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Physics 712 – Electricity and Magnetism Everyone Pick Up: •Syllabus •Two homework passes Eric Carlson “Eric” “Professor Carlson” Olin 306 Office Hours always 758-4994 (o) 407-6528 (c) [email protected] Materials •Classical Electrodynamics by John David Jackson •Calculator •Pencils or pens, paper •Symbolic manipulation http://users.wfu.edu/ecarlson/eandm/index.html 1/13 Dr. Carlson’s Approximate Schedule 9:00 10:00 11:00 12:00 1:00 2:00 3:00 4:00 5:00 Monday PHY 712 PHY 742 PHY 109 research Tuesday research office hour office hour office hour Wednesday PHY 712 Thursday research office hour Friday PHY 742 PHY 109 office hour office hour PHY 712 PHY 742 PHY 109 research research research Free food Collins Hall colloquium Collins Hall Lab Meeting •I will try to be in my office Tues. Thurs. 10-12 •When in doubt, call/email first Reading Assignments / The Text Classical Electrodynamics by J.D. Jackson, 3rd Edition • I’m not writing my own textbook • Contains useful information • Reading assignments every day ASSIGNMENTS Day Read Homework Today none none Friday 1.1-1.4 0.1 Wednesday 1.5-1.8 1.1, 1.2 Homework http://users.wfu.edu/ecarlson/quantum • • • • • • • • Problems assigned as we go along About one problems due every day Homework is due at 10:00 on day it is due Late homework penalty 20% per day Two homework passes per semester Working with other student is allowed Seek my help when stuck You should understand anything you turn in ASSIGNMENTS Day Read Homework Today none none Friday 1.1-1.4 0.1 Wednesday 1.5-1.8 1.1, 1.2 Attendance and Tests Attendance •I do not grade on attendance •Attendance is expected •Class participation is expected •I take attendance every day Tests •Midterm will be from 10-12 on March 4 •Final will be from 9-12 on April 29 Grades, Pandemic Plans Percentage Breakdown: Homework 50% Midterm 20% Final 30% Grade Assigned 94% A 77% C+ 90% A73% C 87% B+ 70% C83% B <70% F 80% B- Pandemic Plans •If there is a catastrophic closing of the university, we will attempt to continue the class: •Some curving possible Emergency contacts: Web page email Cell: 336-407-6528 0A. Math Coordinate Systems • • • • • We will generally work in three dimensions A general coordinate in 3d will be denoted x A general vector will be shown in bold face v x xxˆ yyˆ zzˆ We will often work in Cartesian coordinates 2 2 2 Sometimes in r x y z x r sin cos spherical coordinates, cos 1 z r y r sin sin related to Cartesian by z r cos tan 1 y x • Coordinate x is x rrˆ • Sometimes in x2 y 2 x cos cylindrical coordinates, y sin tan 1 y x related to Cartesian by zz zz • Coordinate x is x ρˆ zzˆ Vector Identities • Vectors can be combined using dot a b axbx a y by az bz products to make a scalar • Vectors can be combined a b xˆ a y bz az by yˆ az bx ax bz zˆ ax by a y bx using cross-products to make a vector • We often abbreviate a2 a a ax2 ay2 az2 • Some vector identities: ab b a • Symmetry/antisymmetry: aa 0 a b b a • Triple scalar product: a b c b c a c a b • Double cross-product: a b c b a c c a b • These and many others in Jackson front cover Derivatives in 3D • The vector derivative can be used for several types of derivatives: • Gradient turns a scalar function f f f ˆ ˆ ˆ f x y z into a vector function x y z • Divergence turns a vector Ax Ay Az function into a scalar function A • Curl turns a vector x y z function into a Az Ay Ax Az Ay Ax vector function A xˆ yˆ zˆ z x x y z y • There is also the Laplacian, a second 2 f 2 f 2 f 2 derivative that can act on f 2 2 2 x y z scalar or vector functions • Each of these has more complicated forms in non-Cartesian coordinates • See QM lecture notes or inside back cover of Jackson Derivatives Product rule • The product rule for derivatives: d df dg fg g f dx dx dx • Numerous 3D equivalents for products • Gradient of product of scalars fg f g f g • Divergence of scalar times vector fA f A f A • Curl of scalar times vector • Divergence of a cross product fA f A f A A B A B A B • Each of these and many more in Jackson, front cover Integrals in 3D 3 f x d x • 3d integrals of vector scalar functions will be common • You should know how to handle these in any coordinate system • Cartesian: 3 f x d x dx dy dz f x • Spherical: 2 0 2 0 1 0 1 f x d x d d 3 f x r 2 sin dr 0 d d cos r 2 f x dr • Cylindrical: f x d x 3 2 0 0 0 d dz f x d Fundamental Theorem of Calculus in 3D • Fundamental theorem of calculus says: b a df dx f b f a dx In general, in 3d, this theorem lets you do one integral whenever you have an integral in 3d: b • Line integral: f dl f b f a a • Stokes’ Theorem: A nˆ da A dl Divergence Theorem: A d x A nˆ da Another theorem: f d x fnˆ da S • C 3 V • • And another theorem: S 3 V S V A d 3x nˆ A da S • All these and more can be found on inside front cover of Jackson Sample Problem 0.1 3. Consider the vector function x/|x| • Work in x rrˆ rˆ (a) Find the divergence for x 0. 3 spherical 3 2 r r x (b) By integrating over a sphere coordinates ˆ r 1 • For x 0, 2 r 2 2 0 around the origin, show that the r r r divergence does not vanish there. take divergence • Tricky at x = 0 because everything is infinite there! • Integrate over a sphere of radius R using the divergence theorem: R 1 rˆ rˆ 1 3 2 ˆ da d x r da 4 R 4 2 S V r 2 S r 2 2 R R • Since the integral is zero except at the origin, we must have x3 4 3 x x • You can generalize this where x is replaced by the x x 3 difference from an arbitrarily chosen origin x': 4 x x 3 x x 0B. Units • Units is one of the most messed-up topics in electricity and magnetism • We will use SI units throughout • Fundamental units: Distance Time Mass Charge meter second kilogram coulomb m s kg C • From these are derived lots of non fundamental units: • Equations of E and M sometimes depends on choice of units! Frequency hertz Force newton Energy joule Power watt Current amp Potential volt Resistance ohm Magnetic induction tesla Magnetic flux weber Inductance Henry Hz N J W A V s–1 kgm/s2 Nm J/s C/s J/C V/A T Wb H kg/C/s Tm2 Vs/A 1. Electrostatics 1A. Coulombs Law, El. Field, and Gauss’s Law Coulomb’s Law • Charges are measured in units called Coulombs k1qqrˆ • The force on a charge q at x from another charge q' at x': F 2 x x • The unit vector points from x' to x rˆ x x • We rewrite the unit vector as x x 1 • For reasons that will make some sense later, we rewrite constant k1 as k1 4 0 • So we have Coulomb’s Law: qq x x F • For complicated reasons having 4 0 x x 3 to do with unit definitions, the constant 0 is known exactly: 0 8.854 1012 N C2 /m 2 • This constant is called the permittivity of free space Multiple Charges, and the Electric Field • If there are several charges q'i, you can add the forces: • If you have a continuous distribution of charges (x), you can integrate: qqi x xi F 3 4 i 0 x xi F q 4 0 qq x x F 4 0 x x 3 3 x d x x x 3 x x • In the modern view, such “action at a distance” seems unnatural • Instead, we claim that there is an 1 x x 3 E x x d x 3 electric field caused by the other charges 4 0 xx – Electric field has units N/C or V/m • It is the electric field that then causes the forces F qE Gauss’s Law: Differential Version x x E x d x 3 4 0 x x 1 3 • Let’s find the divergence of the electric field: x x E x x d x 3 4 0 xx • From four slides ago: x x 3 4 x x 3 x x 1 3 1 3 3 • We therefore have: E x 1 x d x 4 x x x 4 0 0 • Gauss’s Law (differential version): • Notice that this equation is local 0 E x x Gauss’s Law: Integral Version 0 E x x • Integrate this formula over E x d 3x x d 3x 0 V V an arbitrary volume • Use the divergence theorem: 0 nˆ E x da q V S – q(V) is the charge inside the volume V • Integral of electric field over area is called electric flux Why is it true? • Consider a charge in a region • Electric field from a charge inside a region produces electric field lines q • All the field lines “escape” the region somewhere • Hence the total electric flux escaping must be proportional to amount of charge in the region Sample Problem 1.1 (1) A charge q is at the center of a cylinder of radius r and height 2h. Find the electric flux out of all sides of the cylinder, and check that it satisfies Gauss’s Law • Let’s work in cylindrical coordinates • Electric field is: q ρˆ zzˆ q x q ρˆ zzˆ 0E 3 3 4 2 z 2 3/2 4 x 4 ρˆ zzˆ E q • Do integral over top surface: 0 E nˆ da top 4 q 4 0 2 r d 0 0 h d h 2 2 3/2 • By symmetry, the integral over the bottom surface is the same 2 qh 4 2 h 2 r 0 ρ̂ z da 2 z h h q h z r q h 1 2 2 2 r h 0 E nˆ da 0 E nˆ da top 2 3/2 bot E ẑ q qh 2 2 r 2 h2 r Sample Problem 1.1 (2) A charge q is at the center of a cylinder of radius r and height 2h. Find the electric flux out of all sides of the cylinder, and check that it satisfies Gauss’s Law q ρˆ zzˆ 0E 4 2 z 2 3/2 q qh ˆ 0 E nda top 2 2 r 2 h2 • Do integral over lateral surface: q da q 0 E nˆ da lat 4 2 z 2 3/2 4 2 q z 4 r 2 z 2 2 h rd 0 h r h h 0 E nˆ da lat • Add in the top and bottom surfaces: 0 E nˆ da 0 E nˆ da 2 0 E nˆ da tot lat top r 2 z E ρ̂ 2 3/2 E ẑ h h q h z r dz qh r 2 h2 qh r h 2 2 r q qh r h 2 2 q Using Gauss’s Law in Problems 0 nˆ E x da q V S Gauss’s Law can be used to solve three types of problems • Total electric flux out of an enclosed region – Simply calculate the total charge inside • Electric flux out of one side of a symmetrical region – Must first argue that the flux out of each side is the same • Electric field in a highly symmetrical problem – Must deduce direction and symmetry of electric field from other arguments – Must define a Gaussian Surface to perform the calculation – Generally use boxes, cylinders or spheres Sample Problem 1.2 A line with uniform charge per unit length passes through the long diagonal of a cube of side a. What is the electric flux out of one face of the cube? • The long diagonal of d a2 a2 a 2 a 3 the cube has a length • The charge inside the cube is therefore Q d a 3 • The total electric flux nˆ E da Q a 3 0 out of the cube is S • If we rotate the cube 120 around the axis, the three faces at one end will interchange – So they must all have the same flux around them • If we rotate the line of charge, the three faces at one end will interchange with the three faces in back – So front and back must be the same a 3 1 • Therefore, all six nˆ E da nˆ E da one face faces have the same flux 6 0 6 S Sample Problem 1.3 A sphere of radius R with total charge Q has its charge spread uniformly over its volume. What is the electric field everywhere? • By symmetry, electric field points directly away from the center • By symmetry, electric field depends only on distance from origin Outside the sphere: E x E r rˆ • Draw a larger sphere of radius r • Charge inside this sphere is q(r) = Q 2 E r da ˆ 4 r 0E r Q E x n da 0 S • By Gauss’s Law, 0 S Inside the sphere: Q E r • Draw a smaller sphere of radius r q r Q V r Qr 3 R3 4 0 r 2 V R • Charge inside this sphere is only • By Gauss’s Law, Qr 3 R3 0 E x nˆ da 0 E r da 4 r 2 0 E r S S • Final answer: Qrˆ E 4 0 r 2 if r R, 3 if r R . rR Qr E r 4 0 R 3 1B. Electric Potential Curl of the Electric Field x 0 • From homework problem 0.1: 3 x • Generalize to origin at x': x x 0 3 x x • Consider the curl of the electric field: 1 x x 1 x x 3 3 E x d x x d x 3 3 4 0 4 0 x x xx • Using Stokes’ theorem, we can get an integral version of this equation: E dl 0 3 E d x0 S E 0 Electric Field: Discontinuity at a Boundary Consider a surface (locally flat) with a surface charge • How does electric field change across the boundary? A • Consider a small thin box of area A crossing the boundary • Since it is small, assume E is constant over top surface and bottom surface • Charge inside the box is A • Use Gauss’s Law A q E x nˆ da Et nˆ t A Eb nˆ b A A Et Eb nˆ t on this small box 0 0 ˆ E n • Consider a small loop of length L penetrating the surface 0 • Use the identity 0 E dl Et xˆ L Eb xˆ L • Ends are short, so E xˆ 0 only include the lateral part E nˆ 0 • So the change in E across the boundary is The Electric Potential • In general, any function that has curl zero can be written as a gradient E 0 – Proven using Stokes’ Theorem E • We therefore write: – is the potential (or electrostatic potential) – Unit is volts (V) It isn’t hard to find an expression for : 1 x x 1 1 rˆ x • First note that 2 3 3 x x • Generalize by shifting: x r x x r x • If we write: x 1 4 0 x d 3x x x • Then it follows that: 1 1 1 x x 3 3 x x d x x d x 3 E x 4 0 x x 4 0 x x Working with the Potential E Why is potential useful? • It is a scalar quantity – easier to work with • It is useful when thinking about energy – To be dealt with later x 1 4 0 How can we compute it? • Direct integration of charge density when possible • We can integrate the electric field • It satisfies the following differential equation: 0 E 0 2 0 Solving this equation is one of the main goals of the next couple chapters x d 3x x x