Topic 7. Electric flux Gauss s Law Divergence of E Application of Gauss Law Curl of E
|
|
- Ethelbert Carter
- 5 years ago
- Views:
Transcription
1 Topic 7 Electric flux Gauss s Law Divergence of E Application of Gauss Law Curl of E
2 urface enclosing an electric dipole. urface enclosing charges 2q and q.
3 Electric flux Flux density : The number of field lines per unit area, perpendicular to the field direction (a vector quantity). This quantity is proportional to electric field strength E. Flux (symbol ϕ) : is the total number of field lines passing through a particular area (a scalar quantity). φ E = EA = E A = EAcos θ ( )% N m 2 / C & ' (
4 Area elements We assumed a uniform area and a flat surface area in the previous slide. If we take an infinitesimally small area element, da, the field will be uniform and the area element flat. E da The flux through the area element is dφ E = E da Total flux through the entire surface is φ E = E da
5 A 2 E Gauss s Law r A 1 da q E da = 1 4π % q r ˆr ( ' * r 2 sinθdθdφ ˆr & 2 ) ( ) = 1 q The same number of field lines pass through surface A 2 as through the spherical surface A 1. ince the flux through a surface is proportional to the number of field lines through it, the flux through both surfaces is the same: E da = E da A 2 = q A 1
6 q 1 q 2 Multiple charges q 3 uppose now that we have a bunch of charges scattered around. The total field is the vector sum of all the individual fields: E = n i=1 E i (Principle of superposition) The flux through a surface that encloses them all: E da = For any closed surface, then n ( E i da) = i=1 n i=1 % 1 ( ' q i * & ) E da = 1 Q enc
7 Differential form Apply divergence theorem E da = V ( E) dτ Rewrite Q enc in terms of charge density ρ Q enc = ρdτ V ( E) dτ = V V ' ) ( ρ *, dτ ince this holds for any volume the integrands must be equal E = ρ
8 Example calculation uppose an electric field (in spherical coordinates) in some region is found to be E = kr 3ˆr in spherical coordinates (k is some constant). (a) Find the charged density ρ. ρ = 1 r 2 ρ = E E = 1 ( r 2 r r2 E r ) r r2 kr 3 ( ) = 1 r 2 k 5r 4 ( ) = 5 kr 2 1 rsinθ θ sinθε θ ( ) (b) Find the total charge contained in a sphere of radius R, centered at the origin. 1 Ε φ rsinθ φ Gauss Law ( )( ) = 4π kr 5 Q enc = E da = kr 3 4πR 2 Integration: Q enc = R ρ dτ = 5 kr 2 0 4πr 2 R ( dr) = 20π k r 4 dr = 4π kr 5 0
9 Divergence of E E( r) = 1 4π E = 1 4π where # 1 % $ r - r' 2 allspace ρ ( r' ) r - r' 2 # 1 % $ % r - r' 2 ( r - r' ) r - r' dτ' ( r - r' )& ( r - r' '( ρ ( r' )dτ' ( r - r' )& r - r' ( = 4πδ 3 r - r' ' ( ) E da = $ 1 R ˆr ' & ) % 2 ( E = 1 4πδ 3 ( r - r' ) ρ ( r' )dτ' = 1 ρ r 4π ( Unit vector = r - r' ) r - r' eparation vector = ( r - r' ) Magnitude = r - r' V ( E) dτ ( R 2 sinθdθdφ ˆr ) = 1 r 2 $ 1 ' & r % r2 ) dτ r 2 ( δ( r)dτ = δ( x)δ( y)δ( z) dxdydz = 1 all space Integral form - integrate over volume and apply divergence theorem: 1 Edτ = E da = V V ( ) ρ dτ = 1 Q enc
10 Application of Gauss s Law uppose we have a uniformly charged sphere of radius R possessing a total charge Q. Find the electric field outside the shell. r R A Choose our Gaussian sphere (r > R) so it is concentric with the shell A. E da = 1 Q enc E da = E da E da = E da = E 4πr 2 E 4πr 2 = 1 Q E = 1 Q 4πr 2 r ˆr 2 The field outside is the same as that for a point charge of the same magnitude located at the center!
11 Example. uppose we have a thin spherical shell of radius R possessing a total charge σ that is uniformly distributed over the surface. Find the electric field outside the shell. r R A Choose our Gaussian sphere (r > R) so it is concentric with the shell A. E da = E ( 4πr 2 ) = 1 σ ( 4π R 2 ) dq = σ da = σ R 2 sinθ dθ dφ E = 1 σ ( 4π R 2 ) 4πr 2 ε ˆr = σ R2 0 r 2 ˆr The field outside is the same as if all charge was concentrated at the middle!
12 What is the field inside the shell (r < R)? Q enc = 0 R r E da = E ( 4πr 2 ) = 1 Q enc = 0 What happens to a uniformly charged solid spherical conductor?
13 What is the field inside the sphere? r R A Choose our Gaussian surface of radius r (r < R) so it is concentric with the shell A. Q enc is only a portion of the total charge Q Define charge density as charge per unit volume ρ E = dq dv Charge enclosed in our Gaussian surface is! 4 πr 3 ρ Qenc = # 3 E # 4 " π R 3 ρ 3 E Therefore from Gauss s law: $ & & Q = r3 % R Q 3 E da = E( 4πr 2 ) = Q E ( 4πr 2 ) = Q encl = r3 R Q E = Q 3 4π R r 3 Lets plot this!
14 E Q E = 4π R 2 r=r r Magnitude of the electric field as a function of the distance r from the center of a uniformly charged solid sphere.
15 Example: Cylindrical symmetry A long cylinder carries a charge density that is proportional to the distance from the axis: ρ = ks, for some constant k. Find the electric field inside the cylinder. E Gaussian surface a l E Gauss s Law states : E da = 1 Q enc
16 The enclosed charge is Q enc = ρ dτ Q enc = ( ks) ( sdsdφdz) z s P ẑ ŝ ˆφ Q enc = 2πkl Volume element for cylindrical coordinates 0 a ( s 2 ds) = 2 3 πkla3 x ϕ y ymmetry dictates that E must point radially outward, so for the curved part of the Gaussian surface we have: E da = Eda = E da = E 2πal ince E is perpendicular to da, the two ends contribute nothing. E 2πal = πkla3 E = 1 3 ka 2 ŝ
17 Example: Plane symmetry uppose we want to find the electric field of an infinite plane carrying a uniform surface charge σ. E Draw Gaussian pillbox, extending equal distances above and below the plane. Apply Gauss s Law E da = 1 Q enc E The enclosed charge is Q enc = σ A By symmetry E points away from the plane, with the sides contributing nothing, so that the top and bottom surfaces yield E da = 2A E 2A E = 1 σ A E = σ 2 ˆn
18 ummary Electric field of a sphere falls off as 1/r 2 Electric field of an infinite line as 1/r An infinite plane does not fall off at all
19 Curl of E What is the curl of the field in this figure with a point charge at the center? E = 1 4π q r 2 ˆr uppose we have to calculate the line integral of the field from point a to point b. a b E dl z b In spherical coordinates r b dl = drˆrrdθˆθ rsin dφˆφ q E dl = 1 4π q r 2 dr x a r a y
20 b E dl = 1 a 4π b q r dr = 1 r q b = 1 a 2 4π r ra 4π & q ( - q ' r a r b ) * The integral around a closed path is zero (as then r a = r b ) Applying tokes theorem ( v ) E dl = 0 da = v di P E = 0 From Principle of uperposition E = E 1 E 2 E 3. E = ( E 1 E 2 E 3... ) = 0
Chapter 2. Electrostatics. Introduction to Electrodynamics, 3 rd or 4 rd Edition, David J. Griffiths
Chapter 2. Electrostatics Introduction to Electrodynamics, 3 rd or 4 rd Edition, David J. Griffiths 2.1 The Electric Field Test charge 2.1.1 Introduction Source charges The fundamental problem that electromagnetic
More informationGauss s Law. Chapter 22. Electric Flux Gauss s Law: Definition. Applications of Gauss s Law
Electric Flux Gauss s Law: Definition Chapter 22 Gauss s Law Applications of Gauss s Law Uniform Charged Sphere Infinite Line of Charge Infinite Sheet of Charge Two infinite sheets of charge Phys 2435:
More informationIMPORTANT: LABS START NEXT WEEK
Chapter 21: Gauss law Thursday September 8 th IMPORTANT: LABS START NEXT WEEK Gauss law The flux of a vector field Electric flux and field lines Gauss law for a point charge The shell theorem Examples
More informationChapter 21: Gauss law Tuesday September 13 th. Gauss law and conductors Electrostatic potential energy (more likely on Thu.)
Chapter 21: Gauss law Tuesday September 13 th LABS START THIS WEEK Quick review of Gauss law The flux of a vector field The shell theorem Gauss law for other symmetries A uniformly charged sheet A uniformly
More informationSummary: Applications of Gauss Law
Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 1 Summary: Applications of Gauss Law 1. Field outside of a uniformly charged sphere of radius a: 2. An infinite, uniformly charged plane
More informationChapter 23. Gauss Law. Copyright 2014 John Wiley & Sons, Inc. All rights reserved.
Chapter 23 Gauss Law Copyright 23-1 Electric Flux Electric field vectors and field lines pierce an imaginary, spherical Gaussian surface that encloses a particle with charge +Q. Now the enclosed particle
More informationPhysics 202, Lecture 3. The Electric Field
Physics 202, Lecture 3 Today s Topics Electric Field (Review) Motion of charged particles in external E field Conductors in Electrostatic Equilibrium (Ch. 21.9) Gauss s Law (Ch. 22) Reminder: HW #1 due
More informationPhys102 General Physics II. Chapter 24: Gauss s Law
Phys102 General Physics II Gauss Law Chapter 24: Gauss s Law Flux Electric Flux Gauss Law Coulombs Law from Gauss Law Isolated conductor and Electric field outside conductor Application of Gauss Law Charged
More information3 Chapter. Gauss s Law
3 Chapter Gauss s Law 3.1 Electric Flux... 3-2 3.2 Gauss s Law (see also Gauss s Law Simulation in Section 3.10)... 3-4 Example 3.1: Infinitely Long Rod of Uniform Charge Density... 3-9 Example 3.2: Infinite
More informationweek 3 chapter 28 - Gauss s Law
week 3 chapter 28 - Gauss s Law Here is the central idea: recall field lines... + + q 2q q (a) (b) (c) q + + q q + +q q/2 + q (d) (e) (f) The number of electric field lines emerging from minus the number
More informationCoordinates 2D and 3D Gauss & Stokes Theorems
Coordinates 2 and 3 Gauss & Stokes Theorems Yi-Zen Chu 1 2 imensions In 2 dimensions, we may use Cartesian coordinates r = (x, y) and the associated infinitesimal area We may also employ polar coordinates
More informationGauss s Law. The first Maxwell Equation A very useful computational technique This is important!
Gauss s Law The first Maxwell quation A very useful computational technique This is important! P05-7 Gauss s Law The Idea The total flux of field lines penetrating any of these surfaces is the same and
More informationGauss s Law. 3.1 Quiz. Conference 3. Physics 102 Conference 3. Physics 102 General Physics II. Monday, February 10th, Problem 3.
Physics 102 Conference 3 Gauss s Law Conference 3 Physics 102 General Physics II Monday, February 10th, 2014 3.1 Quiz Problem 3.1 A spherical shell of radius R has charge Q spread uniformly over its surface.
More informationChapter 22 Gauss s Law. Copyright 2009 Pearson Education, Inc.
Chapter 22 Gauss s Law Electric Flux Gauss s Law Units of Chapter 22 Applications of Gauss s Law Experimental Basis of Gauss s and Coulomb s Laws 22-1 Electric Flux Electric flux: Electric flux through
More informationElectric Field Lines
Electric Field Lines Electric forces Electric fields: - Electric field lines emanate from positive charges - Electric field lines disappear at negative charges If you see a bunch of field lines emanating
More informationFlux. Flux = = va. This is the same as asking What is the flux of water through the rectangle? The answer depends on:
Ch. 22: Gauss s Law Gauss s law is an alternative description of Coulomb s law that allows for an easier method of determining the electric field for situations where the charge distribution contains symmetry.
More informationElectric Flux. To investigate this, we have to understand electric flux.
Problem 21.72 A charge q 1 = +5. nc is placed at the origin of an xy-coordinate system, and a charge q 2 = -2. nc is placed on the positive x-axis at x = 4. cm. (a) If a third charge q 3 = +6. nc is now
More informationLecture 3. Electric Field Flux, Gauss Law. Last Lecture: Electric Field Lines
Lecture 3. Electric Field Flux, Gauss Law Last Lecture: Electric Field Lines 1 iclicker Charged particles are fixed on grids having the same spacing. Each charge has the same magnitude Q with signs given
More informationElectric Field. Electric field direction Same direction as the force on a positive charge Opposite direction to the force on an electron
Electric Field Electric field Space surrounding an electric charge (an energetic aura) Describes electric force Around a charged particle obeys inverse-square law Force per unit charge Electric Field Electric
More informationChapter 22 Gauss s Law. Copyright 2009 Pearson Education, Inc.
Chapter 22 Gauss s Law 22-1 Electric Flux Electric flux: Electric flux through an area is proportional to the total number of field lines crossing the area. 22-1 Electric Flux Example 22-1: Electric flux.
More informationE. not enough information given to decide
Q22.1 A spherical Gaussian surface (#1) encloses and is centered on a point charge +q. A second spherical Gaussian surface (#2) of the same size also encloses the charge but is not centered on it. Compared
More informationChapter 24. Gauss s Law
Chapter 24 Gauss s Law Let s return to the field lines and consider the flux through a surface. The number of lines per unit area is proportional to the magnitude of the electric field. This means that
More informationPHY102 Electricity Topic 3 (Lectures 4 & 5) Gauss s Law
PHY1 Electricity Topic 3 (Lectures 4 & 5) Gauss s Law In this topic, we will cover: 1) Electric Flux ) Gauss s Law, relating flux to enclosed charge 3) Electric Fields and Conductors revisited Reading
More informationChapter 22 Gauss s Law
Chapter 22 Gauss s Law Lecture by Dr. Hebin Li Goals for Chapter 22 To use the electric field at a surface to determine the charge within the surface To learn the meaning of electric flux and how to calculate
More informationChapter 23: Gauss Law. PHY2049: Chapter 23 1
Chapter 23: Gauss Law PHY2049: Chapter 23 1 Two Equivalent Laws for Electricity Coulomb s Law equivalent Gauss Law Derivation given in Sec. 23-5 (Read!) Not derived in this book (Requires vector calculus)
More informationPhys 2102 Spring 2002 Exam 1
Phys 2102 Spring 2002 Exam 1 February 19, 2002 1. When a positively charged conductor touches a neutral conductor, the neutral conductor will: (a) Lose protons (b) Gain electrons (c) Stay neutral (d) Lose
More informationFall Lee - Midterm 2 solutions
Fall 2009 - Lee - Midterm 2 solutions Problem 1 Solutions Part A Because the middle slab is a conductor, the electric field inside of the slab must be 0. Parts B and C Recall that to find the electric
More informationHow to define the direction of A??
Chapter Gauss Law.1 Electric Flu. Gauss Law. A charged Isolated Conductor.4 Applying Gauss Law: Cylindrical Symmetry.5 Applying Gauss Law: Planar Symmetry.6 Applying Gauss Law: Spherical Symmetry You will
More informationGauss s Law & Potential
Gauss s Law & Potential Lecture 7: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Flux of an Electric Field : In this lecture we introduce Gauss s law which happens to
More informationLecture 3. Electric Field Flux, Gauss Law
Lecture 3. Electric Field Flux, Gauss Law Attention: the list of unregistered iclickers will be posted on our Web page after this lecture. From the concept of electric field flux to the calculation of
More informationPhysics 9 WS E3 (rev. 1.0) Page 1
Physics 9 WS E3 (rev. 1.0) Page 1 E-3. Gauss s Law Questions for discussion 1. Consider a pair of point charges ±Q, fixed in place near one another as shown. a) On the diagram above, sketch the field created
More informationLecture 4-1 Physics 219 Question 1 Aug Where (if any) is the net electric field due to the following two charges equal to zero?
Lecture 4-1 Physics 219 Question 1 Aug.31.2016. Where (if any) is the net electric field due to the following two charges equal to zero? y Q Q a x a) at (-a,0) b) at (2a,0) c) at (a/2,0) d) at (0,a) and
More informationEssential University Physics
Essential University Physics Richard Wolfson 21 Gauss s Law PowerPoint Lecture prepared by Richard Wolfson Slide 21-1 In this lecture you ll learn To represent electric fields using field-line diagrams
More informationCH 23. Gauss Law. A. Gauss law relates the electric fields at points on a (closed) Gaussian surface to the net charge enclosed by that surface.
CH 23 Gauss Law [SHIVOK SP212] January 4, 2016 I. Introduction to Gauss Law A. Gauss law relates the electric fields at points on a (closed) Gaussian surface to the net charge enclosed by that surface.
More informationChapter (2) Gauss s Law
Chapter (2) Gauss s Law How you can determine the amount of charge within a closed surface by examining the electric field on the surface! What is meant by electric flux and how you can calculate it. How
More informationChapter 24. Gauss s Law
Chapter 24 Gauss s Law Gauss Law Gauss Law can be used as an alternative procedure for calculating electric fields. Gauss Law is based on the inverse-square behavior of the electric force between point
More informationPH 222-2C Fall Gauss Law. Lectures 3-4. Chapter 23 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)
PH 222-2C Fall 212 Gauss Law Lectures 3-4 Chapter 23 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1 Chapter 23 Gauss Law In this chapter we will introduce the following new concepts:
More informationHomework 4 PHYS 212 Dr. Amir
Homework 4 PHYS Dr. Amir. (I) A uniform electric field of magnitude 5.8 passes through a circle of radius 3 cm. What is the electric flux through the circle when its face is (a) perpendicular to the field
More informationSummary: Curvilinear Coordinates
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 10 1 Summary: Curvilinear Coordinates 1. Summary of Integral Theorems 2. Generalized Coordinates 3. Cartesian Coordinates: Surfaces of Constant
More information24 Gauss s Law. Gauss s Law 87:
Green Items that must be covered for the national test Blue Items from educator.com Red Items from the 8 th edition of Serway 24 Gauss s Law 24.1 Electric Flux 24.2 Gauss s Law 24.3 Application of Gauss
More informationdensity = N A where the vector di erential aread A = ^n da, and ^n is the normaltothat patch of surface. Solid angle
Gauss Law Field lines and Flux Field lines are drawn so that E is tangent to the field line at every point. Field lines give us information about the direction of E, but also about its magnitude, since
More informationWelcome. to Electrostatics
Welcome to Electrostatics Outline 1. Coulomb s Law 2. The Electric Field - Examples 3. Gauss Law - Examples 4. Conductors in Electric Field Coulomb s Law Coulomb s law quantifies the magnitude of the electrostatic
More informationPHYS 1441 Section 002 Lecture #6
PHYS 1441 Section 002 Lecture #6 Monday, Sept. 18, 2017 Chapter 21 Motion of a Charged Particle in an Electric Field Electric Dipoles Chapter 22 Electric Flux Gauss Law with many charges What is Gauss
More informationPotential & Potential Energy
Potential & Potential Energy Lecture 10: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Electrostatic Boundary Conditions : We had seen that electric field has a discontinuity
More informationdt Now we will look at the E&M force on moving charges to explore the momentum conservation law in E&M.
. Momentum Conservation.. Momentum in mechanics In classical mechanics p = m v and nd Newton s law d p F = dt If m is constant with time d v F = m = m a dt Now we will look at the &M force on moving charges
More informationGauss s Law. Phys102 Lecture 4. Key Points. Electric Flux Gauss s Law Applications of Gauss s Law. References. SFU Ed: 22-1,2,3. 6 th Ed: 16-10,+.
Phys102 Lecture 4 Phys102 Lecture 4-1 Gauss s Law Key Points Electric Flux Gauss s Law Applications of Gauss s Law References SFU Ed: 22-1,2,3. 6 th Ed: 16-10,+. Electric Flux Electric flux: The direction
More information1. ELECTRIC CHARGES AND FIELDS
1. ELECTRIC CHARGES AND FIELDS 1. What are point charges? One mark questions with answers A: Charges whose sizes are very small compared to the distance between them are called point charges 2. The net
More informationVector Integrals. Scott N. Walck. October 13, 2016
Vector Integrals cott N. Walck October 13, 16 Contents 1 A Table of Vector Integrals Applications of the Integrals.1 calar Line Integral.........................1.1 Finding Total Charge of a Line Charge..........1.
More informationPHYS 281: Midterm Exam
PHYS 28: Midterm Exam October 28, 200, 8:00-9:20 Last name (print): Initials: No calculator or other aids allowed PHYS 28: Midterm Exam Instructor: B. R. Sutherland Date: October 28, 200 Time: 8:00-9:20am
More informationES.182A Topic 45 Notes Jeremy Orloff
E.8A Topic 45 Notes Jeremy Orloff 45 More surface integrals; divergence theorem Note: Much of these notes are taken directly from the upplementary Notes V0 by Arthur Mattuck. 45. Closed urfaces A closed
More informationFall 12 PHY 122 Homework Solutions #2
Fall 12 PHY 122 Homework Solutions #2 Chapter 21 Problem 40 Two parallel circular rings of radius R have their centers on the x axis separated by a distance l, as shown in Fig. 21 60. If each ring carries
More informationPhysics 114 Exam 1 Spring 2013
Physics 114 Exam 1 Spring 2013 Name: For grading purposes (do not write here): Question 1. 1. 2. 2. 3. 3. Problem Answer each of the following questions and each of the problems. Points for each question
More informationPractice Questions Exam 1/page1. PES Physics 2 Practice Exam 1 Questions. Name: Score: /.
Practice Questions Exam 1/page1 PES 110 - Physics Practice Exam 1 Questions Name: Score: /. Instructions Time allowed for this is exam is 1 hour 15 minutes 5 multiple choice (5 points) 3 to 5 written problems
More informationExam 1 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1
Exam 1 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 A rod of charge per unit length λ is surrounded by a conducting, concentric cylinder
More informationGauss s Law. Name. I. The Law: , where ɛ 0 = C 2 (N?m 2
Name Gauss s Law I. The Law:, where ɛ 0 = 8.8510 12 C 2 (N?m 2 1. Consider a point charge q in three-dimensional space. Symmetry requires the electric field to point directly away from the charge in all
More informationPhysics 202, Exam 1 Review
Physics 202, Exam 1 Review Logistics Topics: Electrostatics (Chapters 21-24.6) Point charges: electric force, field, potential energy, and potential Distributions: electric field, electric potential. Interaction
More informationChapter 22. Dr. Armen Kocharian. Gauss s Law Lecture 4
Chapter 22 Dr. Armen Kocharian Gauss s Law Lecture 4 Field Due to a Plane of Charge E must be perpendicular to the plane and must have the same magnitude at all points equidistant from the plane Choose
More informationElectric Flux. If we know the electric field on a Gaussian surface, we can find the net charge enclosed by the surface.
Chapter 23 Gauss' Law Instead of considering the electric fields of charge elements in a given charge distribution, Gauss' law considers a hypothetical closed surface enclosing the charge distribution.
More information2 Which of the following represents the electric field due to an infinite charged sheet with a uniform charge distribution σ.
Slide 1 / 21 1 closed surface, in the shape of a cylinder of radius R and Length L, is placed in a region with a constant electric field of magnitude. The total electric flux through the cylindrical surface
More informationChapter 22: Gauss s Law
Chapter 22: Gauss s Law How you can determine the amount of charge within a closed surface by examining the electric field on the surface. What is meant by electric flux, and how to calculate it. How Gauss
More informationProblem Set #3: 2.11, 2.15, 2.21, 2.26, 2.40, 2.42, 2.43, 2.46 (Due Thursday Feb. 27th)
Chapter Electrostatics Problem Set #3:.,.5,.,.6,.40,.4,.43,.46 (Due Thursday Feb. 7th). Coulomb s Law Coulomb showed experimentally that for two point charges the force is - proportional to each of the
More informationCh 24 Electric Flux, & Gauss s Law
Ch 24 Electric Flux, & Gauss s Law Electric Flux...is related to the number of field lines penetrating a given surface area. Φ e = E A Φ = phi = electric flux Φ units are N m 2 /C Electric Flux Φ = E A
More informationChapter 24. Gauss s Law
Chapter 24 Gauss s Law Electric Flux Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field Φ E = EA Defining Electric Flux EFM06AN1 Electric
More informationAP Physics C. Gauss s Law. Free Response Problems
AP Physics Gauss s Law Free Response Problems 1. A flat sheet of glass of area 0.4 m 2 is placed in a uniform electric field E = 500 N/. The normal line to the sheet makes an angle θ = 60 ẘith the electric
More informationChapter 4. Electrostatic Fields in Matter
Chapter 4. Electrostatic Fields in Matter 4.1. Polarization 4.2. The Field of a Polarized Object 4.3. The Electric Displacement 4.4. Linear Dielectrics 4.5. Energy in dielectric systems 4.6. Forces on
More informationConductors and Insulators
Conductors and Insulators Lecture 11: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Self Energy of a Charge Distribution : In Lecture 1 we briefly discussed what we called
More informationQuestions Chapter 23 Gauss' Law
Questions Chapter 23 Gauss' Law 23-1 What is Physics? 23-2 Flux 23-3 Flux of an Electric Field 23-4 Gauss' Law 23-5 Gauss' Law and Coulomb's Law 23-6 A Charged Isolated Conductor 23-7 Applying Gauss' Law:
More informationSolution Set One. 4 Problem #4: Force due to Self-Capacitance Charge on Conductors Repulsive Force... 11
: olution et One Northwestern University, Electrodynamics I Wednesday, January 13, 2016 Contents 1 Problem #1: General Forms of Gauss and tokes Theorems. 2 1.1 Gauss Theorem - Ordinary Product............................
More informationSolutions to PS 2 Physics 201
Solutions to PS Physics 1 1. ke dq E = i (1) r = i = i k eλ = i k eλ = i k eλ k e λ xdx () (x x) (x x )dx (x x ) + x dx () (x x ) x ln + x x + x x (4) x + x ln + x (5) x + x To find the field for x, we
More informationChapter 28. Gauss s Law
Chapter 28. Gauss s Law Using Gauss s law, we can deduce electric fields, particularly those with a high degree of symmetry, simply from the shape of the charge distribution. The nearly spherical shape
More informationElectric Flux and Gauss s Law
Electric Flux and Gauss s Law Electric Flux Figure (1) Consider an electric field that is uniform in both magnitude and direction, as shown in Figure 1. The total number of lines penetrating the surface
More informationChapter 21: Gauss s Law
Chapter 21: Gauss s Law Electric field lines Electric field lines provide a convenient and insightful way to represent electric fields. A field line is a curve whose direction at each point is the direction
More informationMP204 Electricity and Magnetism
MATHEMATICAL PHYSICS SEMESTER 2, REPEAT 2016 2017 MP204 Electricity and Magnetism Prof. S. J. Hands, Dr. M. Haque and Dr. J.-I. Skullerud Time allowed: 1 1 2 hours Answer ALL questions MP204, 2016 2017,
More informationxy 2 e 2z dx dy dz = 8 3 (1 e 4 ) = 2.62 mc. 12 x2 y 3 e 2z 2 m 2 m 2 m Figure P4.1: Cube of Problem 4.1.
Problem 4.1 A cube m on a side is located in the first octant in a Cartesian coordinate system, with one of its corners at the origin. Find the total charge contained in the cube if the charge density
More informationChapter 21 Chapter 23 Gauss Law. Copyright 2014 John Wiley & Sons, Inc. All rights reserved.
Chapter 21 Chapter 23 Gauss Law Copyright 23-1 What is Physics? Gauss law relates the electric fields at points on a (closed) Gaussian surface to the net charge enclosed by that surface. Gauss law considers
More informationPhysics Lecture: 09
Physics 2113 Jonathan Dowling Physics 2113 Lecture: 09 Flux Capacitor (Schematic) Gauss Law II Carl Friedrich Gauss 1777 1855 Gauss Law: General Case Consider any ARBITRARY CLOSED surface S -- NOTE: this
More informationPhysics II Fiz Summer 2017
Physics II Fiz138-22 Summer 2017 Instructor: Dr. Mehmet Burak Kaynar Office: H.U. Physics Eng Dept. SNTG Lab. E-mail: bkaynar@hacettepe.edu.tr Office hours: Wednesday 10:00 11:00 Evaluation Attendance:
More informationAmpere s Law. Outline. Objectives. BEE-Lecture Notes Anurag Srivastava 1
Outline Introduce as an analogy to Gauss Law. Define. Applications of. Objectives Recognise to be analogous to Gauss Law. Recognise similar concepts: (1) draw an imaginary shape enclosing the current carrying
More informationUNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 110A. Homework #7. Benjamin Stahl. March 3, 2015
UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS A Homework #7 Benjamin Stahl March 3, 5 GRIFFITHS, 5.34 It will be shown that the magnetic field of a dipole can written in the following
More informationfree space (vacuum) permittivity [ F/m]
Electrostatic Fields Electrostatic fields are static (time-invariant) electric fields produced by static (stationary) charge distributions. The mathematical definition of the electrostatic field is derived
More informationPhysics Lecture 13
Physics 113 Jonathan Dowling Physics 113 Lecture 13 EXAM I: REVIEW A few concepts: electric force, field and potential Gravitational Force What is the force on a mass produced by other masses? Kepler s
More informationExperiment III Electric Flux
Experiment III Electric Flux When a charge distribution is symmetrical, we can use Gauss Law, a special law for electric fields. The Gauss Law method of determining the electric field depends on the idea
More informationFall 2004 Physics 3 Tu-Th Section
Fall 2004 Physics 3 Tu-Th Section Claudio Campagnari Lecture 9: 21 Oct. 2004 Web page: http://hep.ucsb.edu/people/claudio/ph3-04/ 1 Last time: Gauss's Law To formulate Gauss's law, introduced a few new
More informationChapter 24 Gauss Law
Chapter 24 Gauss Law A charge inside a box can be probed with a test charge q o to measure E field outside the box. The volume (V) flow rate (dv/dt) of fluid through the wire rectangle (a) is va when the
More informationAP Physics C - E & M
AP Physics C - E & M Gauss's Law 2017-07-08 www.njctl.org Electric Flux Gauss's Law Sphere Table of Contents: Gauss's Law Click on the topic to go to that section. Infinite Rod of Charge Infinite Plane
More informationGauss Law 1. Name Date Partners GAUSS' LAW. Work together as a group on all questions.
Gauss Law 1 Name Date Partners 1. The statement of Gauss' Law: (a) in words: GAUSS' LAW Work together as a group on all questions. The electric flux through a closed surface is equal to the total charge
More informationElectric Flux and Gauss Law
Electric Flux and Gauss Law Gauss Law can be used to find the electric field of complex charge distribution. Easier than treating it as a collection of point charge and using superposition To use Gauss
More informationUniversity Physics (Prof. David Flory) Chapt_24 Sunday, February 03, 2008 Page 1
University Physics (Prof. David Flory) Chapt_4 Sunday, February 03, 008 Page 1 Name: Date: 1. A point charged particle is placed at the center of a spherical Gaussian surface. The net electric flux Φ net
More information3: Gauss s Law July 7, 2008
3: Gauss s Law July 7, 2008 3.1 Electric Flux In order to understand electric flux, it is helpful to take field lines very seriously. Think of them almost as real things that stream out from positive charges
More informationElectric Field Lines. lecture 4.1.1
Electric Field Lines Two protons, A and B, are in an electric field. Which proton has the larger acceleration? A. Proton A B. Proton B C. Both have the same acceleration. lecture 4.1.1 Electric Field Lines
More informationChapter 22 Gauss s law. Electric charge and flux (sec &.3) Gauss s Law (sec &.5) Charges on conductors (sec. 22.6)
Chapter 22 Gauss s law Electric charge and flux (sec. 22.2 &.3) Gauss s Law (sec. 22.4 &.5) Charges on conductors (sec. 22.6) 1 Learning Goals for CH 22 Determine the amount of charge within a closed surface
More informationElectromagnetism: Worked Examples. University of Oxford Second Year, Part A2
Electromagnetism: Worked Examples University of Oxford Second Year, Part A2 Caroline Terquem Department of Physics caroline.terquem@physics.ox.ac.uk Michaelmas Term 2017 2 Contents 1 Potentials 5 1.1 Potential
More informationMake sure you show all your work and justify your answers in order to get full credit.
PHYSICS 7B, Lectures & 3 Spring 5 Midterm, C. Bordel Monday, April 6, 5 7pm-9pm Make sure you show all your work and justify your answers in order to get full credit. Problem esistance & current ( pts)
More informationElectric Field and Gauss s law. January 17, 2014 Physics for Scientists & Engineers 2, Chapter 22 1
Electric Field and Gauss s law January 17, 2014 Physics for Scientists & Engineers 2, Chapter 22 1 Missing clickers! The following clickers are not yet registered! If your clicker number is in this list,
More informationPhysics 2B. Lecture 24B. Gauss 10 Deutsche Mark
Physics 2B Lecture 24B Gauss 10 Deutsche Mark Electric Flux Flux is the amount of something that flows through a given area. Electric flux, Φ E, measures the amount of electric field lines that passes
More informationWorksheet for Exploration 24.1: Flux and Gauss's Law
Worksheet for Exploration 24.1: Flux and Gauss's Law In this Exploration, we will calculate the flux, Φ, through three Gaussian surfaces: green, red and blue (position is given in meters and electric field
More informationMore Gauss, Less Potential
More Gauss, Less Potential Today: Gauss Law examples Monday: Electrical Potential Energy (Guest Lecturer) new SmartPhysics material Wednesday: Electric Potential new SmartPhysics material Thursday: Midterm
More informationChapter 24. Gauss s Law
Chapter 24 Gauss s Law Gauss Law Gauss Law can be used as an alternative procedure for calculating electric fields. Gauss Law is based on the inverse-square behavior of the electric force between point
More informationPhysics 7B Midterm 2 Solutions - Fall 2017 Professor R. Birgeneau
Problem 1 Physics 7B Midterm 2 Solutions - Fall 217 Professor R. Birgeneau (a) Since the wire is a conductor, the electric field on the inside is simply zero. To find the electric field in the exterior
More informationChapter 2 Gauss Law 1
Chapter 2 Gauss Law 1 . Gauss Law Gauss law relates the electric fields at points on a (closed) Gaussian surface to the net charge enclosed by that surface Consider the flux passing through a closed surface
More information