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
History of quantum field theory wikipedia , lookup
Magnetic monopole wikipedia , lookup
Work (physics) wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Speed of gravity wikipedia , lookup
Fundamental interaction wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Maxwell's equations wikipedia , lookup
Electromagnetism wikipedia , lookup
Anti-gravity wikipedia , lookup
Field (physics) wikipedia , lookup
Lorentz force wikipedia , lookup
Physics 102: Lecture 02 Coulomb’s Law and 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 1 But first, a leftover from last lecture ACT: 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 2 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 3 5 Three Charges • Calculate force on +2mC charge due to other two charges F7 Q=+2.0mC 4m – Calculate force from +7mC charge – Calculate force from –7mC charge – Add (VECTORS!) Q=+7.0mC 6m Q=-7.0 mC 4 10 Three Charges • Calculate force on +2mC charge due to other two charges F+7 Q=+2.0mC 4m – Calculate force from +7mC charge – Calculate force from –7mC charge – Add (VECTORS!) Q=+7.0mC 6m Q=-7.0 mC 5 10 Three Charges • Calculate force on +2mC charge due to other two charges • F = k q1q2 /r2 (9 109 )(2 10 6 )(7 10 6 ) F7 N 25 3 F7 5 10 N (9 109 )(2 106 )(7 106 ) F7 N 25 3 F7 5 10 N F+7 Q=+2.0mC 4m – Calculate force from +7mC charge – Calculate force from –7mC charge – Add (VECTORS!) Q=+7.0mC 6m F-7 Q=-7.0 mC 6 14 Adding Vectors F+7+F-7 F+7 4m Q=+2.0mC Q=+7.0mC 6m F-7 Q=-7.0 mC 14 7 Adding Vectors F+7+F-7 y components cancel x components: F+7 Q=+2.0mC F = F+7,x + F-7,x 4m F+7,x = (3/5)F+7 F-7,x = (3/5)F-7 F F-7 F = (3/5)(5+5)x10-3 N=6 x 10-3 N Q=+7.0mC 6m Q=-7.0 mC 14 8 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 Qp=1.6x10-19 C + r = 1x10-10 m E E = (9109)(1.610-19)/(10-10)2 N = 1.41011 N/C (to the right) 9 21 Preflight 2.3 What is the direction of the electric field at point B? “it is closer to the negative charge, and the field lines point toward negative charges .” Right “B only has the charge from the negative which is pushing away from itself .” 72% 1) Left 16% 2) 9% 3) Zero “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 10 23 ACT: E Field What is the direction of the electric field at point C? A. Left Red is negative B. Right Blue is positive C. Zero Away from positive charge (right) Towards negative charge (right) Net E field is to right. y C x 11 25 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. 12 26 Preflight 2.2 What is the direction of the electric field at point A? 6% 1) Up Red is negative 8% 2) Down Blue is positive 3% 3) Left 58% 4) Right 25% 5) Zero A y B x 17 37 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 Red is negative B. Down Blue is positive C. Right D. Left A y E. Zero B x 18 39 321011 N/C Electric Field of a Point Charge + This is becoming a mess!!! 2.91011 N/C E 19 40 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 - 20 42 Preflight 2.5 X A Charge A is B Y Field lines start on positive charge, end on negative. 1) positive 93% 2) negative 4% 3) unknown 3% 21 44 Preflight 2.6 X A Y B Compare the ratio of charges QA/ QB # lines proportional to Q 1) QA= 0.5QB 12% 2) QA= QB 9% 3) QA= 2 QB 63% 22 45 Preflight 2.8 X A Y B 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 14% 2) False 86% Density of field lines gives E 23 46 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 24 47 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!) 25 48 Preflight 2.10 X A Y B "Charge A" is actually a small, charged metal ball (a conductor). The magnitude of the electric field inside the ball is: (1) Negative 9% (2) Zero 74% (3) Positive 18% 26 50 Recap • E Field has magnitude and direction: – EF/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 27 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! 28