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Physics 1161 Lecture 2 Vectors & Electric Fields Three Charges • Calculate force on +2mC charge due to other two charges – Calculate force from +7mC charge F1,3 3m 6 2 4m 2 2 5.04 10 3 N 6 F2 ,3 3m 2 2.52 10 6 4m 2 2 2.52 10 3 N Q=+2.0mC – Calculate force from –3.5mC charge k 3.5 10 C 2 10 C 3 N – Add (VECTORS!) Q=+7.0mC 3 N 4m k 7 10 C 2 10 C 6 5.04 10 6m Q=-3.5 mC Three Charges • Resolve each force into x and y components 3 N cos 53 3.03 10 3 F1,3 y 5.04 10 3 N sin 53 4.03 10 3 F2 ,3 x 2 .5 2 1 0 3 N co s 3 0 7 1 .5 1 1 0 F1,3 x 5.04 10 F2 ,3 y 2 .5 2 1 0 3 o o o N N 3 N N sin 3 0 7 1 .9 9 7 1 0 o 3 5.04 10 • Add the x-components & the y-comp. F x 3.03 10 N 1.51 10 F y 4.03 10 3 N 1.997 10 N 4.54 10 3 3 N 2.03 10 N 3 2.52 10 3 53o 53o N 4m 3 N N Q=+2.0mC 3 3 N • Use Pyth. Theorem & Trigonometry to express in R,θ notation Q=+7.0mC 6m Q=-3.5 mC Three Charges • Use Pyth. Theorem & Trigonometry to express in R,θ notation F 2.03 10 3 N 4.54 10 F 4.9 10 F 3 3 N 2 2.03 10 3 N φ 4.54 10 3 N Fy 2.03 10 3 N o arctan 24 arctan 3 4.54 10 N Fx Since resultant is in first quadrant, θ = φ 24 o N 2 Electric Force on Electron by Proton • What are the magnitude and direction of the force on the electron by the proton? q=1.6x10-19 C e- + F kq1 q 2 r F r = 1x10-10 m 2 9 10 9 N m 2 1.6 10 2 C 10 8 F 2.30 10 N 10 19 m C 1.6 10 19 C 2 Toward the left 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 = qE • Both are vectors, with magnitude and direction Electric Field • Charged particles create electric fields. – Direction is the same as for the force that a + charge would feel at that location. E F/q – Magnitude given by: • Field at A due to proton? E kq r E q=1.6x10-19 C + 2 9 10 1.6 10 10 m 9 N m 2 E 1.44 10 C 2 10 11 N C 19 C r = 1x10-10 m 2 Toward the right A What is the direction of the electric field at point A, if the two positive charges have equal magnitude? 1. 2. 3. 4. 5. Up Down Right Left Zero A y 0% 1 + + B x 0% 0% 2 3 0% 0% 4 5 What is the direction of the electric field at point A, if the two positive charges have equal magnitude? 1. 2. 3. 4. 5. Up Down Right Left Zero A y 0% 1 + + B x 0% 0% 2 3 0% 0% 4 5 Two Charges Checkpoint1 What is the direction of the electric field at point A? 1) Up 2) Down A y 3) Left 4) Right 5) Zero + B x Two Charges Checkpoint 2 What is the direction of the electric field at point B? 1) up 2) down 3) Left 4) Right A 5) Zero + y B x What is the direction of the electric field at point C? 1. Left 2. Right 3. zero y + C - x 0% 1 0% 2 0% 3 Electric Field Applet • http://www.cco.caltech.edu/~phys1/java/phys 1/EField/EField.html Checkpoint X A Charge A is Y B Field lines start on positive charge, end on negative. 1) positive 2) negative 3) unknown Checkpoint X A Y B Compare the ratio of charges QA/ QB # lines proportional to |Q| 1) QA= 0.5QB 2) QA= QB 3) QA= 2 QB 4) can’t say Checkpoint 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 2) False Density of field lines gives E Compare the magnitude of the electric field at point A and B 1. EA> EB 2. EA= EB 3. EA< EB B A E inside of conductor • Conductor electrons free to move – Electrons feels electric force - will move until they feel no more force (F=0) – F=qE: if F=0 then E=0 • E=0 inside a conductor (in electrostatics) Checkpoint 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 (2) Zero (3) Positive E inside of conductor in electrostatics • Conductor electrons free to move – Electrons feel electric force - will move until they feel no more force (F=0) – F=qE: if F=0 then E=0 • E=0 inside a conductor (in a static situation) 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 To Do • • • • • Read sections 19-6 & 19-7 Watch Prelecture 3 by 6am 1/15 Complete Checkpoint 3 by 6 am 1/15 Complete Homework 1 by 11pm 1/15 Complete Homework 2 by 11pm 1/17