sphere. 3) Repelled from charged sphere. Van de Graaff demo

3. Recall Coulomb’s Law • Magnitude of the force between charges q1 and q2 separated a distance r: F = k q1q2/r2 k = 9x109 Nm2/C2 • Force is • attractive if q1 and q2 have opposite sign • repulsive if q1 and q2 have same sign • Units: • q’s have units of Coulombs (C) • charge on proton is 1.6 x 10-19 C • r has units of m • F has units of N 5

4. F7 1 2 3 4 5 6 7 8 9 10 11 12 Example Three Charges • Calculate force on +2mC charge due to other two charges • Calculate force from +7mC charge • Calculate force from –7mC charge • Add (VECTORS!) Q=+2.0mC 4 m 6 m Q=+7.0mC Q=-7.0 mC 10

5. F+7 Example Three Charges • Calculate force on +2mC charge due to other two charges • Calculate force from +7mC charge • Calculate force from –7mC charge • Add (VECTORS!) Q=+2.0mC 4 m 6 m Q=+7.0mC Q=-7.0 mC 10

6. Example Three Charges • Calculate force on +2mC charge due to other two charges • Calculate force from +7mC charge • Calculate force from –7mC charge • Add (VECTORS!) F+7 Q=+2.0mC • F = k q1q2/r2 F-7 5 m 4 m 6 m Q=+7.0mC Q=-7.0 mC 14

7. F+7 Q=+2.0mC F-7 5 m 4 m 6 m Q=+7.0mC Q=-7.0 mC 14 Example Adding Vectors F+7+F-7

8. F+7 Q=+2.0mC F-7 5 m 4 m 6 m Q=+7.0mC Q=-7.0 mC 14 Example Adding Vectors F+7+F-7 y components cancel x components: F = F+7,x + F-7,x F+7,x = (3/5)F+7 F-7,x = (3/5)F-7 F = (3/5)(5+5)x10-3 N=6 x 10-3 N F

9. Qp=1.6x10-19 C + r = 1x10-10 m Electric Field • Charged particles create electric fields. • Direction is the same as for the force that a + charge would feel at that location. • Magnitude given by: E  F/q = kq/r2 Example E • E = (9109)(1.610-19)/(10-10)2 N = 1.41011 N/C (to the right) 21

10. Preflight 2.3 • What is the direction of the electric field at point B? • Left • Right • Zero 72% 16% 9% “it is closer to the negative charge, and the field lines point toward negative charges .” “B only has the charge from the negative which is pushing away from itself .” “electric fields of equal magnitudes but opposite directions are present due to the positive and negative charges .” Since charges have equal magnitude, and point B is closer to the negative charge net electric field is to the left y A B x 23

11. ACT: E Field • What is the direction of the electric field at point C? • Left • Right • Zero Red is negative Blue is positive Away from positive charge (right) Towards negative charge (right) Net E field is to right. y C x 25

12. Comparison:Electric Force vs. Electric Field • Electric Force (F) - the actual force felt by a charge at some location. • Electric Field (E) - found for a location only – tells what the electric force would be if a charge were located there: F = Eq • Both are vectors, with magnitude and direction. Add x & y components. Direction determines sign. 26

13. Preflight 2.2 • What is the direction of the electric field at point A? • Up • Down • Left • Right • Zero Red is negative Blue is positive 6% 8% 3% 58% 25% y A B x 37

14. ACT: E Field II • What is the direction of the electric field at point A, if the two positive charges have equal magnitude? • Up • Down • Right • Left • Zero Red is negative Blue is positive y A B x 39

15. Electric Field of a Point Charge 0.81011 N/C Example 321011 N/C 251011 N/C 2.91011 N/C + E This is becoming a mess!!! 40

16. Electric Field Lines • Closeness of lines shows field strength • - lines never cross • # lines at surface  Q • Arrow gives direction of E • - Start on +, end on - 42

17. X A B Y Preflight 2.5 Charge A is 1) positive 2) negative 3) unknown Field lines start on positive charge, end on negative. 93% 4% 3% 44

18. Preflight 2.6 X A B Y Compare the ratio of charges QA/ QB 1) QA= 0.5QB 2) QA= QB 3) QA= 2 QB # lines proportional to Q 12% 9% 63% 45

19. X A B Y Preflight 2.8 The electric field is stronger when the lines are located closer to one another. The magnitude of the electric field at point X is greater than at point Y 1) True 2) False Density of field lines gives E 14% 86% 46

20. ACT: E Field Lines B A Compare the magnitude of the electric field at point A and B 1) EA>EB 2) EA=EB 3) EA<EB 47

21. E inside of conductor • Conductor  electrons free to move • Electrons feels electric force - will move until they feel no more force (F=0) • F=Eq: if F=0 then E=0 • E=0 inside a conductor (Always!) 48

22. X A B Y Preflight 2.10 "Charge A" is actually a small, charged metal ball (a conductor). The magnitude of the electric field inside the ball is: (1) Negative (2) Zero (3) Positive 9% 74% 18% 50

23. Recap • E Field has magnitude and direction: • EF/q • Calculate just like Coulomb’s law • Careful when adding vectors • Electric Field Lines • Density gives strength (# proportional to charge.) • Arrow gives direction (Start + end on -) • Conductors • Electrons free to move  E=0

24. To Do • Read Sections 17.1-17.3 • Extra problems from book Ch 16: • Concepts 9-15 • Problems 11, 15, 23, 27, 29, 39 • Do your preflight by 6:00 AM Wednesday. Have a great weekend. No classes Monday. See you next Wednesday!