, the action per unit length. We use g = 1 and will use the function. gψd 2 x = A 36. Ψ 2 d 2 x = A2 45

Size: px
Start display at page:

Download ", the action per unit length. We use g = 1 and will use the function. gψd 2 x = A 36. Ψ 2 d 2 x = A2 45"

Transcription

1 Gbriel Brello - Clssicl Electrodynmics.. For this problem, we compute A L z, the ction per unit length. We use g = nd will use the function Ψx, y = Ax x y y s the form of our pproximte solution. First we compute Ψ = A x y y x y y, y x x y x x, Ψ = A y y x + x x y And note tht gψ = Ψ. The integrls compute to be v v gψd x = A 36 Ψ d x = A 45 Then the minimum of the difference of these is given when A = 5 4. Indeed, the second derivtive is positive t this vlue of A, so it is indeed minimum if tht wsn t obvious from the form. b. Figure : This is comprison of the true potentil in blue nd the pproximte potentil dshed The left plot shows y =.5 nd the right shows y =.5. The verticl xis is in units of ev 4πɛ

2 .4 Since this setup is π/ rdins rottionlly symmetric ech of the corner points nd ech of the edge points re equivlent, so we relly only need to compute one corner, one edge nd the middle. Let us lbel these points {, b, e} respectively. Recll the best-verge formul for the poisson eqution Φ p = Φ + h 5 g + h g c where Φ = 4 5 Φ c + 4 Φ s And the c nd s stnd for cross nd squre verge. the cross verge is the one tht mkes plus sign on the grid. Note, too tht h = /6. Now we re redy to get things lined up. Note the following tble: ij i+,j i,j+ i-,j i,j- i+,j+ i-,j- i+,j- i-,j+ b b e b e b b e b b b b After some fussing, the itertive equtions we get re: i+ = e i 6 + 4b i b i+ = i 5 + e i 5 + b i e i+ = 4b i Using the jcobin method with ɛ Φ = on ll sites initilly gives Note tht my nswers differ by fctor of 4π from the book s exct solution, becuse I computed Φ ɛ wheres Jckson gve 4πɛ Φ.

3 b. After ten itertions of the guss-seidel method using the sme initil vlues we get c. Itertion b e We fctored out /ɛ t the begining, so we re ctully reporting the vlue ɛ Φ. The following plot compres our pproximte solution to the exct one Figure : This is log plot of the proximte solution using the jcobin method blue nd the guss seidel method red compred to the exct solution lines. Unfortuntely the lowest exct vlue is obscured by the log plot, but it is there, I promise. The verticl xis is in units of Φ ɛ. As you cn see the guss-seidel method converges fster nd is more ccurte fter itertions. 3

4 .. The proprite imge chrge is q t position d,. Then the electric field t point, y is E, y = q dˆx + yŷ dˆx yŷ 4πɛ d + y 3 = q d ˆx πɛ d + y 3 This is relted to the surfce chrge by the eqution σ = ɛ E. So, Here is plot σ = qd d + y 3 π Figure 3: This shows the nondimentionlized chrge density with d = b. The force on the chrge is the sme s the force tht would result from the presence of the imge chrge. c. F = q 4πɛ d = q 6πɛ d The totl force cting on the plne cn be computed by computing the integrl F p = qd 8π ɛ = qd 8πɛ π d = 6πɛ q d u 3 du d + r 3 rdrdθ u d + r 4

5 d. The work is given by the integrl e. The potentil energy is given by the integrl W = F dl = q 6πɛ = q 6πɛ d d x dx d PE = F dl = q 4πɛ = q 8πɛ d d x dx This is fctor of two lrger, of course this is becuse s the chrge moves wy from the plne the chrge on the plne disperses, s if the imge chrge were moving wy t the sme rte. We usully define potentil energy by holding one chrge fixed while moving the other chrge in from infinity. In this cse, it is s if we re moving both chrges in from infinity t the sme time, resulting in fctor of two difference. One could lso think of it in terms of the fields. Potentil energy is stored in the field, but in the current sitution, only hlf the field ctully exists the other hlf is cncelled out by the conductor so there is only hlf s much energy stored in the field s in the corresponding sitution with rel chrge insted of conducting plne. f. q = e =.6 9 C d = A = m ɛ = F/m W = ev..9 The rection of the sphere rdius to uniform electric field E = Eẑ cn be modeled using imge chrges of chrge ± Q R t positions ± R with Q πɛ R = E nd tking the limit R. In the limit, this becomes n electric dipole, with dipole moment p = Q3 R ẑ = 4πɛ 3 Eẑ. Which gives surfce chrge density σ = 3ɛ E cosθ Unchrged σ q = σ + q 4π Chrged with chrge q Where θ is mesured from the positive z xis. Finlly, recll tht the force per unit re on differentil element is: F /A = σ ɛ 3 To find the force on ech hemisphere we must only integrte the bove formul over ech hemisphere. Note tht the force on the hemisperes will be equl nd opposite, since the chrge densities re equl nd opposite. Furthermore, note tht the component of force on surfce element not in the ẑ direction will cncel due to the zimuthl symmetry so we cn integrte insted only the z components of the force. Applying these observtions, the force required to keep the hemispheres from seprting is given by F = π π σ cosθ sinθdθ 4 ɛ 5

6 With 4 nd, we cn clculte the force for the two scenrios given:. F = 9πE ɛ π = 9 4 πe ɛ cosθ 3 sinθdθ b. For chrged sphere we get F q = π π = F + π π = F + q 4ɛ = F + q 3ɛ E cosθ + ɛ π q 4π 6ɛ E cosθ ɛ 6ɛ E cosθ + q 4ɛ 8π ɛ E cosθ sinθdθ q 4π + q 6π 4 cosθ sinθdθ q cosθ sinθdθ 4π And the totl force is given by F q = F + q q 4ɛ 8π ɛ E 5.7. The free spce greens function for three dimentions prt from n rbitrry multiplictive constnt is Gx, y, z; x, y, z = x x = x x + y y + z z By defining = x x + y x we cn write the -dimentionl free-spce greens function s Gx, y; x, y = + x dx x = z z Which hs the well-known solution prt from n inessentil, though infinite constnt. This constnt is hnd-wved wy by insted integrting to lrge vlue Z insted of to infinity, but I think this will be good enough. b. Where the lst line follows from the lw of cosines. Gx, y; x, y = logx x + y y = log + cosφ φ The generl solution to lplce s eqution is given by free-spce solutions for > nd < which re then ptched together t =. To this end, we define two new functions, R nd Φ, by the ssumption G = R, Φφ, φ. Let us compute the fourier coefficients of the Φ function for the lplce eqution sourced by delt functions. R, π e imφ Φφ, φ dφ R, π m e imφ Φφ, φ dφ = 4π π δ e imφ 6

7 The surfce terms in the second term of the l.h.s. cnceled becuse of the cyclic boundry conditions on Φ. We cn divide by e imφ nd pull out n overll fourier integrl to rrive t π R, m R, e imφ φ Φφ, φ dφ = 4πδ The fourier-trnsform fctor is constnt w.r.t. nd likewise the rest is constnt w.r.t. φ so w.l.o.g. we cn set the fourier trnsform fctor equl to constnt, we choose, so tht π Thus we rrive t the generl solution in the form e imφ φ Φφ, φ dφ = π Φφ, φ = R, m R, = 4π δ π eimφ φ where R m stisfies the bove eqution. G = π e imφ φ R m, c. To complete the solution we solve the bove eqution in the homogenious cse nd then ptch together the solution t =. The generl solution of the homogenious eqution is given in Jckson s R, = m m m Z\{} R, = + b log m = We need two solutions one for > nd one for <, we will distinguish them with subscript > nd < respectively on the coefficients. Then, the two full solutions re give by R <, = < + b < log + R >, = > + b > log + m Z\{} m Z\{} m< m m> m There re few conditions we cn impose right off the bt. Nmely, the solution needs to be well-behved t zero, nd be logrithmic s. With this we cn determine tht b < =, nd tht R < hs no negtive powers of nd R > hs no positive powers of. This leves us with R <, = < + m< m m> R >, = > + b > log + m> m m< Next, the solution must be symmetric. Tht is to sy tht if < then R <, = R >, nd likewise, for >, R >, = R <,. These conditions stte tht < + m< m = > + b > log + m> m m> m< > + b > log + m> m = < + m< m m< m> 7

8 Note tht the solutions for vrious m must independently obey this symmetry, since they re ll linerly independent. Then this tells us tht < = A log, > = nd b > = A, with A yet to be determined constnt. In ddition it must be tht m> = A m m nd m< = A m m with A m constnts tht will be determined by considering the derivtive discontinuity. So tht we finlly hve the solutions, us to n overll constnt in ech term R <, = A log + m> R >, = A log + m> A m m A m m The lst considertion to mke is tht the difference in the derivtives t = should be 4π/. evluted with respect to ner = re The derivtive R <,, = m> A m m R >,, = A m> A m m Ech term stisfies the discontinuity condition individully so we get tht A = 4π nd tht A m = π m. So tht the two solutions cn be written in the form Which cn be conveniently expressed s R <, = π log + 4π m m> R >, = π log + 4π m m> R <, = π log > + 4π m> m m m m < > With < > the lesser greter of the two rdii. Lstly we cn plug this bck into our originl solution. Note tht the full solution for ech m of the rdil eqution hs positive nd negtive power of m. We threw out the poorly-behved powers of m, but for the well behved ones we get contributions, one from the positive m in the series, nd one from the negtive m in the bove series. Thus this mmounts to dding the positive-m term to the negtive m term, cusing the imginry prts of the exponentil to cncel s needs to be the cse to itnerprit this s potentil, leving fctor of cos, so in the end we get G = log > + m < > m cosmφ φ.. Using the greens function of problem.7 the potentil is given s σ, φ Φ, φ = G, φ;, φ ddφ 4πɛ = λ 3 n log 4πɛ > + n= m= m < > m cosmφ nπ/ 8

9 Where > < is the greter lesser of nd, but since the log term hs no φ dependence it cncels when the sum over n is performed, thus we cn write the solution s Φ, φ = λ πɛ m= m < > m 3 n cosmφ nπ/ n= for moment lets focus on the sum of cosines. Writing this out in full nd mnipulting we find tht cosmφ cosmφ mπ/ + cosmφ mπ/ cosmφ 3mπ/ = cosmφ cosmφ mπ/ + m So tht we cn write the solution s = { cosmφ if m = 4k +, k Z else Φ, φ = λ πɛ k= k + < > 4k+ cos4k + φ b. By mnipulting our solution for Φ we quickly rrive t Φ, φ = λ 4k+ > Re e iφ πɛ k + < Note the identity pge 75 of jckson Where for our cse we hve Z = >e iφ < Φ = nodd so tht k= Z n n = + Z log Z λ + > Re log < e iφ πɛ > < e iφ = λ Re log πɛ i > < e iφ + i > < e iφ > < e iφ + > < e iφ In the interior we hve tht < = nd > = nd the solution tkes the form Φ = λ z iz + i Re log = Rewz πɛ z z + In the exterior < = nd > = nd we hve Φ = λ iz + iz Re log πɛ z + z But we cn multiply the interior by i = without chnging the rel prt since the rel prt of log depends only on the mgnitude of z. 9

10 Φ = λ i + zi z Re log πɛ z + z = λ z ii + z Re log πɛ z z + = Rewz Thus, we cn identify Φ = Rewz So tht the potentil is exctly the rel prt of the funtion wz defined in Jckson. Now, s is stted in problem.3 of Jckson, the potentil of n infinite line chrge is the logrithm of the distnce. This solution reflects tht beutifully by converting the crtesin components of position into complex vrible nd then tking the rel prt. The four fctors in the rgument of teh logrithm correspond exctly to two positive line chrges t ±i, or ±ŷ, nd two negtive line chrges t ±, or ±ˆx, where x nd y hve been identified with the rel nd imginry xes respectively. c. The crtesin components of the field re given by Note tht for < E x = cosθ Φ sin θ Φ φ E y = sinθ Φ + cos θ Φ φ So the full expressions re E x = λ πɛ k= = λ πɛ k= E y = λ πɛ k= = λ πɛ k= Φ = λ πɛ Φ = λ πɛ k= 4k+ cos4k + φ k= 4k+ sin4k + φ 4k+ cos4k + φ cosθ + sinθ sin4k + φ 4k+ cos4k + φ 4k+ cos4k + φ sinθ cosθ sin4k + φ 4k+ sin4k + 3φ Keeping only the first two terms, we get E x = λ πɛ E y = λ πɛ cosφ + sin3φ + 5 cos5φ 5 sin7φ

11 When y =, φ = or π nd the x component of the electric field is given by E x = ± λ πɛ ± 5 with the ± corresponding to the negtive or positive xis respectively. The reltive strength of the k = contribution nd k = contribution is k = k = = ± 4.6 Consider the -dimentionl region, φ bounded by conducting surfces held t zero potentil.the lrge vlue of the potentil Φ is determined by some unknown chrge distribution t >> nd thus this my be considered s lplce problem with dirichlet boundry conditions.. The generl solution to lplce s eqution in polr coordines is given, fter seprtion of vribles, by Rp = ν + b ν, Ψφ = A cosνφ + B sinνφ ν > Rp = + b log, Ψφ = A + B φ ν = Given the boundry conditions we cn conclude tht A = B = A = nd tht ν = n π. So, only the sin term of the ν > solution will pper. Lstly we hve the condition tht Φ, φ = which gives us the condition b. Which gives us the generl form of the solution Φ, φ = n Z Keeping only the lowest nonvnishing term we get ν + b ν = b = ν C n nπ Φ, φ = C π nπ π nπ sin nπ φ π sin π φ By which we cn clculte the electric fields t the surfces vi the eqution Eˆn = Φ ˆn to be E, φ = C π π π + E φ, φ = Φ ˆn = Φ φ = C π π π + sin π φ π π cos π φ

12 With E zero t φ =, nd E φ = t =. Then, the surfce chrge density is given by σ = ɛ E, or c. πɛ σ = = C π sin π φ πɛ σ φ= = C π π Consider = π. The liner chrge density looks like this π = σ φ= Figure 4: This plot shows the liner chrge density. Note tht the chrge density on the cylinder goes s sinφ but since it is hlf-circle, it works out tht the liner chrge density is constnt! Of course, this is plot of the non-dimentionlized function with = Now lets convert the polr vectors into crtesin vectors! I will use ˆx prllel to the plne nd ŷ perpendiculr to the plne. Then the electric field in x nd y coordintes, for lrge s is required to be fr wy from the cylinder, is given by Ex E y = E cosφ E φ sinφ E sinφ + E φ cosφ sinφ cosφ cosφ sinφ = C cosφ + sinφ = E So, indeed the field is constnt nd perpendiculr to the plne fr from the cylinder. Now, we hve the chrge density on the cylinder, we just need to integrte over φ. This works out to be φ σ = = E ɛ sinφ = 4E ɛ. Now, the surfce chrge density on n infinite conducting plne is σ = E ɛ so indeed the surfce chrge on the cylinder is twice tht contined on strip on width on n infinite conducting plne. Of course, ll the surfce chrge densities re ctully chrge density per unit length in the z direction. Next we need to show tht the extr chrge density is tken from the surrounding re. To show this we will tke the difference between the surfce chrge density of this system nd tht of n infinite conducting plne, nd integrte this from = to, nd multiply by, of course. This works out to be:

13 σ φ σ d = E ɛ = E ɛ d Exctly wht we needed! Now, just for completeness ske, note tht the difference between these two surfce chrges goes to zero for lrge so indeed the sttment tht the chrge is tken from nerby points on the plne is ccurte. This lso nswers the lst question regrding the surfce chrge on wide strip being the sme wether or not the bump is present. Since the difference in surfce chrge densities between the two cses goes to zero s becomes lrge, the totl surfce chrge within tht lrge strip will be the sme between the two cses, since the totl chrge density on the two surfces is the sme, nd the surfce chrge density t lrge is the sme. 3

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The two-dimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero

More information

Jackson 2.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jackson 2.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell Jckson.7 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: Consider potentil problem in the hlf-spce defined by, with Dirichlet boundry conditions on the plne

More information

Summary: Method of Separation of Variables

Summary: Method of Separation of Variables Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section

More information

MATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The half-angle formula cos 2 θ = 1 2

MATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The half-angle formula cos 2 θ = 1 2 MATH 53 WORKSHEET MORE INTEGRATION IN POLAR COORDINATES ) Find the volume of the solid lying bove the xy-plne, below the prboloid x + y nd inside the cylinder x ) + y. ) We found lst time the set of points

More information

Problems for HW X. C. Gwinn. November 30, 2009

Problems for HW X. C. Gwinn. November 30, 2009 Problems for HW X C. Gwinn November 30, 2009 These problems will not be grded. 1 HWX Problem 1 Suppose thn n object is composed of liner dielectric mteril, with constnt reltive permittivity ɛ r. The object

More information

Problem Set 3 Solutions

Problem Set 3 Solutions Msschusetts Institute of Technology Deprtment of Physics Physics 8.07 Fll 2005 Problem Set 3 Solutions Problem 1: Cylindricl Cpcitor Griffiths Problems 2.39: Let the totl chrge per unit length on the inner

More information

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

More information

Lecture 13 - Linking E, ϕ, and ρ

Lecture 13 - Linking E, ϕ, and ρ Lecture 13 - Linking E, ϕ, nd ρ A Puzzle... Inner-Surfce Chrge Density A positive point chrge q is locted off-center inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on

More information

Exam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B

Exam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B PHY 249, Fll 216 Exm 1 Solutions nswer 1 is correct for ll problems. 1. Two uniformly chrged spheres, nd B, re plced t lrge distnce from ech other, with their centers on the x xis. The chrge on sphere

More information

Partial Differential Equations

Partial Differential Equations Prtil Differentil Equtions Notes by Robert Piché, Tmpere University of Technology reen s Functions. reen s Function for One-Dimensionl Eqution The reen s function provides complete solution to boundry

More information

Theoretische Physik 2: Elektrodynamik (Prof. A.-S. Smith) Home assignment 4

Theoretische Physik 2: Elektrodynamik (Prof. A.-S. Smith) Home assignment 4 WiSe 1 8.1.1 Prof. Dr. A.-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Mtthis Sb m Lehrstuhl für Theoretische Physik I Deprtment für Physik Friedrich-Alexnder-Universität Erlngen-Nürnberg Theoretische

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

SUMMER KNOWHOW STUDY AND LEARNING CENTRE SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18

More information

Candidates must show on each answer book the type of calculator used.

Candidates must show on each answer book the type of calculator used. UNIVERSITY OF EAST ANGLIA School of Mthemtics My/June UG Exmintion 2007 2008 ELECTRICITY AND MAGNETISM Time llowed: 3 hours Attempt FIVE questions. Cndidtes must show on ech nswer book the type of clcultor

More information

Consequently, the temperature must be the same at each point in the cross section at x. Let:

Consequently, the temperature must be the same at each point in the cross section at x. Let: HW 2 Comments: L1-3. Derive the het eqution for n inhomogeneous rod where the therml coefficients used in the derivtion of the het eqution for homogeneous rod now become functions of position x in the

More information

Physics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016

Physics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016 Physics 333, Fll 16 Problem Set 7 due Oct 14, 16 Reding: Griffiths 4.1 through 4.4.1 1. Electric dipole An electric dipole with p = p ẑ is locted t the origin nd is sitting in n otherwise uniform electric

More information

This final is a three hour open book, open notes exam. Do all four problems.

This final is a three hour open book, open notes exam. Do all four problems. Physics 55 Fll 27 Finl Exm Solutions This finl is three hour open book, open notes exm. Do ll four problems. [25 pts] 1. A point electric dipole with dipole moment p is locted in vcuum pointing wy from

More information

Phys 6321 Final Exam - Solutions May 3, 2013

Phys 6321 Final Exam - Solutions May 3, 2013 Phys 6321 Finl Exm - Solutions My 3, 2013 You my NOT use ny book or notes other thn tht supplied with this test. You will hve 3 hours to finish. DO YOUR OWN WORK. Express your nswers clerly nd concisely

More information

Physics 2135 Exam 1 February 14, 2017

Physics 2135 Exam 1 February 14, 2017 Exm Totl / 200 Physics 215 Exm 1 Ferury 14, 2017 Printed Nme: Rec. Sec. Letter: Five multiple choice questions, 8 points ech. Choose the est or most nerly correct nswer. 1. Two chrges 1 nd 2 re seprted

More information

The Wave Equation I. MA 436 Kurt Bryan

The Wave Equation I. MA 436 Kurt Bryan 1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string

More information

Chapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1

Chapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1 Tor Kjellsson Stockholm University Chpter 5 5. Strting with the following informtion: R = m r + m r m + m, r = r r we wnt to derive: µ = m m m + m r = R + µ m r, r = R µ m r 3 = µ m R + r, = µ m R r. 4

More information

Physics 121 Sample Common Exam 1 NOTE: ANSWERS ARE ON PAGE 8. Instructions:

Physics 121 Sample Common Exam 1 NOTE: ANSWERS ARE ON PAGE 8. Instructions: Physics 121 Smple Common Exm 1 NOTE: ANSWERS ARE ON PAGE 8 Nme (Print): 4 Digit ID: Section: Instructions: Answer ll questions. uestions 1 through 16 re multiple choice questions worth 5 points ech. You

More information

Homework Assignment 3 Solution Set

Homework Assignment 3 Solution Set Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution.

More information

) 4n+2 sin[(4n + 2)φ] n=0. a n ρ n sin(nφ + α n ) + b n ρ n sin(nφ + β n ) n=1. n=1. [A k ρ k cos(kφ) + B k ρ k sin(kφ)] (1) 2 + k=1

) 4n+2 sin[(4n + 2)φ] n=0. a n ρ n sin(nφ + α n ) + b n ρ n sin(nφ + β n ) n=1. n=1. [A k ρ k cos(kφ) + B k ρ k sin(kφ)] (1) 2 + k=1 Physics 505 Fll 2007 Homework Assignment #3 Solutions Textbook problems: Ch. 2: 2.4, 2.5, 2.22, 2.23 2.4 A vrint of the preceeding two-dimensionl problem is long hollow conducting cylinder of rdius b tht

More information

3 Mathematics of the Poisson Equation

3 Mathematics of the Poisson Equation 3 Mthemtics of the Poisson Eqution 3. Green functions nd the Poisson eqution () The Dirichlet Green function stisfies the Poisson eqution with delt-function chrge 2 G D (r, r o ) = δ 3 (r r o ) (3.) nd

More information

Density of Energy Stored in the Electric Field

Density of Energy Stored in the Electric Field Density of Energy Stored in the Electric Field Deprtment of Physics, Cornell University c Tomás A. Aris October 14, 01 Figure 1: Digrm of Crtesin vortices from René Descrtes Principi philosophie, published

More information

Name Solutions to Test 3 November 8, 2017

Name Solutions to Test 3 November 8, 2017 Nme Solutions to Test 3 November 8, 07 This test consists of three prts. Plese note tht in prts II nd III, you cn skip one question of those offered. Some possibly useful formuls cn be found below. Brrier

More information

Bernoulli Numbers Jeff Morton

Bernoulli Numbers Jeff Morton Bernoulli Numbers Jeff Morton. We re interested in the opertor e t k d k t k, which is to sy k tk. Applying this to some function f E to get e t f d k k tk d k f f + d k k tk dk f, we note tht since f

More information

Prof. Anchordoqui. Problems set # 4 Physics 169 March 3, 2015

Prof. Anchordoqui. Problems set # 4 Physics 169 March 3, 2015 Prof. Anchordoui Problems set # 4 Physics 169 Mrch 3, 15 1. (i) Eight eul chrges re locted t corners of cube of side s, s shown in Fig. 1. Find electric potentil t one corner, tking zero potentil to be

More information

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors

Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector

More information

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1 The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

More information

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral

f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

More information

Physics 9 Fall 2011 Homework 2 - Solutions Friday September 2, 2011

Physics 9 Fall 2011 Homework 2 - Solutions Friday September 2, 2011 Physics 9 Fll 0 Homework - s Fridy September, 0 Mke sure your nme is on your homework, nd plese box your finl nswer. Becuse we will be giving prtil credit, be sure to ttempt ll the problems, even if you

More information

Homework Assignment 5 Solution Set

Homework Assignment 5 Solution Set Homework Assignment 5 Solution Set PHYCS 44 3 Februry, 4 Problem Griffiths 3.8 The first imge chrge gurntees potentil of zero on the surfce. The secon imge chrge won t chnge the contribution to the potentil

More information

Improper Integrals, and Differential Equations

Improper Integrals, and Differential Equations Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

More information

Physics Jonathan Dowling. Lecture 9 FIRST MIDTERM REVIEW

Physics Jonathan Dowling. Lecture 9 FIRST MIDTERM REVIEW Physics 10 Jonthn Dowling Physics 10 ecture 9 FIRST MIDTERM REVIEW A few concepts: electric force, field nd potentil Electric force: Wht is the force on chrge produced by other chrges? Wht is the force

More information

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

f(a+h) f(a) x a h 0. This is the rate at which

f(a+h) f(a) x a h 0. This is the rate at which M408S Concept Inventory smple nswers These questions re open-ended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnk-out-n-nswer problems! (There re plenty of those in the

More information

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: Volumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

More information

APPLICATIONS OF THE DEFINITE INTEGRAL

APPLICATIONS OF THE DEFINITE INTEGRAL APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its cross-section in plne pssing through

More information

Phys 4321 Final Exam December 14, 2009

Phys 4321 Final Exam December 14, 2009 Phys 4321 Finl Exm December 14, 2009 You my NOT use the text book or notes to complete this exm. You nd my not receive ny id from nyone other tht the instructor. You will hve 3 hours to finish. DO YOUR

More information

IMPORTANT. Read these directions carefully:

IMPORTANT. Read these directions carefully: Physics 208: Electricity nd Mgnetism Finl Exm, Secs. 506 510. 7 My. 2004 Instructor: Dr. George R. Welch, 415 Engineering-Physics, 845-7737 Print your nme netly: Lst nme: First nme: Sign your nme: Plese

More information

Practice Problem Set 3

Practice Problem Set 3 Prctice Problem Set 3 #1. A dipole nd chrge A dipole with chrges ±q nd seprtion 2 is locted distnce from point chrge Q, oriented s shown in Figure 20.32 of the tetbook (reproduced here for convenience.

More information

ES.182A Topic 32 Notes Jeremy Orloff

ES.182A Topic 32 Notes Jeremy Orloff ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In

More information

Method of Localisation and Controlled Ejection of Swarms of Likely Charged Particles

Method of Localisation and Controlled Ejection of Swarms of Likely Charged Particles Method of Loclistion nd Controlled Ejection of Swrms of Likely Chrged Prticles I. N. Tukev July 3, 17 Astrct This work considers Coulom forces cting on chrged point prticle locted etween the two coxil,

More information

221B Lecture Notes WKB Method

221B Lecture Notes WKB Method Clssicl Limit B Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using

More information

We divide the interval [a, b] into subintervals of equal length x = b a n

We divide the interval [a, b] into subintervals of equal length x = b a n Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:

More information

Conducting Ellipsoid and Circular Disk

Conducting Ellipsoid and Circular Disk 1 Problem Conducting Ellipsoid nd Circulr Disk Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 (September 1, 00) Show tht the surfce chrge density σ on conducting ellipsoid,

More information

Integration Techniques

Integration Techniques Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u

More information

Physics 712 Electricity and Magnetism Solutions to Final Exam, Spring 2016

Physics 712 Electricity and Magnetism Solutions to Final Exam, Spring 2016 Physics 7 Electricity nd Mgnetism Solutions to Finl Em, Spring 6 Plese note tht some possibly helpful formuls pper on the second pge The number of points on ech problem nd prt is mrked in squre brckets

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

Chapter 6 Electrostatic Boundary Value Problems. Dr. Talal Skaik

Chapter 6 Electrostatic Boundary Value Problems. Dr. Talal Skaik Chpter 6 Electrosttic Boundry lue Problems Dr. Tll Skik 1 1 Introduction In previous chpters, E ws determined by coulombs lw or Guss lw when chrge distribution is known, or potentil is known throughout

More information

ragsdale (zdr82) HW2 ditmire (58335) 1

ragsdale (zdr82) HW2 ditmire (58335) 1 rgsdle (zdr82) HW2 ditmire (58335) This print-out should hve 22 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. 00 0.0 points A chrge of 8. µc

More information

Elliptic Equations. Laplace equation on bounded domains Circular Domains

Elliptic Equations. Laplace equation on bounded domains Circular Domains Elliptic Equtions Lplce eqution on bounded domins Sections 7.7.2, 7.7.3, An Introduction to Prtil Differentil Equtions, Pinchover nd Rubinstein 1.2 Circulr Domins We study the two-dimensionl Lplce eqution

More information

The usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1

The usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1 Mth50 Introduction to Differentil Equtions Brief Review of Complex Numbers Complex Numbers No rel number stisfies the eqution x =, since the squre of ny rel number hs to be non-negtive. By introducing

More information

Problem Set 3 Solutions

Problem Set 3 Solutions Chemistry 36 Dr Jen M Stndrd Problem Set 3 Solutions 1 Verify for the prticle in one-dimensionl box by explicit integrtion tht the wvefunction ψ ( x) π x is normlized To verify tht ψ ( x) is normlized,

More information

The Regulated and Riemann Integrals

The Regulated and Riemann Integrals Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

Physics 24 Exam 1 February 18, 2014

Physics 24 Exam 1 February 18, 2014 Exm Totl / 200 Physics 24 Exm 1 Februry 18, 2014 Printed Nme: Rec. Sec. Letter: Five multiple choice questions, 8 points ech. Choose the best or most nerly correct nswer. 1. The totl electric flux pssing

More information

Math 32B Discussion Session Session 7 Notes August 28, 2018

Math 32B Discussion Session Session 7 Notes August 28, 2018 Mth 32B iscussion ession ession 7 Notes August 28, 28 In tody s discussion we ll tlk bout surfce integrls both of sclr functions nd of vector fields nd we ll try to relte these to the mny other integrls

More information

Practice final exam solutions

Practice final exam solutions University of Pennsylvni Deprtment of Mthemtics Mth 26 Honors Clculus II Spring Semester 29 Prof. Grssi, T.A. Asher Auel Prctice finl exm solutions 1. Let F : 2 2 be defined by F (x, y (x + y, x y. If

More information

7.2 The Definite Integral

7.2 The Definite Integral 7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

More information

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems

More information

Chapter 2. Vectors. 2.1 Vectors Scalars and Vectors

Chapter 2. Vectors. 2.1 Vectors Scalars and Vectors Chpter 2 Vectors 2.1 Vectors 2.1.1 Sclrs nd Vectors A vector is quntity hving both mgnitude nd direction. Emples of vector quntities re velocity, force nd position. One cn represent vector in n-dimensionl

More information

Polynomials and Division Theory

Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

More information

13.4 Work done by Constant Forces

13.4 Work done by Constant Forces 13.4 Work done by Constnt Forces We will begin our discussion of the concept of work by nlyzing the motion of n object in one dimension cted on by constnt forces. Let s consider the following exmple: push

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: olumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

More information

Chapter 0. What is the Lebesgue integral about?

Chapter 0. What is the Lebesgue integral about? Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

More information

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

More information

Math 3B Final Review

Math 3B Final Review Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:45-10:45m SH 1607 Mth Lb Hours: TR 1-2pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems

More information

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.) MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

More information

4.4 Areas, Integrals and Antiderivatives

4.4 Areas, Integrals and Antiderivatives . res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014 SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.

More information

p(t) dt + i 1 re it ireit dt =

p(t) dt + i 1 re it ireit dt = Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)

More information

Best Approximation. Chapter The General Case

Best Approximation. Chapter The General Case Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given

More information

Definite integral. Mathematics FRDIS MENDELU

Definite integral. Mathematics FRDIS MENDELU Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

More information

Math 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions

Math 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,

More information

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019

ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019 ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil

More information

1 Probability Density Functions

1 Probability Density Functions Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our

More information

Trignometric Substitution

Trignometric Substitution Trignometric Substitution Trigonometric substitution refers simply to substitutions of the form x sinu or x tnu or x secu It is generlly used in conjunction with the trignometric identities to sin θ+cos

More information

1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.

1.2. Linear Variable Coefficient Equations. y + b ! = a y + b  Remark: The case b = 0 and a non-constant can be solved with the same idea as above. 1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt

More information

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30 Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

More information

4. Calculus of Variations

4. Calculus of Variations 4. Clculus of Vritions Introduction - Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the

More information

Things to Memorize: A Partial List. January 27, 2017

Things to Memorize: A Partial List. January 27, 2017 Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved

More information

P812 Midterm Examination February Solutions

P812 Midterm Examination February Solutions P8 Midterm Exmintion Februry s. A one dimensionl chin of chrges consist of e nd e lterntively plced with neighbouring distnce. Show tht the potentil energy of ech chrge is given by U = ln. 4πε Explin qulittively

More information

The Velocity Factor of an Insulated Two-Wire Transmission Line

The Velocity Factor of an Insulated Two-Wire Transmission Line The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the

More information

Sections 5.2: The Definite Integral

Sections 5.2: The Definite Integral Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)

More information

221A Lecture Notes WKB Method

221A Lecture Notes WKB Method A Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using ψ x, t = e

More information

We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.

We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b. Mth 255 - Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn

More information

Physics 1402: Lecture 7 Today s Agenda

Physics 1402: Lecture 7 Today s Agenda 1 Physics 1402: Lecture 7 Tody s gend nnouncements: Lectures posted on: www.phys.uconn.edu/~rcote/ HW ssignments, solutions etc. Homework #2: On Msterphysics tody: due Fridy Go to msteringphysics.com Ls:

More information

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1 Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the

More information

Patch Antennas. Chapter Resonant Cavity Analysis

Patch Antennas. Chapter Resonant Cavity Analysis Chpter 4 Ptch Antenns A ptch ntenn is low-profile ntenn consisting of metl lyer over dielectric sustrte nd ground plne. Typiclly, ptch ntenn is fed y microstrip trnsmission line, ut other feed lines such

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Statistical Physics I Spring Term Solutions to Problem Set #1

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Statistical Physics I Spring Term Solutions to Problem Set #1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Deprtment 8.044 Sttisticl Physics I Spring Term 03 Problem : Doping Semiconductor Solutions to Problem Set # ) Mentlly integrte the function p(x) given in

More information

Math 0230 Calculus 2 Lectures

Math 0230 Calculus 2 Lectures Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two

More information

Line Integrals. Chapter Definition

Line Integrals. Chapter Definition hpter 2 Line Integrls 2.1 Definition When we re integrting function of one vrible, we integrte long n intervl on one of the xes. We now generlize this ide by integrting long ny curve in the xy-plne. It

More information

Note 16. Stokes theorem Differential Geometry, 2005

Note 16. Stokes theorem Differential Geometry, 2005 Note 16. Stokes theorem ifferentil Geometry, 2005 Stokes theorem is the centrl result in the theory of integrtion on mnifolds. It gives the reltion between exterior differentition (see Note 14) nd integrtion

More information

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp. MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O 1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

More information