1. Physics 102: Lecture 2, Slide 1 Coulomb’s Law and Electric Fields Physics 102: Lecture 02 Today we will … get some practice using Coulomb’s Law learn the concept of an Electric Field

2. Physics 102: Lecture 2, Slide 2 Recall Coulomb’s Law Magnitude of the force between charges q 1 and q 2 separated a distance r: F = k q 1 q 2 /r 2 k = 9x10 9 Nm 2 /C 2 Force is –attractive if q 1 and q 2 have opposite sign –repulsive if q 1 and q 2 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

3. Physics 102: Lecture 2, Slide 3 Coulomb Law practice: Three Charges Q=-7.0  C Q=+2.0  C 6 m 4 m Calculate force on +2  C charge due to other two charges –Draw forces –Calculate force from +7  C charge –Calculate force from –7  C charge –Add (VECTORS!) 1 2 3 4 5 6 7 8 9 10 11 12 F +7 Q=+7.0  C F -7

4. Physics 102: Lecture 2, Slide 4 Three Charges – Calculate forces Q=-7.0  C Q=+2.0  C 6 m 4 m Calculate force on +2  C charge due to other two charges –Draw forces –Calculate force from +7  C charge –Calculate force from –7  C charge –Add (VECTORS!) Calculate magnitudes F +7 Q=+7.0  C F -7 5 m

5. Physics 102: Lecture 2, Slide 5 Q=-7.0  C Q=+2.0  C 6 m 4 m F +7 Q=+7.0  C F -7 5 m Three charges – Adding Vectors F +7 +F -7 Calculate components of vectors F +7 and F -7 :

6. Physics 102: Lecture 2, Slide 6 Q=-7.0  C Q=+2.0  C 6 m 4 m F +7 Q=+7.0  C F -7 5 m Three charges – Adding Vectors F +7 +F -7 Add like components of vectors F +7 and F -7 : F Final vector F has magnitude and direction Double-check with drawing

7. Physics 102: Lecture 2, Slide 7 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/r 2 + r = 1x10 -10 m Q p =1.6x10 -19 C E E = (9  10 9 )(1.6  10 -19 )/(10 -10 ) 2 N = 1.4  10 11 N/C (to the right)

8. Physics 102: Lecture 2, Slide 8 Preflight 2.3 What is the direction of the electric field at point B? 1)Left 2)Right 3) Zero x y A B 71% 15% 13% “B only has the charge from the negative which is pushing away from itself.” “it is closer to the negative charge, and the field lines point toward negative charges.” “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

9. Physics 102: Lecture 2, Slide 9 Preflight 2.2 What is the direction of the electric field at point A? 1)Up 2)Down 3)Left 4)Right 5)Zero x y A B 7% 2% 53% 32%

10. Physics 102: Lecture 2, Slide 10 ACT: E Field What is the direction of the electric field at point C? A.Left B.Right C. Zero x y C Away from positive charge (right) Towards negative charge (right) Net E field is to right. A B

11. Physics 102: Lecture 2, Slide 11 E Field from 2 Charges Q = –3.5  C A 6 m 4 m Calculate electric field at point A due to two unequal charges –Draw electric fields –Calculate E from +7  C charge –Calculate E from –3.5  C charge –Add (VECTORS!) 1 2 3 4 5 6 7 8 9 10 11 12 Note: this is similar to (but a bit harder than) my earlier example. E x = 2.25  10 +3 N/C E y = 1.0  10 +3 N/C Q = +7.0  C You try this at home!

12. Physics 102: Lecture 2, Slide 12 E Field from 2 Charges Q=-3.5  C Q=+7.0  C 6 m 4 m Calculate electric field at point A due to charges –Calculate E from +7  C charge –Calculate E from –3.5  C charge –Add* E7E7 5 m E = k q/r 2 A 32 E3E3

13. Physics 102: Lecture 2, Slide 13 Adding Vectors E 7 +E 3 Q=-3.5  C Q=+7.0  C 6 m 4 m Decompose into x and y components. E7E7 E 7x E 7y =E 7 (3/5) =E 7 (4/5) 5   A 4 m 34

14. Physics 102: Lecture 2, Slide 14 Adding Vectors E 7 +E 3 Q=-3.5  C Q=+7.0  C 6 m 4 m Decompose into x and y components. Add components. E7E7 E3E3 E total E x = 2.25  10 +3 N/C E y = 1.0  10 +3 N/C A 35

15. Physics 102: Lecture 2, Slide 15 Comparison: Electric Force vs. Electric Field Electric Force (F) – the actual force felt by a real charge at some location Electric Field (E) – found for a location only (any location) – tells what the electric force would be if a charge were located there: F = Eq Both are vectors, with magnitude and direction. Ok, what is E actually good for?

16. Physics 102: Lecture 2, Slide 16 x y C A B Electric field defined at any location (we did three: A, B, C) Electric Field Map

17. Physics 102: Lecture 2, Slide 17 Electric Field Lines Closeness of lines shows field strength (lines never cross) Number of lines at surface  Q Arrow gives direction of E (Start on +, end on –) This is becoming a mess!!!

18. Physics 102: Lecture 2, Slide 18 AB X Y Preflight 2.5 Charge A is 1) positive2) negative3) unknown Field lines start on positive charge, end on negative. 93% 4% 3%

19. Physics 102: Lecture 2, Slide 19 AB X Y Preflight 2.6 Compare the ratio of charges Q A / Q B 1) Q A = 0.5Q B 2) Q A = Q B 3) Q A = 2 Q B # lines proportional to Q 17%9%61% AB X Y

20. Physics 102: Lecture 2, Slide 20 AB X Y Preflight 2.8 The magnitude of the electric field at point X is greater than at point Y 1) True2) False Density of field lines gives E 14% 86% The electric field is stronger when the lines are located closer to one another.

21. Physics 102: Lecture 2, Slide 21 ACT: E Field Lines Compare the magnitude of the electric field at point A and B 1) E A >E B 2) E A =E B 3) E A <E B A B

22. Physics 102: Lecture 2, Slide 22 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!)

23. Physics 102: Lecture 2, Slide 23 AB X 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% 68% 23%

24. Physics 102: Lecture 2, Slide 24 Demo: E-field from dipole x y C A B

25. Physics 102: Lecture 2, Slide 25 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

26. Physics 102: Lecture 2, Slide 26 To Do Do your preflight by 6:00 AM Wednesday. Homework 1 due Tuesday, Jan 25 8AM!