Physics 9 Fall 2009 Homework 3 - Solutions
|
|
- April Chase
- 7 years ago
- Views:
Transcription
1 1 Chapter 28 - Exercise 8 Physics 9 Fall 2009 Homework 3 - s The cube in the figure contains no net charge The electric field is constant over each face of the cube Does the missing electric field vector on the front face point in or out? What is the field strength? Because the electric field is constant, and because the cube contains no net charge, the total flux through the cube has to be zero The flux into the cube is 15 A + 20 A = 35A, where A is the area of the cubes face The flux out of the the cube is 10 A + 15 A + 15 A = 40A This means that more flux is leaving the cube than entering it Because the fluxes in have to balance the fluxes out, we see that the flux on the front face has to point in, and have a magnitude of 5 N/C 1
2 2 Chapter 28 - Exercise 11 The electric flux through the surface shown in the figure is 25 Nm 2 /C What is the electric field strength? Since the electric field is constant, the net flux through the surface is Φ E = E A = EA cos θ The electric field makes an angle of 60 with the surface of the area, meaning that it makes an angle of θ = 30 with respect to the normal of the surface, which is what we want Now, Φ E = EA cos θ E = Φ E /A cos θ With numbers we find E = Φ E A cos θ = 25 (01) 2 = 2890 N/C cos 30 2
3 3 Chapter 28 - Exercise 13 A 20 cm 30 cm rectangle lies in the xz plane What is the electric flux through the rectangle if (a) E ( ) = 50î + 100ˆk N/C? (b) E ( ) = 50î + 100ĵ N/C? The rectangle lies in the xz plane, and so its normal points along the y direction So, A = A y ĵ, where A y = = m 2 Thus, A = ĵ (a) Since the flux is Φ E = E A, and since î ĵ = î ˆk = 0, then the net flux in part (a) is zero! (b) Again, Φ E = E A, and since î ĵ = 0, and ĵ ĵ = 1, we have Φ E = E A = E y A y = = 006 N/m 2 3
4 4 Chapter 28 - Exercise 29 The figure shows four sides of a 30 cm 30 cm 30 cm cube (a) What are the electric fluxes Φ 1 to Φ 4 through sides 1 to 4? (b) What is the net flux through these four sides? (a) Again, the flux is Φ E = AE cos θ On side 1, the angle between E and the normal is θ 1 = 150 For side 2, it s θ 2 = 60, for side 3 it s θ 3 = 30, and for side 4 it s θ 4 = 120 Thus, Φ 1 = EA cos θ 1 = (500) (03) 2 cos 150 = 039 N/Cm 2 Φ 2 = EA cos θ 2 = (500) (03) 2 cos 60 = 0225 N/Cm 2 Φ 3 = EA cos θ 3 = (500) (03) 2 cos 30 = 039 N/Cm 2 Φ 4 = EA cos θ 4 = (500) (03) 2 cos 120 = 0225 N/Cm 2 (b) The net flux through the four sides is just the sum of the individual fluxes, Φ = Φ 1 + Φ 2 + Φ 3 + Φ 4 = 0, which we know has to be the case, since there is no enclosed net charge 4
5 5 Chapter 28 - Problem 39 A hollow metal sphere has inner radius a and outer radius b The hollow sphere has charge +2Q A point charge +Q sits at the center of the hollow sphere (a) Determine the electric fields in the three regions r a, a < r < b, and r b (b) How much charge is on the inside surface of the hollow sphere? On the exterior surface? The setup is seen in the figure to the right There is a single point charge, Q at the center of the sphere The inner radius of the sphere is a, and the outer radius is b We need to determine the fields in each of the regions using Gauss s law (a) Gauss s law says that E d A = Q encl Inside the sphere, for r a, the net enclosed charge is entirely from the point charge at the center So, Q encl = +Q Now, the spherical symmetry of the charge says that we should take a sphere of radius r < a as our Gaussian surface Everywhere on that sphere, the electric field is constant Furthermore, the direction of the electric field is perpendicular to the direction of the normal to the Gaussian surface (both point radially) So, E d A = E da = E da = EA, where A is the area of the surface Since the surface area of sphere is 4πr 2, we find that E (r) = 1 Q r 2, as expected Next, let s look outside the sphere, where r b In this case the net enclosed charge is +3Q, from the center point charge, and the charged sphere We again want a spherical Gaussian surface Proceeding as before we find E (r) = 1 3Q r 2 Finally, what happens inside the metal itself? This is a static collection of charges The charges inside the metal sphere feel the positive charge inside the shell They move around due to the electrical force until they have reached a configuration where all the force cancels out in other words, until the electric field is zero What happens? Since there is a positive charge inside, negative charges in the 5
6 metal shell are attracted to it and move to coat the inner surface with a negative charge, Q, which is just enough to cancel the positive point charge This leaves a net positive charge in the shell which then pushes away from itself, and onto the surface of the sphere So, this charge is now on the surface, in addition to the +2Q that was there before! Thus, the net charge on the surface of the shell is +3Q! So, to recap, 1 Q r a r 2 E (r) = 0 a < r < b 1 3Q r 2 r b (b) As discussed above, there is Q smeared over the inside surface, and +3Q on the outside 6
7 6 Chapter 28 - Problem 44 A positive point charge q sits at the center of a hollow spherical shell The shell, with radius R and negligible thickness, has net charge 2q Find an expression for the electric field strength (a) inside the sphere, r < R, and (b) outside the sphere, r < R In what direction does the electric field point in each case? (a) Again, let s start by drawing a picture, seen to the right Taking a Gaussian surface inside the shell, r < R, encloses only the positive charge +q So, by Gauss s law, E da = E ( 4πr 2) = q, which says that E = 1 q ˆr, for r 2 r R, and points radially outward (b) Now, outside the shell, a Gaussian surface encloses both charges, and so the net enclosed charge is Q encl = 2q+q = q So, Gauss s theorem says that E da = q Evaluating the integral as before gives E = 1 q ˆr, for r R, and points r 2 radially inward 7
8 7 Chapter 28 - Problem 45 Find the electric field inside and outside a hollow plastic ball of radius R that has charge Q uniformly distributed on its outer surface Inside the sphere the field is zero there s no charge inside the ball; it s all on the surface So, inside E = 0, for r < R Outside the ball, we do see a net charge of Q Since the ball is spherically symmetric, we choose a spherical Gaussian surface of radius r As before, the electric field is constant everywhere on the surface because of the symmetry So, as we ve seen before, E d A = EA = Q encl = Q, and so we just find the ordinary Coulomb s law, E = 1 like an ordinary point charge, as we ve seen before Q r 2 So, the field looks just 8
9 8 Chapter 28 - Problem 53 A spherical shell has inner radius R in and outer radius R out The shell contains total charge Q, uniformly distributed The interior of the shell is empty of charge and matter (a) Find the electric field outside of the shell, r R out (b) Find the electric field in the interior of the shell, r R in (c) Find the electric field within the shell, R in r R out (d) Show that your solutions match at both the inner and outer boundaries (e) Draw a graph of E versus r Once again, we will use Gauss s Law to find the electric field in each region (a) Outside of the shell, when r R out, the field looks like that of a point charge, since any spherical Gaussian surface encloses the full charge So, E = 1 Q ˆr r 2 (b) Inside the shell, where r R in, there is no charge and hence nothing to source the electric field So, the electric field is zero, E = 0 (c) Now, this gets a bit tougher Now, we want to use a Gaussian surface of radius R in r R out ; ie, the surface is inside the metal itself Since the total charge on the shell is uniformly distributed, the Gaussian surface encloses a certain amount of charge The charge density is constant, so ρ = Q/V, where V is the volume of the charge distribution The total volume of the sphere is just the difference between two spheres of radii R out and R in, V = 4 3 π (R3 out R 3 in) Thus, ρ = 3Q 4π(R 3 out R3 in) Now the Gaussian surface encloses a volume ( V encl ) = 4π 3 (r3 Rin), 3 which gives an r enclosed charge of Q encl = ρv encl = Q 3 Rin 3 Again, using Gauss s law we Rout 3 R3 in find that E da = EA = E (4πr 2 ) = Q encl, and so E = 1 ( ) Q r 3 Rin 3 ˆr r 2 Rout 3 Rin 3 Again, we find the following field values 0 ( ) r R in 1 Q r E (r) = 3 R 3 r 2 in ˆr R Rout 3 in r R out R3 in 1 Q ˆr r R r 2 out 9
10 (d) Now, the boundaries are at the values r = R in and R out We want to look at the field inside( the metal, ) when R in r R out When r = R in we get E (R in ) = 1 Q R 3 in R 3 Rin 2 in ˆr = 0 This matches the value inside the sphere On Rout 3 R3 in the other boundary, when r = R out, then E ( ) = 1 R 3 out Rin 3 ˆr = 1 Q ˆr, Rout 2 (e) Q R 2 out R 3 out R3 in which exactly matches our solution for r R out on the surface So, we can trust our solutions! The electric field is plotted in the graph to the right The field starts out at zero at the origin, and remains so until it reaches the inner surface at r = R in Then the field begins to rise, not quite linearly, going as r r 2, until it reaches the outer surface, r = R out This is where it reaches its maximum, after which it begins to fall off with the usual r 2 Coulomb law 10
11 9 Chapter 28 - Problem 54 An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge Ze Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere (a) Show that the electric field inside this atom is E in = Ze ( 1 r 2 r R 3 (b) What is E at the surface of the atom? Is this the expected value? Explain (c) A uranium atom has Z = 92 and R = 010 nm What is the electric field strength at r = 1 2 R? ) The atom is seen in the figure to the right It contains a positively charged nucleus of charge +Ze, which is embedded inside a uniform sphere of charge Ze We are interested in the field inside the sphere, so we take a spherical gaussian surface of radius r (a) Since the electric field is radial, and everywhere constant on the Gaussian surface, Gauss s law gives the usual result that E da = EA = E (4πr 2 ) = Q encl Now, what s Q encl? This problem is very similar to the last one, except with the addition of +Ze at the center So, the total enclosed charge is Q encl = +Ze negative charge ( The negative charge is just the charge density, times the enclosed volume, 4π 3 r3 = Ze r3 Thus, R 3 ) Ze 4 3 πr3 ) Q encl = Ze (1 r3, R 3 which, using Gauss s law as discussed above, gives E in = Ze ( 1 r r ) 2 R 3 (b) On the surface of the atom, r = R, and so the electric field is zero This is to be expected, since on the surface the enclosed charge is zero (+Ze Ze), and so the electric flux is zero from Gauss s law The atom looks neutral outside the surface 11
12 (c) When r = R/2, then E ( ) R = Ze ( 4 2 R 1 ) 2 2R 2 = 7 Ze 2 R 2 Plugging in the numbers gives ( ) R E = 7 Ze 2 2 R = ( ) 2 = N/C So, E = N/C, which is a very strong electric field! 12
13 10 Chapter 28 - Problem 58 A sphere of radius R has total charge Q The volume charge density (C/m 3 ) within the sphere is ( ρ = ρ 0 1 r ) R This charge density decreases linearly from ρ 0 at the center to zero at the edge of the sphere (a) Show that ρ 0 = 3Q/πR 2 Hint: You ll need to do a volume integral (b) Show that the electric field inside the sphere points radially outward with magnitude E = Qr (4 3 r ) R 3 R (c) Show that your result of part b has the expected value at r = R (a) Now, the charge density is not constant! So, we can t just say that Q = ρv Instead, we have to integrate, Q = dq = ρdv As we ve seen before, for a sphere dv = 4πr 2 dr, as you can see by imagining a series of thin shells of thickness dr piled on top of each other So, R ( Q = ρdv = 4πρ 0 1 r ) [ ] r r 2 3 dr = 4πρ 0 R 3 r4 R = π 4R 3 ρ 0R 3, 0 and so ρ 0 = 3Q πr 3 (b) Now, we take a Gaussian surface of radius r inside the sphere From Gauss s law, E da = Q encl, the left-hand side is E (4πr 2 ) because the field is radial inside the sphere What s Q encl? Again, we integrate as before, only now out to a radius r, corresponding to our surface r Q encl = ρdv = 4πρ 0 Plugging for ρ 0 gives Thus, from Gauss s law 0 ( 1 r R Q encl = 4πρ 0 3 E (r) = ) r 2 dr = 4πρ 0 [ r 3 ] 3 r4 4R ] [r 3 3r4 = Q [4 r3 3 r ] 4R R 3 R Qr (4 3 r R 3 R ) 0 = 4πρ 0 3 ] [r 3 3r4 4R ( ) (c) At r = R, then E (R) = QR R 4 3 R 3 R = Q So, the electric field at r = R R 2 is that of a point charge, which is exactly we should expect! 13
Chapter 22: Electric Flux and Gauss s Law
22.1 ntroduction We have seen in chapter 21 that determining the electric field of a continuous charge distribution can become very complicated for some charge distributions. t would be desirable if we
More informationCHAPTER 24 GAUSS S LAW
CHAPTER 4 GAUSS S LAW 4. The net charge shown in Fig. 4-40 is Q. Identify each of the charges A, B, C shown. A B C FIGURE 4-40 4. From the direction of the lines of force (away from positive and toward
More informationPhysics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings
1 of 11 9/7/2012 1:06 PM Logged in as Julie Alexander, Instructor Help Log Out Physics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings Course Home Assignments Roster Gradebook Item Library
More informationExam 1 Practice Problems Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8 Spring 13 Exam 1 Practice Problems Solutions Part I: Short Questions and Concept Questions Problem 1: Spark Plug Pictured at right is a typical
More informationChapter 22: The Electric Field. Read Chapter 22 Do Ch. 22 Questions 3, 5, 7, 9 Do Ch. 22 Problems 5, 19, 24
Chapter : The Electric Field Read Chapter Do Ch. Questions 3, 5, 7, 9 Do Ch. Problems 5, 19, 4 The Electric Field Replaces action-at-a-distance Instead of Q 1 exerting a force directly on Q at a distance,
More informationChapter 18. Electric Forces and Electric Fields
My lecture slides may be found on my website at http://www.physics.ohio-state.edu/~humanic/ ------------------------------------------------------------------- Chapter 18 Electric Forces and Electric Fields
More informationChapter 4. Electrostatic Fields in Matter
Chapter 4. Electrostatic Fields in Matter 4.1. Polarization A neutral atom, placed in an external electric field, will experience no net force. However, even though the atom as a whole is neutral, the
More informationThe Electric Field. Electric Charge, Electric Field and a Goofy Analogy
. The Electric Field Concepts and Principles Electric Charge, Electric Field and a Goofy Analogy We all know that electrons and protons have electric charge. But what is electric charge and what does it
More informationHW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case.
HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 22.P.053 The figure below shows a portion of an infinitely
More informationAs customary, choice (a) is the correct answer in all the following problems.
PHY2049 Summer 2012 Instructor: Francisco Rojas Exam 1 As customary, choice (a) is the correct answer in all the following problems. Problem 1 A uniformly charge (thin) non-conucting ro is locate on the
More informationSolution. Problem. Solution. Problem. Solution
4. A 2-g ping-pong ball rubbed against a wool jacket acquires a net positive charge of 1 µc. Estimate the fraction of the ball s electrons that have been removed. If half the ball s mass is protons, their
More informationPhysics 202, Lecture 3. The Electric Field
Physics 202, Lecture 3 Today s Topics Electric Field Quick Review Motion of Charged Particles in an Electric Field Gauss s Law (Ch. 24, Serway) Conductors in Electrostatic Equilibrium (Ch. 24) Homework
More informationVector surface area Differentials in an OCS
Calculus and Coordinate systems EE 311 - Lecture 17 1. Calculus and coordinate systems 2. Cartesian system 3. Cylindrical system 4. Spherical system In electromagnetics, we will often need to perform integrals
More informationCHAPTER 26 ELECTROSTATIC ENERGY AND CAPACITORS
CHAPTER 6 ELECTROSTATIC ENERGY AND CAPACITORS. Three point charges, each of +q, are moved from infinity to the vertices of an equilateral triangle of side l. How much work is required? The sentence preceding
More informationExam 2 Practice Problems Part 1 Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Exam Practice Problems Part 1 Solutions Problem 1 Electric Field and Charge Distributions from Electric Potential An electric potential V ( z
More informationExercises on Voltage, Capacitance and Circuits. A d = (8.85 10 12 ) π(0.05)2 = 6.95 10 11 F
Exercises on Voltage, Capacitance and Circuits Exercise 1.1 Instead of buying a capacitor, you decide to make one. Your capacitor consists of two circular metal plates, each with a radius of 5 cm. The
More informationLecture 5. Electric Flux and Flux Density, Gauss Law in Integral Form
Lecture 5 Electric Flux and Flux ensity, Gauss Law in Integral Form ections: 3.1, 3., 3.3 Homework: ee homework file LECTURE 5 slide 1 Faraday s Experiment (1837), Flux charge transfer from inner to outer
More informationElectric Fields in Dielectrics
Electric Fields in Dielectrics Any kind of matter is full of positive and negative electric charges. In a dielectric, these charges cannot move separately from each other through any macroscopic distance,
More informationElectromagnetism Laws and Equations
Electromagnetism Laws and Equations Andrew McHutchon Michaelmas 203 Contents Electrostatics. Electric E- and D-fields............................................. Electrostatic Force............................................2
More informationProblem 1 (25 points)
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2012 Exam Three Solutions Problem 1 (25 points) Question 1 (5 points) Consider two circular rings of radius R, each perpendicular
More informationELECTRIC FIELD LINES AND EQUIPOTENTIAL SURFACES
ELECTRIC FIELD LINES AND EQUIPOTENTIAL SURFACES The purpose of this lab session is to experimentally investigate the relation between electric field lines of force and equipotential surfaces in two dimensions.
More information( )( 10!12 ( 0.01) 2 2 = 624 ( ) Exam 1 Solutions. Phy 2049 Fall 2011
Phy 49 Fall 11 Solutions 1. Three charges form an equilateral triangle of side length d = 1 cm. The top charge is q = - 4 μc, while the bottom two are q1 = q = +1 μc. What is the magnitude of the net force
More informationReview Questions PHYS 2426 Exam 2
Review Questions PHYS 2426 Exam 2 1. If 4.7 x 10 16 electrons pass a particular point in a wire every second, what is the current in the wire? A) 4.7 ma B) 7.5 A C) 2.9 A D) 7.5 ma E) 0.29 A Ans: D 2.
More informationChapter 6. Current and Resistance
6 6 6-0 Chapter 6 Current and Resistance 6.1 Electric Current... 6-2 6.1.1 Current Density... 6-2 6.2 Ohm s Law... 6-5 6.3 Summary... 6-8 6.4 Solved Problems... 6-9 6.4.1 Resistivity of a Cable... 6-9
More informationGauss Formulation of the gravitational forces
Chapter 1 Gauss Formulation of the gravitational forces 1.1 ome theoretical background We have seen in class the Newton s formulation of the gravitational law. Often it is interesting to describe a conservative
More informationElectrostatic Fields: Coulomb s Law & the Electric Field Intensity
Electrostatic Fields: Coulomb s Law & the Electric Field Intensity EE 141 Lecture Notes Topic 1 Professor K. E. Oughstun School of Engineering College of Engineering & Mathematical Sciences University
More informationHow To Understand The Theory Of Gravity
Newton s Law of Gravity and Kepler s Laws Michael Fowler Phys 142E Lec 9 2/6/09. These notes are partly adapted from my Physics 152 lectures, where more mathematical details can be found. The Universal
More informationChapter 27 Magnetic Field and Magnetic Forces
Chapter 27 Magnetic Field and Magnetic Forces - Magnetism - Magnetic Field - Magnetic Field Lines and Magnetic Flux - Motion of Charged Particles in a Magnetic Field - Applications of Motion of Charged
More information1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 80 µ T at the loop center. What is the loop radius?
CHAPTER 3 SOURCES O THE MAGNETC ELD 1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 8 µ T at the loop center. What is the loop radius? Equation 3-3, with
More informationSURFACE TENSION. Definition
SURFACE TENSION Definition In the fall a fisherman s boat is often surrounded by fallen leaves that are lying on the water. The boat floats, because it is partially immersed in the water and the resulting
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. 8.02 Spring 2013 Conflict Exam Two Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 802 Spring 2013 Conflict Exam Two Solutions Problem 1 (25 points): answers without work shown will not be given any credit A uniformly charged
More informationChapter 19. Mensuration of Sphere
8 Chapter 19 19.1 Sphere: A sphere is a solid bounded by a closed surface every point of which is equidistant from a fixed point called the centre. Most familiar examples of a sphere are baseball, tennis
More informationCalculation of gravitational forces of a sphere and a plane
Sphere and plane 1 Calculation of gravitational forces of a sphere and a plane A paper by: Dipl. Ing. Matthias Krause, (CID) Cosmological Independent Department, Germany, 2007 Objective The purpose of
More informationCopyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass
Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of
More informationǫ 0 = 8.85419 10 12 C 2 /N m 2,
Version 001 review unit chiu 58655 1 This print-out should have 45 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The long negatively charged
More informationPhysics 112 Homework 5 (solutions) (2004 Fall) Solutions to Homework Questions 5
Solutions to Homework Questions 5 Chapt19, Problem-2: (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields in Figure P19.2, as shown. (b) Repeat
More informationChapter 9 Circular Motion Dynamics
Chapter 9 Circular Motion Dynamics 9. Introduction Newton s Second Law and Circular Motion... 9. Universal Law of Gravitation and the Circular Orbit of the Moon... 9.. Universal Law of Gravitation... 3
More informationFORCE ON A CURRENT IN A MAGNETIC FIELD
7/16 Force current 1/8 FORCE ON A CURRENT IN A MAGNETIC FIELD PURPOSE: To study the force exerted on an electric current by a magnetic field. BACKGROUND: When an electric charge moves with a velocity v
More informationVOLUME AND SURFACE AREAS OF SOLIDS
VOLUME AND SURFACE AREAS OF SOLIDS Q.1. Find the total surface area and volume of a rectangular solid (cuboid) measuring 1 m by 50 cm by 0.5 m. 50 1 Ans. Length of cuboid l = 1 m, Breadth of cuboid, b
More informationAwell-known lecture demonstration1
Acceleration of a Pulled Spool Carl E. Mungan, Physics Department, U.S. Naval Academy, Annapolis, MD 40-506; mungan@usna.edu Awell-known lecture demonstration consists of pulling a spool by the free end
More informationBasic Nuclear Concepts
Section 7: In this section, we present a basic description of atomic nuclei, the stored energy contained within them, their occurrence and stability Basic Nuclear Concepts EARLY DISCOVERIES [see also Section
More informationThis makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5
1. (Line integrals Using parametrization. Two types and the flux integral) Formulas: ds = x (t) dt, d x = x (t)dt and d x = T ds since T = x (t)/ x (t). Another one is Nds = T ds ẑ = (dx, dy) ẑ = (dy,
More informationSolutions to Homework 10
Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x
More informationCHAPTER 30 GAUSS S LAW
CHPTER GUSS S LW. Given : E /5 E î /5 E ĵ E. N/C The pane is parae to yz-pane. Hence ony /5 E î passes perpendicuar to the pane whereas /5 E ĵ goes parae. rea.m (given Fux E /5.. Nm /c Nm /c. Given ength
More informationElectromagnetism Extra Study Questions Short Answer
Electromagnetism Extra Study Questions Short Answer 1. The electrostatic force between two small charged objects is 5.0 10 5 N. What effect would each of the following changes have on the magnitude of
More informationChapter 19. General Matrices. An n m matrix is an array. a 11 a 12 a 1m a 21 a 22 a 2m A = a n1 a n2 a nm. The matrix A has n row vectors
Chapter 9. General Matrices An n m matrix is an array a a a m a a a m... = [a ij]. a n a n a nm The matrix A has n row vectors and m column vectors row i (A) = [a i, a i,..., a im ] R m a j a j a nj col
More informationElectromagnetism - Lecture 2. Electric Fields
Electromagnetism - Lecture 2 Electric Fields Review of Vector Calculus Differential form of Gauss s Law Poisson s and Laplace s Equations Solutions of Poisson s Equation Methods of Calculating Electric
More informationChapter 10 Rotational Motion. Copyright 2009 Pearson Education, Inc.
Chapter 10 Rotational Motion Angular Quantities Units of Chapter 10 Vector Nature of Angular Quantities Constant Angular Acceleration Torque Rotational Dynamics; Torque and Rotational Inertia Solving Problems
More informationForce on a square loop of current in a uniform B-field.
Force on a square loop of current in a uniform B-field. F top = 0 θ = 0; sinθ = 0; so F B = 0 F bottom = 0 F left = I a B (out of page) F right = I a B (into page) Assume loop is on a frictionless axis
More informationPhysics 201 Homework 8
Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the
More informationM PROOF OF THE DIVERGENCE THEOREM AND STOKES THEOREM
68 Theor Supplement Section M M POOF OF THE DIEGENE THEOEM ND STOKES THEOEM In this section we give proofs of the Divergence Theorem Stokes Theorem using the definitions in artesian coordinates. Proof
More informationEðlisfræði 2, vor 2007
[ Assignment View ] [ Pri Eðlisfræði 2, vor 2007 28. Sources of Magnetic Field Assignment is due at 2:00am on Wednesday, March 7, 2007 Credit for problems submitted late will decrease to 0% after the deadline
More informationLecture 8 : Coordinate Geometry. The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 20
Lecture 8 : Coordinate Geometry The coordinate plane The points on a line can be referenced if we choose an origin and a unit of 0 distance on the axis and give each point an identity on the corresponding
More informationFundamental Theorems of Vector Calculus
Fundamental Theorems of Vector Calculus We have studied the techniques for evaluating integrals over curves and surfaces. In the case of integrating over an interval on the real line, we were able to use
More informationSolution Derivations for Capa #11
Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform
More informationoil liquid water water liquid Answer, Key Homework 2 David McIntyre 1
Answer, Key Homework 2 David McIntyre 1 This print-out should have 14 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making
More informationPhysics 221 Experiment 5: Magnetic Fields
Physics 221 Experiment 5: Magnetic Fields August 25, 2007 ntroduction This experiment will examine the properties of magnetic fields. Magnetic fields can be created in a variety of ways, and are also found
More informationIf Σ is an oriented surface bounded by a curve C, then the orientation of Σ induces an orientation for C, based on the Right-Hand-Rule.
Oriented Surfaces and Flux Integrals Let be a surface that has a tangent plane at each of its nonboundary points. At such a point on the surface two unit normal vectors exist, and they have opposite directions.
More informationAB2.5: Surfaces and Surface Integrals. Divergence Theorem of Gauss
AB2.5: urfaces and urface Integrals. Divergence heorem of Gauss epresentations of surfaces or epresentation of a surface as projections on the xy- and xz-planes, etc. are For example, z = f(x, y), x =
More informationChapter 20 Electrostatics and Coulomb s Law 20.1 Introduction electrostatics. 20.2 Separation of Electric Charge by Rubbing
I wish to give an account of some investigations which have led to the conclusion that the carriers of negative electricity are bodies, which I have called corpuscles, having a mass very much smaller than
More informationShape Dictionary YR to Y6
Shape Dictionary YR to Y6 Guidance Notes The terms in this dictionary are taken from the booklet Mathematical Vocabulary produced by the National Numeracy Strategy. Children need to understand and use
More informationSolutions - Homework sections 17.7-17.9
olutions - Homework sections 7.7-7.9 7.7 6. valuate xy d, where is the triangle with vertices (,, ), (,, ), and (,, ). The three points - and therefore the triangle between them - are on the plane x +
More informationPhys222 Winter 2012 Quiz 4 Chapters 29-31. Name
Name If you think that no correct answer is provided, give your answer, state your reasoning briefly; append additional sheet of paper if necessary. 1. A particle (q = 5.0 nc, m = 3.0 µg) moves in a region
More informationcircular motion & gravitation physics 111N
circular motion & gravitation physics 111N uniform circular motion an object moving around a circle at a constant rate must have an acceleration always perpendicular to the velocity (else the speed would
More information1 Solution of Homework
Math 3181 Dr. Franz Rothe February 4, 2011 Name: 1 Solution of Homework 10 Problem 1.1 (Common tangents of two circles). How many common tangents do two circles have. Informally draw all different cases,
More informationExperiment 5: Magnetic Fields of a Bar Magnet and of the Earth
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2005 Experiment 5: Magnetic Fields of a Bar Magnet and of the Earth OBJECTIVES 1. To examine the magnetic field associated with a
More informationExam 2 Practice Problems Part 2 Solutions
Problem 1: Short Questions MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8. Exam Practice Problems Part Solutions (a) Can a constant magnetic field set into motion an electron, which is initially
More informationChapter 16. Mensuration of Cylinder
335 Chapter 16 16.1 Cylinder: A solid surface generated by a line moving parallel to a fixed line, while its end describes a closed figure in a plane is called a cylinder. A cylinder is the limiting case
More informationMECHANICS OF SOLIDS - BEAMS TUTORIAL 1 STRESSES IN BEAMS DUE TO BENDING. On completion of this tutorial you should be able to do the following.
MECHANICS OF SOLIDS - BEAMS TUTOIAL 1 STESSES IN BEAMS DUE TO BENDING This is the first tutorial on bending of beams designed for anyone wishing to study it at a fairly advanced level. You should judge
More informationMFF 2a: Charged Particle and a Uniform Magnetic Field... 2
MFF 2a: Charged Particle and a Uniform Magnetic Field... 2 MFF2a RT1: Charged Particle and a Uniform Magnetic Field... 3 MFF2a RT2: Charged Particle and a Uniform Magnetic Field... 4 MFF2a RT3: Charged
More informationToday in Physics 217: the method of images
Today in Physics 17: the method of images Solving the Laplace and Poisson euations by sleight of hand Introduction to the method of images Caveats Example: a point charge and a grounded conducting sphere
More informationForce on Moving Charges in a Magnetic Field
[ Assignment View ] [ Eðlisfræði 2, vor 2007 27. Magnetic Field and Magnetic Forces Assignment is due at 2:00am on Wednesday, February 28, 2007 Credit for problems submitted late will decrease to 0% after
More informationThe potential (or voltage) will be introduced through the concept of a gradient. The gradient is another sort of 3-dimensional derivative involving
The potential (or voltage) will be introduced through the concept of a gradient. The gradient is another sort of 3-dimensional derivative involving the vector del except we don t take the dot product as
More informationSolutions for Review Problems
olutions for Review Problems 1. Let be the triangle with vertices A (,, ), B (4,, 1) and C (,, 1). (a) Find the cosine of the angle BAC at vertex A. (b) Find the area of the triangle ABC. (c) Find a vector
More informationNotes: Most of the material in this chapter is taken from Young and Freedman, Chap. 13.
Chapter 5. Gravitation Notes: Most of the material in this chapter is taken from Young and Freedman, Chap. 13. 5.1 Newton s Law of Gravitation We have already studied the effects of gravity through the
More informationMATHEMATICS FOR ENGINEERING BASIC ALGEBRA
MATHEMATICS FOR ENGINEERING BASIC ALGEBRA TUTORIAL 4 AREAS AND VOLUMES This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals.
More informationProblem Set V Solutions
Problem Set V Solutions. Consider masses m, m 2, m 3 at x, x 2, x 3. Find X, the C coordinate by finding X 2, the C of mass of and 2, and combining it with m 3. Show this is gives the same result as 3
More informationMath 1B, lecture 5: area and volume
Math B, lecture 5: area and volume Nathan Pflueger 6 September 2 Introduction This lecture and the next will be concerned with the computation of areas of regions in the plane, and volumes of regions in
More informationWhen the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.
Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs
More informationChapter 7: Polarization
Chapter 7: Polarization Joaquín Bernal Méndez Group 4 1 Index Introduction Polarization Vector The Electric Displacement Vector Constitutive Laws: Linear Dielectrics Energy in Dielectric Systems Forces
More informationLesson 26: Reflection & Mirror Diagrams
Lesson 26: Reflection & Mirror Diagrams The Law of Reflection There is nothing really mysterious about reflection, but some people try to make it more difficult than it really is. All EMR will reflect
More informationPHYS 101-4M, Fall 2005 Exam #3. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
PHYS 101-4M, Fall 2005 Exam #3 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A bicycle wheel rotates uniformly through 2.0 revolutions in
More information6 J - vector electric current density (A/m2 )
Determination of Antenna Radiation Fields Using Potential Functions Sources of Antenna Radiation Fields 6 J - vector electric current density (A/m2 ) M - vector magnetic current density (V/m 2 ) Some problems
More informationDOING PHYSICS WITH MATLAB COMPUTATIONAL OPTICS RAYLEIGH-SOMMERFELD DIFFRACTION INTEGRAL OF THE FIRST KIND
DOING PHYSICS WITH MATLAB COMPUTATIONAL OPTICS RAYLEIGH-SOMMERFELD DIFFRACTION INTEGRAL OF THE FIRST KIND THE THREE-DIMENSIONAL DISTRIBUTION OF THE RADIANT FLUX DENSITY AT THE FOCUS OF A CONVERGENCE BEAM
More informationReflection and Refraction
Equipment Reflection and Refraction Acrylic block set, plane-concave-convex universal mirror, cork board, cork board stand, pins, flashlight, protractor, ruler, mirror worksheet, rectangular block worksheet,
More informationEXPERIMENT 4 The Periodic Table - Atoms and Elements
EXPERIMENT 4 The Periodic Table - Atoms and Elements INTRODUCTION Primary substances, called elements, build all the materials around you. There are more than 109 different elements known today. The elements
More information1051-232 Imaging Systems Laboratory II. Laboratory 4: Basic Lens Design in OSLO April 2 & 4, 2002
05-232 Imaging Systems Laboratory II Laboratory 4: Basic Lens Design in OSLO April 2 & 4, 2002 Abstract: For designing the optics of an imaging system, one of the main types of tools used today is optical
More informationA Survival Guide to Vector Calculus
A Survival Guide to Vector Calculus Aylmer Johnson When I first tried to learn about Vector Calculus, I found it a nightmare. Eventually things became clearer and I discovered that, once I had really understood
More informationExam 1 Review Questions PHY 2425 - Exam 1
Exam 1 Review Questions PHY 2425 - Exam 1 Exam 1H Rev Ques.doc - 1 - Section: 1 7 Topic: General Properties of Vectors Type: Conceptual 1 Given vector A, the vector 3 A A) has a magnitude 3 times that
More informationF = 0. x ψ = y + z (1) y ψ = x + z (2) z ψ = x + y (3)
MATH 255 FINAL NAME: Instructions: You must include all the steps in your derivations/answers. Reduce answers as much as possible, but use exact arithmetic. Write neatly, please, and show all steps. Scientists
More informationLesson 3: Isothermal Hydrostatic Spheres. B68: a self-gravitating stable cloud. Hydrostatic self-gravitating spheres. P = "kt 2.
Lesson 3: Isothermal Hydrostatic Spheres B68: a self-gravitating stable cloud Bok Globule Relatively isolated, hence not many external disturbances Though not main mode of star formation, their isolation
More informationCenter of Gravity. We touched on this briefly in chapter 7! x 2
Center of Gravity We touched on this briefly in chapter 7! x 1 x 2 cm m 1 m 2 This was for what is known as discrete objects. Discrete refers to the fact that the two objects separated and individual.
More informationSection A-3 Polynomials: Factoring APPLICATIONS. A-22 Appendix A A BASIC ALGEBRA REVIEW
A- Appendi A A BASIC ALGEBRA REVIEW C In Problems 53 56, perform the indicated operations and simplify. 53. ( ) 3 ( ) 3( ) 4 54. ( ) 3 ( ) 3( ) 7 55. 3{[ ( )] ( )( 3)} 56. {( 3)( ) [3 ( )]} 57. Show by
More informationMATH 10550, EXAM 2 SOLUTIONS. x 2 + 2xy y 2 + x = 2
MATH 10550, EXAM SOLUTIONS (1) Find an equation for the tangent line to at the point (1, ). + y y + = Solution: The equation of a line requires a point and a slope. The problem gives us the point so we
More informationBegin creating the geometry by defining two Circles for the spherical endcap, and Subtract Areas to create the vessel wall.
ME 477 Pressure Vessel Example 1 ANSYS Example: Axisymmetric Analysis of a Pressure Vessel The pressure vessel shown below is made of cast iron (E = 14.5 Msi, ν = 0.21) and contains an internal pressure
More informationChapter 23 Electric Potential. Copyright 2009 Pearson Education, Inc.
Chapter 23 Electric Potential 23-1 Electrostatic Potential Energy and Potential Difference The electrostatic force is conservative potential energy can be defined. Change in electric potential energy is
More informationAnalysis of Stresses and Strains
Chapter 7 Analysis of Stresses and Strains 7.1 Introduction axial load = P / A torsional load in circular shaft = T / I p bending moment and shear force in beam = M y / I = V Q / I b in this chapter, we
More informationChapter 6 Work and Energy
Chapter 6 WORK AND ENERGY PREVIEW Work is the scalar product of the force acting on an object and the displacement through which it acts. When work is done on or by a system, the energy of that system
More information27.3. Introduction. Prerequisites. Learning Outcomes
olume Integrals 27. Introduction In the previous two sections, surface integrals (or double integrals) were introduced i.e. functions were integrated with respect to one variable and then with respect
More information