Stability investigation for the inverse heat conduction problems Krzysztof Grysa & Artur Maciqg Kielce University of Technology, AL 1000-leciaPP 7,

Size: px
Start display at page:

Download "Stability investigation for the inverse heat conduction problems Krzysztof Grysa & Artur Maciqg Kielce University of Technology, AL 1000-leciaPP 7,"

Transcription

1 Stability investigation for the inverse heat conduction problems Krzysztof Grysa & Artur Maciqg Kielce University of Technology, AL 1000-leciaPP 7, Abstract Discretisation with respect to time changes the heat conduction equation into a system of Helmholtz equations. An error introduced by such a simplification is investigated. For the inverse problems the constrains for a time step are considered in the case of two inaccurate internal responses. Nomographs helping to choose a proper time step for inverse problems with two internal responses are presented. 1 Discretisation Consider the heat conduction equation in dimensionless form: with initial and boundary conditions as follows: or r(*,0) = %(x), xeq (2) r(^,f)=7^,f), J%E6Q, fe(0,co) (3a)

2 94 Advanced Computational Methods in Heat Transfer ~\ <T> (xt,t) = Qt,(x>,t), x>edn, / e (0,«) (3b) CM Approximation of the derivative with respect to time with a first back difference leads to the following formulation of the problem: (x) = /,(4, ^^0, 6 = 1,2,... (4) or where (5a) ^-(xh) = Q,(xi,,kAt), x^dq, k = 1,2,... (5b) CM, A/ - time step, p -, and A/.Y,/X/, A = l,2,... (6) (A-l)Af Solution of such a problem may be expressed in an integral form ^(W/O-W-UW/,), ^50 (7) where $, is a simple layer potential and V^ is a volume potential for the Helmholtz equation. They may be expressed as integrals (8) /r (9) n where h and / stand for densities of the simple layer and volume potentials, respectively. G^ is a fundamental solution of the Helmholtz equation. In the formula (7) the density of the simple layer potential is an unknown function that has to satisfy the boundary condition. Hence, for both boundary conditions, (5a) and (5b), an integral equation has to be solved: for the condition (5a) or S&,p\h) = Tt($-V^,p\h), %e90 (10)

3 Advanced Computational Methods in Heat Transfer 95 for the condition (5b). The form (7) of the solution is convenient for analytical consideration in the case of body with simple geometry (e.g. flat slab, cylindrical or spherical layer, cylinder, sphere). 2 An inverse problem solution If the boundary condition in the problem (1), (2), (3a) or (3b) is replaced by a condition at an inner surface 9Q*eiQ (called an internal response - IR), then we consider an inverse heat conduction problem (IHCP). An integral form of an approximate solution of the IHCP has the form (7), too, and the simple layer potential density has also to satisfy an equation in form (10) (for the temperature IR) or in form (11) (for the heat flux IR), but in both cases jtedq.the IHCPs belong to the class of ill-posed problems. Besides, the Helmholtz model introduces its own constrains for time step magnitude, in particular respecting the possible IR inaccuracy propagation. Subsequently, the constrains for the time step magnitude are discussed. We confine ourselves to an IHCP in a flat slab. However, the conclusions are widened for a cylindrical and spherical layer. 3 The IHCP solution for aflatslab For ID the fundamental solution of Helmholtz equation reads I GI(X- y,p) = exp(-p\x - y) (12) 2f The solution (7) of the inverse problem in the case of two temperature IRs in a flat slab of unique thickness has a form _/ \ sinh[x*f-j;)]r *

4 96 Advanced Computational Methods in Heat Transfer s'm\\[pu] (13) where: u = Xg - x^, x e(0,l), k- 1,2,....,x^,jc^ - coordinates of points with IRs, 7^* ^Tgk ~ ^he temperature IRs at Xg,Xj at the moment kat. When mixed IRs are known (in xj - a temperature T, at ^ - a heat flux?*) then we obtain,-x)] pcos\\[pu], A = 1,2,... (14) The case of two heat flux IRs, ><#,Qgk, leads to a likewise formula. 4 Time step choice for the case of two IRs in flat slab It is obvious that the IRs as functions of time are not known precisely. If measured they depend on sensivity of the measuring device; if given they hardly ever are described as functions of time. Hence, they are always inaccurate. If an IR at x^ is given then the greatest inaccuracy of temperature calculation will appear at the boundary x=0, because between the point x^ and the boundary x=\ an IR at Xg is given. Denote the distance from x=0 to x~ as ifi and let t/^=<0, x >. Likewise, 6 <5 u^ is the distance fromx^to jc=l and U^=<xj,\>.

5 Advanced Computational Methods in Heat Transfer 97 Consider the case of two temperature IRs. With 6^(0) standing for an error of the calculation at x=0 and moment of time equal k&t we find / \ sinh p r / r>\ sinh p- sinh \p[] - u LJ LL g _ (is) where s ^_2 is an average error of the temperature in a slab obtained at ;i( the moment (k-2)kt. It is easy to prove that forp>5 the coefficient at s ^ is greater than the coefficient at s ^_2. Hence, in order to avoid an error propagation in time we require the absolute value of the coefficient at 8 j^_i to be smaller than 1. Finally we arrive to the following inequality: "77 (16) For given u^ the greatest value of p (i.e. the smallest value of the time step A/ denoted as Af %) for which the inequality (16) holds will be found. Likewise, A/ ^ is the smallest value of the time step for a given za Of course, in the case of two temperature IRs A/ ^ = A/ ^. Both, A/ ^ and A/ ^ are called critical time steps. For Af < Af ^ or Af < Af ^ calculations may occur unstable. For mixed IRs and for two heat flux IRs inequalities similar to (16) can be obtained. Then, A/ * and Af ^ can be found. Results concerning the cases of two temperature IRs and mixed IRs are shown on Fig. 1 and Fig. 2, where the nomographs for choosing the time step that ensures no IRerror propagation in time are presented. The hiperbolic lines denote length u of the inverse problem basis (distance from x^ to x ); relation between it, u^ and %/ reads For xj=q is «^=1; «= "n"r T (1?) for x =1 is u^=\. Hence, for each hiperbolic line description is given on the left or lower scale.

6 98 Advanced Computational Methods in Heat Transfer In the case of mixed IRs the time step to be chosen is the greater one (see Fig. 2). The minimal time step for the same length of u^ (z/) depends on the kind of an IR and it is worth to notice that the heat flux IR gives smaller value of the critical time step than the temperature IR. OOOOOOOOOOOOOOOOOOIT) S OOOOOOOOOOOOOOOOOOO 1 no - _[ Ii I i n an U.0 U n An R U OA.4U n n on u.zu n nn \ U Fig.l. Nomograph for two temperature IRs. Upper and right scales show the critical time step A/ ^ and A/ ^ for z/ and %^, respectively. 1.00

7 Advanced Computational Methods in Heat Transfer R U S 8 8 g 8 g OOOOOOOOOOOOOOOOOOCM - OOOOOOOOOOOOOOOOOOO I I I i! i I! I I 2 ' 4 3 i U Fig.2. Nomograph for two mixed IRs (T*^ and Q* scales show Af ^ and A/ ^ for i/ and «^, respectively Upper and right 5 The IHCP solution for a cylindrical and spherical layers Let Q be a spherical or cylindrical layer with an inner radius r and outer radius r+\ (i.e. layer with a unique thickness). Consider the case of two temperature IRs, at r^ and at r^ r<r^ < ^ < 1 +r. For the spherical layer the IHCP solution has a form similar to that one for a flat slab, therefore we do not quote it here. In the case of cylindrical layer we obtain

8 100 Advanced Computational Methods in Heat Transfer (18) where (19) 6/z (20) Formulae desribing solution for mixed IRs and for two heat flux IRs for both, spherical and cylindrical layer, are easy to obtain and we do not quote them. An analysis leading to nomographs likewise to those shown at Fig.l and 2 is not difficult for the case of spherical layer; for two temperature IRs results are likewise to those for a flat slab. In the case of cylindrical layer the modified Bessel functions of the first and second kind make an analysis to be rather complicated. 6 Error introduced by the model itself Consider the following direct (initial-boundary) problem for a flat slab of unique thickness: T(x,Qj=Q, T(Q,t) = \QQ, 7/1,0=0. Comparing the exact solution with its approximate value (from the Helmholtz model) we can notice inaccuracies introduced by the model (see Fig.3). The greatest error appears for the first time step and is generated by the difference between first derivative and the first back diffrence of temperature with respect to time (see Fig.4). For an IHCP the effect of the first time step for solution accuracy is even greater and depends on location of a point with IR. In order to diminish the inaccuracy introduced by the first time step the back difference for the first step is modified as follows: dt Q5M At Geometric interpretation of (20) is presented on Fig.4. A%M±%zA) (21)

9 Advanced Computational Methods in Heat Transfer 101 '00.00 Temp X 1.00 Fig.3. Exact (solid line) and approximate (dotted line) temperature value in a flat slab after dimensionless time equal to: a) 0.01; b) 0.03; c) The parameter 0 can not be too big. For the case shown in Fig.4 0=0.8. For arbitrary jt^e(0,l), x =1 and time step A/, value of 0 can be determined based on analysis of the function sinh[v20p] -100 (22) that describes the difference between exact solution of the direct problem and approximate solution of the IHCP in the case, when the IR is taken from the exact solution. One can prove that if x^ =0 then B(0)=0. Moreover, B(0) is increasing with 0 and for an arbitrary jt^e(0,l) has exactly one solution. However, taking into account physical sense, value of the parameter 0 should be rather close to 0.5 (0e(0,l)).

10 102 Advanced Computational Methods in Heat Transfer Fig.4. Geometric interpretation of the modified first back difference with respect to time (exact solution ybr 7(%,0)=0, 7(0,0=100, 7(l,f)=0 at x=0.13andaf=0.0014). 7 Conclusion The results presented in the paper are obtained in the case of the dimensionless variables. However, in an experiment the IR measurements take place for a real body in time measured in seconds. In order to avoid instable numerical calculation the dimensionless time step should be compared with the results presented in the paper. For instance when considering a steel flat slab of thickness 50 mm, the time step equal to 0,3 s denotes dimensionless time equal to At Fig.l one can check that in the case of two temperature IRs the dimensionless distance between points with the IRs (the base of the IHCP) should be at least about equal to Moreover, regarding the first time step, it is advisable to find the parameter 0 to minimize thefirsttime step error. Such a procedure seems to be a good way to diminish well known problems with stability of the IHCPs.

AP Physics C. Gauss s Law. Free Response Problems

AP 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 information

Chapter 24. Gauss s Law

Chapter 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 information

Transactions on Modelling and Simulation vol 8, 1994 WIT Press, ISSN X

Transactions on Modelling and Simulation vol 8, 1994 WIT Press,   ISSN X Boundary element method for an improperly posed problem in unsteady heat conduction D. Lesnic, L. Elliott & D.B. Ingham Department of Applied Mathematical Studies, University of Leeds, Leeds LS2 9JT, UK

More information

Chapter 24. Gauss s Law

Chapter 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 information

Chapter 4: Transient Heat Conduction. Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University

Chapter 4: Transient Heat Conduction. Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University Chapter 4: Transient Heat Conduction Dr Ali Jawarneh Department of Mechanical Engineering Hashemite University Objectives When you finish studying this chapter, you should be able to: Assess when the spatial

More information

Physics 9 WS E3 (rev. 1.0) Page 1

Physics 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 information

Please fill in your Student ID number (UIN): IMPORTANT. Read these directions carefully:

Please fill in your Student ID number (UIN): IMPORTANT. Read these directions carefully: Physics 208: Electricity and Magnetism. Common Exam 1, 26 September 2016 Printyournameneatly: Last name: First name: Sign your name: Please fill in your Student ID number (UIN): Your classroom instructor:

More information

Chapter 24. Gauss s Law

Chapter 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 information

Chapter 22. Dr. Armen Kocharian. Gauss s Law Lecture 4

Chapter 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 information

Chapter 23 Term083 Term082

Chapter 23 Term083 Term082 Chapter 23 Term083 Q6. Consider two large oppositely charged parallel metal plates, placed close to each other. The plates are square with sides L and carry charges Q and Q. The magnitude of the electric

More information

Chapter 15 Appendix Moment of Inertia of a Spherical Shell

Chapter 15 Appendix Moment of Inertia of a Spherical Shell Chapter 1 Appendix Moment of nertia of a Spherical Shell t is common to regard the rotation of a rigid object with spherical symmetry; we live on one such object (not entirely uniform and not entirely

More information

E. not enough information given to decide

E. 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 information

Counting Dots Kwok-Wai Ng Feb 1, 2007

Counting Dots Kwok-Wai Ng Feb 1, 2007 Counting Dots Kwok-Wai Ng Feb 1, 007 This sounds so easy (indeed it is not difficult), yet so simple that we never think about it carefully. When we are asked to do it, suddenly we do not know what to

More information

Chapter 22 Gauss s Law. Copyright 2009 Pearson Education, Inc.

Chapter 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 information

Fokker-Planck Equation with Detailed Balance

Fokker-Planck Equation with Detailed Balance Appendix E Fokker-Planck Equation with Detailed Balance A stochastic process is simply a function of two variables, one is the time, the other is a stochastic variable X, defined by specifying: a: the

More information

Chapter 22 Gauss s Law. Copyright 2009 Pearson Education, Inc.

Chapter 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 information

2 Which of the following represents the electric field due to an infinite charged sheet with a uniform charge distribution σ.

2 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 information

. (70.1) r r. / r. Substituting, we have the following equation for f:

. (70.1) r r. / r. Substituting, we have the following equation for f: 7 Spherical waves Let us consider a sound wave in which the distribution of densit velocit etc, depends only on the distance from some point, ie, is spherically symmetrical Such a wave is called a spherical

More information

1. Short Answer (25 points total)

1. Short Answer (25 points total) Physics 116b First Practice Examination Due September 19, 2001 Name: Please circle your section: Section 1 Section 2 Section 3 Section 4 I nstructions This is practice for a one hour, closed book examination.

More information

Gauss s Law. Name. I. The Law: , where ɛ 0 = C 2 (N?m 2

Gauss 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 information

Chapter 4. Electrostatic Fields in Matter

Chapter 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 information

Physics 208, Spring 2015 Exam #1

Physics 208, Spring 2015 Exam #1 Physics 208, Spring 2015 Exam #1 A Name (Last, First): ID #: Section #: You have 75 minutes to complete the exam. Formulae are provided on a separate colored sheet. You may NOT use any other formula sheet.

More information

Session 5 Heat Conduction in Cylindrical and Spherical Coordinates I

Session 5 Heat Conduction in Cylindrical and Spherical Coordinates I Session 5 Heat Conduction in Cylindrical and Spherical Coordinates I 1 Introduction The method of separation of variables is also useful in the determination of solutions to heat conduction problems in

More information

Chapter 22 Gauss s Law

Chapter 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 information

Volumes of Solids of Revolution. We revolve this curve about the x-axis and create a solid of revolution.

Volumes of Solids of Revolution. We revolve this curve about the x-axis and create a solid of revolution. Volumes of Solids of Revolution Consider the function ( ) from a = to b = 9. 5 6 7 8 9 We revolve this curve about the x-axis and create a solid of revolution. - 5 6 7 8 9 - - - We want to find the volume

More information

Exam 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 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 information

Fall 2004 Physics 3 Tu-Th Section

Fall 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 information

Chapter 2 Gauss Law 1

Chapter 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

Chapter (2) Gauss s Law

Chapter (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 information

MURI teleconference 28 May Optical Antimatter. John Pendry and Sebastien Guenneau Imperial College London. 24 May 2004 page 1

MURI teleconference 28 May Optical Antimatter. John Pendry and Sebastien Guenneau Imperial College London. 24 May 2004 page 1 24 May 2004 page 1 MURI teleconference 28 May 2004 Optical Antimatter John Pendry and Sebastien Guenneau Imperial College London 05 March 2004 page 2 A Conventional Lens Contributions of the far field

More information

Sample Question: A point in empty space is near 3 charges as shown. The distances from the point to each charge are identical.

Sample Question: A point in empty space is near 3 charges as shown. The distances from the point to each charge are identical. A point in empty space is near 3 charges as shown. The distances from the point to each charge are identical. A. Draw a vector showing the direction the electric field points. y +2Q x B. What is the angle

More information

Gauss s Law. 3.1 Quiz. Conference 3. Physics 102 Conference 3. Physics 102 General Physics II. Monday, February 10th, Problem 3.

Gauss 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 information

free space (vacuum) permittivity [ F/m]

free 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 information

2. Equations of Stellar Structure

2. Equations of Stellar Structure 2. Equations of Stellar Structure We already discussed that the structure of stars is basically governed by three simple laws, namely hyostatic equilibrium, energy transport and energy generation. In this

More information

Department of Physics Preliminary Exam January 2 5, 2013

Department of Physics Preliminary Exam January 2 5, 2013 Department of Physics Preliminary Exam January 2 5, 2013 Day 2: Electricity, Magnetism and Optics Thursday, January 3, 2013 9:00 a.m. 12:00 p.m. Instructions: 1. Write the answer to each question on a

More information

Questions Chapter 23 Gauss' Law

Questions 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 information

Heat and Mass Transfer Prof. S.P. Sukhatme Department of Mechanical Engineering Indian Institute of Technology, Bombay

Heat and Mass Transfer Prof. S.P. Sukhatme Department of Mechanical Engineering Indian Institute of Technology, Bombay Heat and Mass Transfer Prof. S.P. Sukhatme Department of Mechanical Engineering Indian Institute of Technology, Bombay Lecture No. 18 Forced Convection-1 Welcome. We now begin our study of forced convection

More information

Phy207 Exam I (Form1) Professor Zuo Fall Semester Signature: Name:

Phy207 Exam I (Form1) Professor Zuo Fall Semester Signature: Name: Phy207 Exam I (Form1) Professor Zuo Fall Semester 2015 On my honor, I have neither received nor given aid on this examination Signature: Name: ID number: Enter your name and Form 1 (FM1) in the scantron

More information

University of Rome Tor Vergata

University of Rome Tor Vergata University of Rome Tor Vergata Faculty of Engineering Department of Industrial Engineering THERMODYNAMIC AND HEAT TRANSFER HEAT TRANSFER dr. G. Bovesecchi gianluigi.bovesecchi@gmail.com 06-7259-727 (7249)

More information

Notes on multivariable calculus

Notes on multivariable calculus Notes on multivariable calculus Jonathan Wise February 2, 2010 1 Review of trigonometry Trigonometry is essentially the study of the relationship between polar coordinates and Cartesian coordinates in

More information

Angular Momentum. 1. Object. 2. Apparatus. 3. Theory

Angular Momentum. 1. Object. 2. Apparatus. 3. Theory ngular Momentum. Object To verify conservation of angular momentum, determine the moment of inertia for various objects and look at the exchange of angular momentum in different situations.. pparatus rotational

More information

Homework 4 PHYS 212 Dr. Amir

Homework 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 information

University Physics (Prof. David Flory) Chapt_24 Sunday, February 03, 2008 Page 1

University 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 information

PHY102 Electricity Topic 3 (Lectures 4 & 5) Gauss s Law

PHY102 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 information

Gauss s Law. Chapter 22. Electric Flux Gauss s Law: Definition. Applications of Gauss s Law

Gauss 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 information

CHAPTER 3 CYLINDRICAL WAVE PROPAGATION

CHAPTER 3 CYLINDRICAL WAVE PROPAGATION 77 CHAPTER 3 CYLINDRICAL WAVE PROPAGATION 3.1 INTRODUCTION The phase and amplitude of light propagating from cylindrical surface varies in space (with time) in an entirely different fashion compared to

More information

Chapter 10: Steady Heat Conduction

Chapter 10: Steady Heat Conduction Chapter 0: Steady Heat Conduction In thermodynamics, we considered the amount of heat transfer as a system undergoes a process from one equilibrium state to another hermodynamics gives no indication of

More information

Chapter 24. Gauss s Law

Chapter 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 information

Chapter 21: Gauss s Law

Chapter 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 information

Electricity & Magnetism Qualifier

Electricity & Magnetism Qualifier Electricity & Magnetism Qualifier For each problem state what system of units you are using. 1. Imagine that a spherical balloon is being filled with a charged gas in such a way that the rate of charge

More information

CHAPTER 4 ANALYSIS AND DESIGN OF THE DUAL INVERTED-F ANTENNA

CHAPTER 4 ANALYSIS AND DESIGN OF THE DUAL INVERTED-F ANTENNA CHAPTER 4 ANALYSIS AND DESIGN OF THE DUAL INVERTED-F ANTENNA 4.1. Introduction The previous chapter presented the Inverted-F Antenna (IFA) and its variations as antenna designs suitable for use in hand-held

More information

Gauss Law. Challenge Problems

Gauss Law. Challenge Problems Gauss Law Challenge Problems Problem 1: The grass seeds figure below shows the electric field of three charges with charges +1, +1, and -1, The Gaussian surface in the figure is a sphere containing two

More information

Flux. Flux = = va. This is the same as asking What is the flux of water through the rectangle? The answer depends on:

Flux. 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 information

Essential University Physics

Essential 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 information

1 Introduction. Green s function notes 2018

1 Introduction. Green s function notes 2018 Green s function notes 8 Introduction Back in the "formal" notes, we derived the potential in terms of the Green s function. Dirichlet problem: Equation (7) in "formal" notes is Φ () Z ( ) ( ) 3 Z Φ (

More information

Chapter 23. Gauss Law. Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

Chapter 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 information

1. Electricity and Magnetism (Fall 1995, Part 1) A metal sphere has a radius R and a charge Q.

1. Electricity and Magnetism (Fall 1995, Part 1) A metal sphere has a radius R and a charge Q. 1. Electricity and Magnetism (Fall 1995, Part 1) A metal sphere has a radius R and a charge Q. (a) Compute the electric part of the Maxwell stress tensor T ij (r) = 1 {E i E j 12 } 4π E2 δ ij both inside

More information

Electric flux. Electric Fields and Gauss s Law. Electric flux. Flux through an arbitrary surface

Electric flux. Electric Fields and Gauss s Law. Electric flux. Flux through an arbitrary surface Electric flux Electric Fields and Gauss s Law Electric flux is a measure of the number of field lines passing through a surface. The flux is the product of the magnitude of the electric field and the surface

More information

Lecture 3. Electric Field Flux, Gauss Law

Lecture 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 information

PHYSICS - CLUTCH 1E CH 12: TORQUE & ROTATIONAL DYNAMICS.

PHYSICS - CLUTCH 1E CH 12: TORQUE & ROTATIONAL DYNAMICS. !! www.clutchprep.com INTRO TO TORQUE TORQUE is a twist that a Force gives an object around an axis of rotation. - For example, when you push on a door, it rotates around its hinges. - When a Force acts

More information

PHYSICS - CLUTCH CH 12: TORQUE & ROTATIONAL DYNAMICS.

PHYSICS - CLUTCH CH 12: TORQUE & ROTATIONAL DYNAMICS. !! www.clutchprep.com TORQUE & ACCELERATION (ROTATIONAL DYNAMICS) When a Force causes rotation, it produces a Torque. Think of TORQUE as the equivalent of FORCE! FORCE (F) TORQUE (τ) - Causes linear acceleration

More information

TRANSIENT HEAT CONDUCTION

TRANSIENT HEAT CONDUCTION TRANSIENT HEAT CONDUCTION Many heat conduction problems encountered in engineering applications involve time as in independent variable. This is transient or Unsteady State Heat Conduction. The goal of

More information

Experiment III Electric Flux

Experiment 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 information

Chapter 21 Chapter 23 Gauss Law. Copyright 2014 John Wiley & Sons, Inc. All rights reserved.

Chapter 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 information

TALLINN UNIVERSITY OF TECHNOLOGY, DIVISION OF PHYSICS 13. STOKES METHOD

TALLINN UNIVERSITY OF TECHNOLOGY, DIVISION OF PHYSICS 13. STOKES METHOD 13. STOKES METHOD 1. Objective To determine the coefficient of viscosity of a known fluid using Stokes method.. Equipment needed A glass vessel with glycerine, micrometer calliper, stopwatch, ruler. 3.

More information

The temperature of a body, in general, varies with time as well

The temperature of a body, in general, varies with time as well cen58933_ch04.qd 9/10/2002 9:12 AM Page 209 TRANSIENT HEAT CONDUCTION CHAPTER 4 The temperature of a body, in general, varies with time as well as position. In rectangular coordinates, this variation is

More information

CHAPTER 8 CONSERVATION LAWS

CHAPTER 8 CONSERVATION LAWS CHAPTER 8 CONSERVATION LAWS Outlines 1. Charge and Energy 2. The Poynting s Theorem 3. Momentum 4. Angular Momentum 2 Conservation of charge and energy The net amount of charges in a volume V is given

More information

A A + B. ra + A + 1. We now want to solve the Einstein equations in the following cases:

A A + B. ra + A + 1. We now want to solve the Einstein equations in the following cases: Lecture 29: Cosmology Cosmology Reading: Weinberg, Ch A metric tensor appropriate to infalling matter In general (see, eg, Weinberg, Ch ) we may write a spherically symmetric, time-dependent metric in

More information

Lesson 9: Multiplying Media (Reactors)

Lesson 9: Multiplying Media (Reactors) Lesson 9: Multiplying Media (Reactors) Laboratory for Reactor Physics and Systems Behaviour Multiplication Factors Reactor Equation for a Bare, Homogeneous Reactor Geometrical, Material Buckling Spherical,

More information

PHYS102 EXAM #1 February 17, MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

PHYS102 EXAM #1 February 17, MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. PHYS02 EXAM # February 7, 2005 Last Name First Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. ) A spherical metallic shell carries a charge

More information

Chapter 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. 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 information

Steady and unsteady diffusion

Steady and unsteady diffusion Chapter 5 Steady and unsteady diffusion In this chapter, we solve the diffusion and forced convection equations, in which it is necessary to evaluate the temperature or concentration fields when the velocity

More information

1. Mathematical Tools

1. Mathematical Tools 1. Mathematical Tools 1.1 Coordinate Systems Suppose u 1, u 2, and u 3 are the coordinates of a general coordinate coordinate system in which the (ê 1, ê 2, ê 3 ) unit or basis vectors specify the directions

More information

Numerical Solution of Integral Equations in Solidification and Melting with Spherical Symmetry

Numerical Solution of Integral Equations in Solidification and Melting with Spherical Symmetry Numerical Solution of Integral Equations in Solidification and Melting with Spherical Symmetry V. S. Ajaev and J. Tausch 2 Southern Methodist University ajaev@smu.edu 2 Southern Methodist University tausch@smu.edu

More information

Massachusetts Institute of Technology Department of Materials Science and Engineering

Massachusetts Institute of Technology Department of Materials Science and Engineering Massachusetts Institute of Technology Department of Materials Science and Engineering 3.05 Thermodynamics and Kinetics of Materials Fall 003 November 7, 003 We are looking at the incorporation of Al 3

More information

Physics 210: Worksheet 26 Name:

Physics 210: Worksheet 26 Name: (1) A all is floating in. If the density of the all is 0.95x10 kg/m, what percentage of the all is aove the? (2) An with a density of 19.x10 kg/m and a mass m50 kg is placed into a cylinder which contains

More information

Physics 1302W.500 Lecture 9 Introductory Physics for Scientists and Engineering II

Physics 1302W.500 Lecture 9 Introductory Physics for Scientists and Engineering II Physics 1302W.500 Lecture 9 Introductory Physics for Scientists and Engineering II In today s lecture, we will finish our discussion of Gauss law. Slide 25-1 Applying Gauss s law Procedure: Calculating

More information

( ) 5. Bearing internal load distribution and displacement. 5.1 Bearing internal load distribution

( ) 5. Bearing internal load distribution and displacement. 5.1 Bearing internal load distribution 5. internal load distribution and displacement 5. internal load distribution This section will begin by examing the effect of a radial load F r and an axial load F a applied on a single-row bearing with

More information

Phys102 Second Major-181 Zero Version Coordinator: Kunwar, S Monday, November 19, 2018 Page: 1

Phys102 Second Major-181 Zero Version Coordinator: Kunwar, S Monday, November 19, 2018 Page: 1 Coordinator: Kunwar, S Monday, November 19, 2018 Page: 1 Q1. A neutral metal ball is suspended by a vertical string. When a positively charged insulating rod is placed near the ball (without touching),

More information

APPLICATIONS OF GAUSS S LAW

APPLICATIONS OF GAUSS S LAW APPLICATIONS OF GAUSS S LAW Although Gauss s Law is always correct it is generally only useful in cases with strong symmetries. The basic problem is that it gives the integral of E rather than E itself.

More information

Columbia University Department of Physics QUALIFYING EXAMINATION

Columbia University Department of Physics QUALIFYING EXAMINATION Columbia University Department of Physics QUALIFYING EXAMINATION Wednesday, January 15, 2014 3:10PM to 5:10PM Modern Physics Section 4. Relativity and Applied Quantum Mechanics Two hours are permitted

More information

Electric flux. You must be able to calculate the electric flux through a surface.

Electric flux. You must be able to calculate the electric flux through a surface. Today s agenda: Announcements. lectric field lines. You must be able to draw electric field lines, and interpret diagrams that show electric field lines. A dipole in an external electric field. You must

More information

Problem Solving 3: Calculating the Electric Field of Highly Symmetric Distributions of Charge Using Gauss s Law

Problem Solving 3: Calculating the Electric Field of Highly Symmetric Distributions of Charge Using Gauss s Law MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Problem Solving 3: Calculating the Electric Field of Highly Symmetric Distributions of Charge Using Gauss s Law REFERENCE: Section 4.2, 8.02

More information

Separation of Variables in Linear PDE: One-Dimensional Problems

Separation of Variables in Linear PDE: One-Dimensional Problems Separation of Variables in Linear PDE: One-Dimensional Problems Now we apply the theory of Hilbert spaces to linear differential equations with partial derivatives (PDE). We start with a particular example,

More information

Chapter 5. Effects of Photonic Crystal Band Gap on Rotation and Deformation of Hollow Te Rods in Triangular Lattice

Chapter 5. Effects of Photonic Crystal Band Gap on Rotation and Deformation of Hollow Te Rods in Triangular Lattice Chapter 5 Effects of Photonic Crystal Band Gap on Rotation and Deformation of Hollow Te Rods in Triangular Lattice In chapter 3 and 4, we have demonstrated that the deformed rods, rotational rods and perturbation

More information

3 The Friedmann-Robertson-Walker metric

3 The Friedmann-Robertson-Walker metric 3 The Friedmann-Robertson-Walker metric 3.1 Three dimensions The most general isotropic and homogeneous metric in three dimensions is similar to the two dimensional result of eq. (43): ( ) dr ds 2 = a

More information

Version: A. Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p = kg

Version: A. Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p = kg PHYS 2212 G & J Quiz and Exam Formulæ & Constants Fall 2017 Fundamental Charge e = 1.602 10 19 C Mass of an Electron m e = 9.109 10 31 kg Earth s gravitational field g = 9.81 N/kg Mass of a Proton m p

More information

Methods in Computer Vision: Introduction to Matrix Lie Groups

Methods in Computer Vision: Introduction to Matrix Lie Groups Methods in Computer Vision: Introduction to Matrix Lie Groups Oren Freifeld Computer Science, Ben-Gurion University June 14, 2017 June 14, 2017 1 / 46 Definition and Basic Properties Definition (Matrix

More information

Consider a point P on the line joining the two charges, as shown in the given figure.

Consider a point P on the line joining the two charges, as shown in the given figure. Question 2.1: Two charges 5 10 8 C and 3 10 8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

More information

A cylinder in a magnetic field (Jackson)

A cylinder in a magnetic field (Jackson) Problem 1. A cylinder in a magnetic field (Jackson) A very long hollow cylinder of inner radius a and outer radius b of permeability µ is placed in an initially uniform magnetic field B o at right angles

More information

PHYSICS GRADUATE SCHOOL QUALIFYING EXAMINATION

PHYSICS GRADUATE SCHOOL QUALIFYING EXAMINATION PHYSICS GRADUATE SCHOOL QUALIFYING EXAMINATION Partl INSTRUCTIONS: Work all problems. This is a closed book examination. Start each problem on a new pack of yellow paper and use only one side of each sheet.

More information

Chapter 8. Rotational Equilibrium and Rotational Dynamics. 1. Torque. 2. Torque and Equilibrium. 3. Center of Mass and Center of Gravity

Chapter 8. Rotational Equilibrium and Rotational Dynamics. 1. Torque. 2. Torque and Equilibrium. 3. Center of Mass and Center of Gravity Chapter 8 Rotational Equilibrium and Rotational Dynamics 1. Torque 2. Torque and Equilibrium 3. Center of Mass and Center of Gravity 4. Torque and angular acceleration 5. Rotational Kinetic energy 6. Angular

More information

Moment of inertia. Contents. 1 Introduction and simple cases. January 15, Introduction. 1.2 Examples

Moment of inertia. Contents. 1 Introduction and simple cases. January 15, Introduction. 1.2 Examples Moment of inertia January 15, 016 A systematic account is given of the concept and the properties of the moment of inertia. Contents 1 Introduction and simple cases 1 1.1 Introduction.............. 1 1.

More information

Gauss s Law. Lecture 3. Chapter Course website:

Gauss s Law. Lecture 3. Chapter Course website: Lecture 3 Chapter 24 Gauss s Law 95.144 Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsii Today we are going to discuss: Chapter 24: Section 24.2 Idea of Flux Section 24.3 Electric

More information

Sink particle accretion test

Sink particle accretion test Sink particle accretion test David A. Hubber & Stefanie Walch 1 Objectives Simulate spherically-symmetric Bondi accretion onto a sink particle for an isothermal gas. Calculate the accretion rate onto a

More information

week 3 chapter 28 - Gauss s Law

week 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 information

Physics (

Physics ( Question 2.12: A charge of 8 mc is located at the origin. Calculate the work done in taking a small charge of 2 10 9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

More information

PHY103A: Lecture # 4

PHY103A: Lecture # 4 Semester II, 2017-18 Department of Physics, IIT Kanpur PHY103A: Lecture # 4 (Text Book: Intro to Electrodynamics by Griffiths, 3 rd Ed.) Anand Kumar Jha 10-Jan-2018 Notes The Solutions to HW # 1 have been

More information

Electric Flux and Gauss Law

Electric 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 information

Electricity. Revision Notes. R.D.Pilkington

Electricity. Revision Notes. R.D.Pilkington Electricity Revision Notes R.D.Pilkington DIRECT CURRENTS Introduction Current: Rate of charge flow, I = dq/dt Units: amps Potential and potential difference: work done to move unit +ve charge from point

More information