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 Note the change in the Daily Planner

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

3 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 C –r has units of m –F has units of N 5

4 Three Charges Q=-7.0  C Q=+7.0  C Q=+2.0  C 6 m 4 m Calculate force on +2  C charge due to other two charges –Calculate force from +7  C charge –Calculate force from –7  C charge –Add (VECTORS!) F7F7

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

6 Q=-7.0  C Q=+7.0  C Q=+2.0  C 6 m 4 m Calculate force on +2  C charge due to other two charges –Calculate force from +7  C charge –Calculate force from –7  C charge –Add (VECTORS!) F +7 5 m F = k q 1 q 2 /r 2 14 Three Charges F -7

7 Q=-7.0  C Q=+7.0  C Q=+2.0  C 6 m 4 m F +7 5 m 14 F -7 Adding Vectors F +7 +F -7

8 Q=-7.0  C Q=+7.0  C Q=+2.0  C 6 m 4 m F +7 5 m 14 F -7 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 N F

9 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 = 1x m Q p =1.6x C E E = (9  10 9 )(1.6  )/( ) 2 N = 1.4  N/C (to the right) 21

10 Preflight 2.3 What is the direction of the electric field at point B? 1)Left 2)Right 3) Zero x y A B 72% 16% 9% “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.” 23 Since charges have equal magnitude, and point B is closer to the negative charge net electric field is to the left

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

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 E Field from 2 Charges Q=-3.5  C Q=+7.0  C A 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 (VECTORS!) Note: this is similar to (but a bit harder than) my earlier example.

14 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

15 Adding Vectors E 7 +E 3 Q=-3.5  C Q=+7.0  C Q=+2.0  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

16 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  N/C E y = 1.0  N/C A 35

17 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 6% 8% 3% 58% 25% 37 Red is negative Blue is positive

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

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

20 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

21 AB X Y Preflight 2.5 Charge A is 1) positive2) negative3) unknown Field lines start on positive charge, end on negative % 4% 3%

22 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 12%9%63% 45

23 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 46 14% 86% The electric field is stronger when the lines are located closer to one another.

24 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 47

25 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

26 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 50 9% 74% 18%

27 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

28 To Do Read Sections 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!