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Lecture Notes

by: Michael M

Lecture Notes 122

Michael M
Physics - Electromagnetism
Nikolai Tolich

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About this Document

Physics 122 Lecture Notes for 1st and 2nd weeks.
Physics - Electromagnetism
Nikolai Tolich
Class Notes
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This 15 page Class Notes was uploaded by Michael M on Friday October 9, 2015. The Class Notes belongs to 122 at University of Washington taught by Nikolai Tolich in Fall 2015. Since its upload, it has received 6 views. For similar materials see Physics - Electromagnetism in Physics 2 at University of Washington.


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Date Created: 10/09/15
Physics 122 Documentation Release 01 Nikolai Tolich October 09 2015 CONTENTS CHAPTER ONE LECTURES 11 Lecture 1 Powerpoint Slides 111 Charge 0 There are two types of charge which we denote and 0 Charge is quantized in values of e e 16 X 10190 this means all charges are integer multiples of e 0 Each electron in an atom has a charge of e and each proton e Neutrons are uncharged Note NOT ON EXAM quarks can have fractional charges but never occur as free particles They re only seen grouped together in particles that have appropriately quantized charges Total charge is always conserved but it can be moved around Example 11 ttl h 12 1090 0 a C arge 5 X 78 X 1011 electrons l t 6 CC Ions electron charge 15 X 10 110 Charge can move around freely inside of a conductor but cannot move freely in an insulator 12 Lecture 2 Warning Missed rst 15 minutes of class 121 Coulomb s Law 4 1912 F12 T712 712 Physics 122 Documentation Release 01 122 Total Force The net force on any given thing is the vector sum of all forces acting on it from anywhere superposition Example 2 Calculate F12 and F13 using Coulomb s Law 0 Calculate total force 151 F12 F13 0 Consider horizontal force by summing X components of those vectors 533N 0 Consider vertical force by rumming y components 0 Get answer I71 2 583Ni 650Nj Clicker Question 22 Correct answer is A at least one sphere must be charged This happens because a charged object near a conductor causes a charge separation in the conductor then because the charges that attract the other charged object are closer in the conductor they have a greater effect than the other farther away repelling charges which has a net effect of attraction 13 Lecture 3 Powerpoint Slides Warning Missed rst 15 minutes of class 131 Electric Field Lines 0 We can use them to visualize electric elds 0 The direction of the eld line at any point indicates the direction of the total electric eld 0 eld lines extend radially outward from a positive charge and radially inward from a negative charge 0 the number of lines leaving a charge is proportional to its chatge and the space between the lines is proportional to the electric eld s strength Dipoles 0 A dipole is something where the total charge is 0 this does not mean there is no eld they don t cancel entirely 0 A positive charge and negatice charge are separated by a distance L 0 the electric eld on the dipole aXis at a point a long way away from the dipole is in the same direction as the dipole and magnitude is E 0 We write vectors as pointing from the positive charge to the negative charge 2 Chapter 1 Lectures Physics 122 Documentation Release 01 Clicker Question 32 Correct answer is D because the lines pointing radially outward are much denser than those pointing straight up Because of this the vector actually sums to point more right than up though there are still components of both right and up 132 Charge in an Electric Field THe force on a charge in an E eld is Example 2 Initally 3903 U Uy 0 The electric eld is upward positive y Pg 2 qE From Newton s second Law we know that Pg 2 may qE E The time taken to cross the eld is given by v that is distance travelled divided by time taken equals the velocity The y velocity after time t is given by Ew vfyzviyayt0 so our nal answer can be written IE w UFZ U LI m U 133 Dipoles in an Electric Field 0 The torque on a dipole is given by 7quot 13 X E 0 The magnitude of the torque is given by W W sin 6 Microwaves This is how a microwave oven works in fact This dipole rotation creates a rapidly rotating eld Dipoles in NonUniform E fields Because of the torque mentioned above dipoles align themselves with electric elds Then the dipole can actually be attracted to a nonuniform electric eld This in turn can result in weirdness like wood a dipole being attracted to a charged rod because the water molecules in the wood can rotate and attract to a charged rod 13 Lecture 3 3 Physics 122 Documentation Release 01 14 Lecture 4 Powerpoint Slides Warning Missed rst 10 minutes of class 141 Continuous Charge Distribution Clicker Question 42 Correct answer is A Continuous Charge Distribution in 1D A is the charge density Continuous Charge Distribution in 2D 0 is the area charge density Continuous Charge Distribution in 3D V is the area charge density math rho fracqdV text 142 Symmetries An object has 0 Rotational Invariance if it does not change under and arbitrary rotation 0 Translational Invariance if it does not change under arbitrary translation 0 Re ection Invariance if it does not change under a re ection Example A circle has rotational invariance about the center The same circle has rotational invariance about the center though a triangle does not Note This means that the electric eld for an object that has one of these invariances must have the same invariance This is purely because of mathematics Clicker Question 43 Correct Answer is A eld must go straight out from the center of the sphere Apparently the dot was supposed to be in the center of the circle 4 Chapter 1 Lectures Physics 122 Documentation Release 01 143 E Fields E Field for a Continuous Charge The eletric eld at position 77 from a small charge element dq is given by d z qr 7quot So we can gure out the electric eld for a continuous charge distribution by integrating a kd E 2q7quot 7quot E Field for a Ring The magnitude of the E eld a distance 2 along the axes of a circular charge ring with radiaus a and linear charge density A is given by 27rkAaz 22 a2 E Field From a Disk Acts like a series of rings Todo Fill this in from slides E Field Infinite Plane For charge density a is given by E 27rk0 in the direction normal to the plane Clicker Question 44 Correct answer is D down The left and right forces cancel leaving only the down Example 42 Part a ATq 14 Lecture 4 5 Physics 122 Documentation Release 01 Part b E eld will point in the X direction and the E eld can be calculated with the integral L kAd Ex 2 0 0 a L x2 kqL dm L 0 aL e kg 1 1 L a aL kq aaL 15 Lecture 5 Electric Flux and Gauss Law Powerpoint Slides 151 NonUniform Charge Distributions In a nonuniform charge distributions the following equations are still true 0 1D dq Adl 0 2D dq adA 0 3D dq pdV Example 51 Calculate the total charge on a thin rod starting at 1 0 and nishing at 1 L that has a linear charge density of Mm ax3 Solution 152 Electric Flux The number of electric eld lines passing through a surface 6 Chapter 1 Lectures Physics 122 Documentation Release 01 Clicker Question 51 Correct answer is C Field lines are just a representation of charge because they re just proportional to charge A2 has greater ux because it s larger Basically ux depends on area with some caveats Clicker Question 52 Correct Answer is A The number of eld lines through each is the same because everything that enters one side in the direction of the eld also leaves The formula given A2 cos 6 gives that A1 is smaller Basically area and angle are important Clicker Question 53 Correct Answer is B More lines pass through A1 than A2 because the angle of the eld even though the areas are the same Basically the strength of the electric eld determines the ux 153 Electric Flux Equation The electric ux through a surface depends on The 0 surface area 0 the angle of the electric eld with respect tot he normal to the surfaceface 0 the strength of the electric eld The Formula ltpSEadA Where 7 is the unit vector normal to the surface and the integral is over the surface Example 52 A non uniform electric eld given by E 30x 40 NC1 pierces the cube shown below What is the electric ux through the a right face and b top face Solution E 303 403 15 Lecture 5 Electric Flux and Gauss Law 7 Physics 122 Documentation Release 01 For the right face 73 E E a 301 403 17 3031 5 4003 17 30x lt1gt mdA E s 230di s 3 0x30dA S 290dA S 90xA I 36Nm2C 1 2 For the top face 73 E a 30m 539 4013 40 154 Flux Through a Closed Surface A closed surface is asurface from which there is no way out The total uX through a closed surface is given by 15 f dA For nonclosed surface there are two choice for the direction of 73 For closed surfaces convention for the direction of 73 is that it pioints away from the closed side 155 Gauss Law The Law net 73 dA Qenclosed 6O 1 where 60 4 885 X 10 1212N 1m 2 7TH Clicker Question 54 D is the correct answer because Gauss Law That s the only charge enclosed by the line Chapter 1 Lectures Physics 122 Documentation Release 01 Clicker Question 55 A is the correct answer because Gauss Law That s the only charge enclosed by the line Even though the eld changes the ux does stay the same because of Gauss Law 15 Lecture 5 Electric Flux and Gauss Law 9 Physics 122 Documentation Release 01 10 Chapter 1 Lectures CHAPTER TWO COURSE INFO Nikolai Tolich lt ntolichuwedu gt Schedule https canvasuWeducourses 99239 1 les 3 27 5026 1downloadwrap 1 21 Grading 0 Labs are not curved 0 Do get credit for clicker responses in the other section MWF 230320 in PAA A102 0 Midterms are best 2 of the 3 0 Clickers get 80 for participation 20 for correctness and at the end 100 is scaled down to 80 So as long as you re always there you get 100 regardless 22 General Notes 0 Lectures are on canvas at httpscanvasuweducourses992391 lesfolderLectures 0 WebAssign eXpects 1 accuracy and is due at 1159pm on Wednesday 0 Clicker channel for the classroom is 1 the intro slide is incorrect 0 Graphing calculator is OK on the exams just don t store text 23 Office Hours 0 In the Physics Study Center 0 Monday 1200pm 100pm 0 Friday 930am 1030am 24 lndices and tables 0 genindeX 0 modindeX 0 search 11


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