Physics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems.

Size: px
Start display at page:

Download "Physics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems."

Transcription

1 Physics 55 Fll 5 Midtem Solutions This midtem is two hou open ook, open notes exm. Do ll thee polems. [35 pts] 1. A ectngul ox hs sides of lengths, nd c z x c [1] ) Fo the Diichlet polem in the inteio of the ox, the Geen s function my e expnded s G(x, y, z; x, y, z ) = m=1 n=1 g mn (z, z ) mπx y mπx nπy nπy Wite down the ppopite diffeentil eqution tht g mn (z, z ) must stisfy. Note tht kx stisfies the completeness eltion m=1 mπx Hence the Geen s function eqution mπx = δ(x x ) x G( x, x ) = δ 3 ( x x ) hs n expnsion ( x g mn (z, z ) mπx mπx nπy ) nπy = δ(z z ) 4 mπx mπx nπy nπy Woking out the x nd y deivtives on the left-hnd side yields [ d ( mπ ) ( nπ dz ) ] g mn (z, z ) mπx mπx nπy nπy = 16π δ(z z ) mπx mπx nπy nπy

2 Howeve, ce kx foms n othogonl sis, ech tem in this sum must vnish y itself. This esults in the diffeentil eqution ( d dz γ mn ) g mn (z, z ) = 16π δ(z z ) (1) whee γ mn = π (m/) + (n/) is given in pt c). Note tht the Fouie e expnsion utomticlly stisfies Diichlet oundy conditions fo x nd y. The emining oundy condition is tht g mn (z, z ) vnishes wheneve z o z is equl to o c. [15] ) Solve the Geen s function eqution fo g mn (z, z ) suject to Diichlet oundy conditions nd wite down the esult fo G(x, y, z; x, y, z ). We my solve the Geen s function eqution (1) y fist noting tht the homogeneous eqution is of the fom g (z ) γ mng(z ) = This is second-ode line eqution with constnt coefficients dmitting the fmili solution g(z ) = Ae γ mnz + Be γ mnz Howeve, we wnt g(z ) = when z = o z = c. This motivtes us to wite out the solutions u(z ) = h γ mn z v(z ) = h[γ mn (c z )] < z < z z < z < c As we hve seen, the full Geen s function solution is then given y g mn (z, z ) = 16π A u(z <)v(z > ) whee A is given y the Wonskin, W (u, v) = A/p(z ) whee the Stum-Liouville opeto hs the fom d dz p(z ) d dz + q(z ). In this cse p(z ) = 1, nd it is esy to see tht W = u v u v = γ mn [h γ mn z cosh γ mn (c z ) + cosh γ mn z h γ mn (c z )] Hence = γ mn h γ mn c g mn (z, z ) = 16π γ mn h γ mn c h γ mnz < h γ mn [(c z > )]

3 nd G( x, x ) = 16π 1 mπx γ mn h γ mn c mπx nπy nπy h γ mn z < h γ mn [(c z > )] () [1] c) Conside the oundy vlue polem whee the potentil on top of the ox is Φ(x, y, c) = V (x, y) while the potentil on the othe five sides vnish. Ug the Geens function otined ove, show tht the potentil my e witten s Φ(x, y, z) = m=1 n=1 whee γ mn = π (m/) + (n/) nd A mn = 4 h γ mn c A mn mπx dx nπy h γ mn z dy V (x, y) mπx nπy Since we only hve to woy out the potentil on the top of the ox (nd ce we ssume thee is no chge inside the ox), the Geen s function solution my e witten Φ( x ) = 1 Φ( x ) G( x, x ) n d = 1 V (x, y ) G( x, x ) n dz dy Note tht the outwd-pointing noml ˆn on the top of the ox is in the +ẑ diection, we compute the noml deivtive of () G( x, x ) n = G( x, x ) z = 16π 1 mπx γ mn h γ mn c mπx nπy nπy = ( h γ mn z γ mn cosh γ mn [(c z )]) 4 mπx h γ mn c mπx nπy h γ mn z nπy Inseting this into (3) then stightfowdly gives the desied esult. Note tht the pimes my e dopped fom the doule integl A mn once it hs een isolted fom the est of the expession. (3)

4 [3 pts]. The potentil on the sufce of sphee of dius is specified y β {, θ < β V β V (θ, φ) = V, β θ π β, π β < θ π Thee e no othe chges in this polem. [] ) Show tht the potentil outside the sphee my e expessed s Φ(, θ, φ) = ( ) l+1 V [P l+1 (cos β) P l 1 (cos β)] Pl (cos θ) l=,4,6,... whee we tke P 1 (x) =. Note tht Legende polynomils stisfy the eltion (l + 1)P l (x) = P l+1 (x) P l 1 (x). Thee e sevel wys of solving this polem. Pehps the most stightfowd is to elize fom zimuthl symmety tht the potentil necessily dmits Legende expnsion Φ( x ) = ( ) l+1 α l Pl (cos θ) l The oundy conditions t = gives V (θ) = l α l P l (cos θ) This is clely Legende expnsion fo V (θ). The legende othogonlity eltion llows us to wite the expnsion coefficients α l s α l = l Fo the specified potentil, this ecomes α l = l + 1 cos β cos β V (cos θ) P l (cos θ) d(cos θ) V P l (cos θ) d(cos θ) = V cos β cos β Ug the Legende eltion given in the polem, we otin α l = V = V cos β cos β [P l+1(x) P l 1(x)] dx (l + 1)P l (x) dx [ P l+1 (cos β) P l 1 (cos β) P l+1 ( cos β) + P l 1 ( cos β) = V (1 + ( )l )[P l+1 (cos β) P l 1 (cos β)] ]

5 This vnishes unless l is even (which should e ovious fom the z z symmety of the polem). The esult is then Φ(, θ) = ( ) l+1 V [P l+1 (cos β) P l 1 (cos β)] Pl (cos θ) l even Note tht l = is llowed, nd gives the monopole contiution. (Excluding l = fom the sum ws mistke.) This polem could lso hve een solved y ug the Diichlet Geen s function outside sphee G( x, x ) = ) 1 ( l< l+1 1 l + 1 l+1 l,m < > l+1 Ylm(Ω )Y lm (Ω) Afte slight mnipultion, we would hve ended up with simil Legende polynomil integtion. [1] ) Fo fixed V, wht ngle β mximizes the qudupole moment? The qudupole is given y l =. Hence the qudupole moment is elted to the α tem in the expnsion. Since we do not ce out nomliztion (ny only to mximize the moment) it is sufficient to wite q, α P 3 (cos β) P 1 (cos β) We extemize this quntity y tking deivtive with espect to β nd setting the esult to zeo. Noting tht tking deivtive simply undoes the integtion, we end up solving dq, dβ P (cos β) β = 1 (3 cos β 1) β = This is solved y β = nd β = cos 1 (±1/ 3). Clely the qudupole moment vnishes when β = (s the entie sphee is t constnt zeo potentil). Hence the mximum is when β = cos 1 (1/ 3), povided we estict β π/. Note tht β > π/ does not elly mke sense, except in foml mnne whee V V (coesponding to intechnging the integtion limits). Techniclly, the polem should hve sked to mximize the mgnitude of the qudupole moment insted of to mke q, s positive s possile (which would depend on the sign of V ). [35 pts] 3. A line chge on the z xis extends fom z = to z = + nd hs line chge density vying s x λ z α λ z α y λ(z) = { λ z α, < z λ z α, z <

6 whee α is positive constnt. The totl chge on the < z segment is Q (nd the chge on the z < segment is Q). [] ) Clculte ll of the multipole moments of the chge distiution. Mke sue to indicte which moments e non-vnishing. Noting tht unifomly chged line chge on the +z xis hs chge density we see tht vying line chge yields ρ = λ δ(cosθ 1) π ρ = λ() δ(cos θ 1) π This my e checked y oseving dq = ρ d 3 x = ρ d dφ d(cos θ) = λ()d dφ π cos θ=1 (Note tht thee is no distinction etween nd z fo cos θ = 1.) Hence, fo the positive nd negtive line chge, we hve ρ = λ α [δ(cos θ 1) δ(cos θ + 1)] π We my nomlize λ y integting fom to to otin the totl chge Q Q = λ(z) dz = λ z α dz = λ α+1 α + 1 Hence ρ = Q α + 1 ( ) α [δ(cos θ 1) δ(cos θ + 1)] π The multipole moments e then q lm = l Ylm(Ω)ρ d dω = Q α + 1 ( l ) α d Y lm(ω)[δ(cos θ 1) δ(cos θ + 1)] dω π By zimuthl symmety, only the m = moments e non-vnishing q l, = Q α + 1 l+1 1 α + l = Q l α + 1 l + 1 α + l + 1 [P l(1) P l ( 1)] = Q l α + 1 l + 1 α + l + 1 [1 ( )l ] l + 1 P l(cos θ)[δ(cos θ 1) δ(cos θ + 1)]d(cos θ)

7 Hence ll moments vnish unless l is odd nd m is zeo. Then q l, = Q l α + 1 l + 1 α + l + 1 π l odd [1] ) Wite down the multipole expnsion fo the potentil in explicit fom up to the fist two non-vnishing tems. The multipole expnsion gives Φ = 1 ɛ l,m = 1 ɛ l odd l + 1 q lm = Q ɛ l odd = Q ( (α + 1) ɛ α + = Q ɛ ( (α + 1) α + l + 1 q l, Y lm (θ, φ) l+1 α + 1 α + l + 1 ( P l (cos θ) l+1 ) l Pl (cos θ) P l (cos θ) (α + 1)3 P 3 (cos θ) + α cos θ (α + 1)3 5 cos 3 θ 3 cos θ + (α + 4) 4 + [5] c) Wht is the dipole moment p in tems of Q, nd α? Note tht the dipole tem in (4) hs the fom Φ = 1 Q(α + 1) z ɛ α + 3 Comping this with the dipole expession gives φ = 1 p x ɛ 3 p = Q(α + 1) ẑ α + Altentively, one could compute diectly p = xρ d 3 x = xq(α + 1) z α sgn(z) dz α+1 The z component is the only non-vnishing component x=y= ) ) (4) p z = Q(α + 1) z α+1 Q(α + 1) dz = α+1 α +

Physics 604 Problem Set 1 Due Sept 16, 2010

Physics 604 Problem Set 1 Due Sept 16, 2010 Physics 64 Polem et 1 Due ept 16 1 1) ) Inside good conducto the electic field is eo (electons in the conducto ecuse they e fee to move move in wy to cncel ny electic field impessed on the conducto inside

More information

U>, and is negative. Electric Potential Energy

U>, and is negative. Electric Potential Energy Electic Potentil Enegy Think of gvittionl potentil enegy. When the lock is moved veticlly up ginst gvity, the gvittionl foce does negtive wok (you do positive wok), nd the potentil enegy (U) inceses. When

More information

FI 2201 Electromagnetism

FI 2201 Electromagnetism FI 1 Electomgnetism Alexnde A. Isknd, Ph.D. Physics of Mgnetism nd Photonics Resech Goup Electosttics ELECTRIC PTENTIALS 1 Recll tht we e inteested to clculte the electic field of some chge distiution.

More information

Lecture 11: Potential Gradient and Capacitor Review:

Lecture 11: Potential Gradient and Capacitor Review: Lectue 11: Potentil Gdient nd Cpcito Review: Two wys to find t ny point in spce: Sum o Integte ove chges: q 1 1 q 2 2 3 P i 1 q i i dq q 3 P 1 dq xmple of integting ove distiution: line of chge ing of

More information

Radial geodesics in Schwarzschild spacetime

Radial geodesics in Schwarzschild spacetime Rdil geodesics in Schwzschild spcetime Spheiclly symmetic solutions to the Einstein eqution tke the fom ds dt d dθ sin θdϕ whee is constnt. We lso hve the connection components, which now tke the fom using

More information

10 Statistical Distributions Solutions

10 Statistical Distributions Solutions Communictions Engineeing MSc - Peliminy Reding 1 Sttisticl Distiutions Solutions 1) Pove tht the vince of unifom distiution with minimum vlue nd mximum vlue ( is ) 1. The vince is the men of the sques

More information

Electric Potential. and Equipotentials

Electric Potential. and Equipotentials Electic Potentil nd Euipotentils U Electicl Potentil Review: W wok done y foce in going fom to long pth. l d E dl F W dl F θ Δ l d E W U U U Δ Δ l d E W U U U U potentil enegy electic potentil Potentil

More information

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface Genel Physics II Chpte 3: Guss w We now wnt to quickly discuss one of the moe useful tools fo clculting the electic field, nmely Guss lw. In ode to undestnd Guss s lw, it seems we need to know the concept

More information

π,π is the angle FROM a! TO b

π,π is the angle FROM a! TO b Mth 151: 1.2 The Dot Poduct We hve scled vectos (o, multiplied vectos y el nume clled scl) nd dded vectos (in ectngul component fom). Cn we multiply vectos togethe? The nswe is YES! In fct, thee e two

More information

u(r, θ) = 1 + 3a r n=1

u(r, θ) = 1 + 3a r n=1 Mth 45 / AMCS 55. etuck Assignment 8 ue Tuesdy, Apil, 6 Topics fo this week Convegence of Fouie seies; Lplce s eqution nd hmonic functions: bsic popeties, computions on ectngles nd cubes Fouie!, Poisson

More information

Optimization. x = 22 corresponds to local maximum by second derivative test

Optimization. x = 22 corresponds to local maximum by second derivative test Optimiztion Lectue 17 discussed the exteme vlues of functions. This lectue will pply the lesson fom Lectue 17 to wod poblems. In this section, it is impotnt to emembe we e in Clculus I nd e deling one-vible

More information

Fourier-Bessel Expansions with Arbitrary Radial Boundaries

Fourier-Bessel Expansions with Arbitrary Radial Boundaries Applied Mthemtics,,, - doi:./m.. Pulished Online My (http://www.scirp.og/jounl/m) Astct Fouie-Bessel Expnsions with Aity Rdil Boundies Muhmmd A. Mushef P. O. Box, Jeddh, Sudi Ai E-mil: mmushef@yhoo.co.uk

More information

Electric Field F E. q Q R Q. ˆ 4 r r - - Electric field intensity depends on the medium! origin

Electric Field F E. q Q R Q. ˆ 4 r r - - Electric field intensity depends on the medium! origin 1 1 Electic Field + + q F Q R oigin E 0 0 F E ˆ E 4 4 R q Q R Q - - Electic field intensity depends on the medium! Electic Flux Density We intoduce new vecto field D independent of medium. D E So, electic

More information

Answers to test yourself questions

Answers to test yourself questions Answes to test youself questions opic Descibing fields Gm Gm Gm Gm he net field t is: g ( d / ) ( 4d / ) d d Gm Gm Gm Gm Gm Gm b he net potentil t is: V d / 4d / d 4d d d V e 4 7 9 49 J kg 7 7 Gm d b E

More information

(A) 6.32 (B) 9.49 (C) (D) (E) 18.97

(A) 6.32 (B) 9.49 (C) (D) (E) 18.97 Univesity of Bhin Physics 10 Finl Exm Key Fll 004 Deptment of Physics 13/1/005 8:30 10:30 e =1.610 19 C, m e =9.1110 31 Kg, m p =1.6710 7 Kg k=910 9 Nm /C, ε 0 =8.8410 1 C /Nm, µ 0 =4π10 7 T.m/A Pt : 10

More information

Solutions to Midterm Physics 201

Solutions to Midterm Physics 201 Solutions to Midtem Physics. We cn conside this sitution s supeposition of unifomly chged sphee of chge density ρ nd dius R, nd second unifomly chged sphee of chge density ρ nd dius R t the position of

More information

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by Clss Summy.5 Eponentil Functions.6 Invese Functions nd Logithms A function f is ule tht ssigns to ech element D ectly one element, clled f( ), in. Fo emple : function not function Given functions f, g:

More information

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007 School of Electicl nd Compute Engineeing, Conell Univesity ECE 303: Electomgnetic Fields nd Wves Fll 007 Homewok 4 Due on Sep. 1, 007 by 5:00 PM Reding Assignments: i) Review the lectue notes. ii) Relevnt

More information

This immediately suggests an inverse-square law for a "piece" of current along the line.

This immediately suggests an inverse-square law for a piece of current along the line. Electomgnetic Theoy (EMT) Pof Rui, UNC Asheville, doctophys on YouTube Chpte T Notes The iot-svt Lw T nvese-sque Lw fo Mgnetism Compe the mgnitude of the electic field t distnce wy fom n infinite line

More information

Homework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is:

Homework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is: . Homewok 3 MAE 8C Poblems, 5, 7, 0, 4, 5, 8, 3, 30, 3 fom Chpte 5, msh & Btt Point souces emit nuetons/sec t points,,, n 3 fin the flux cuent hlf wy between one sie of the tingle (blck ot). The flux fo

More information

Chapter 28 Sources of Magnetic Field

Chapter 28 Sources of Magnetic Field Chpte 8 Souces of Mgnetic Field - Mgnetic Field of Moving Chge - Mgnetic Field of Cuent Element - Mgnetic Field of Stight Cuent-Cying Conducto - Foce Between Pllel Conductos - Mgnetic Field of Cicul Cuent

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

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007 School of Electicl nd Compute Engineeing, Conell Univesity ECE 303: Electomgnetic Fields nd Wves Fll 007 Homewok 3 Due on Sep. 14, 007 by 5:00 PM Reding Assignments: i) Review the lectue notes. ii) Relevnt

More information

The Formulas of Vector Calculus John Cullinan

The Formulas of Vector Calculus John Cullinan The Fomuls of Vecto lculus John ullinn Anlytic Geomety A vecto v is n n-tuple of el numbes: v = (v 1,..., v n ). Given two vectos v, w n, ddition nd multipliction with scl t e defined by Hee is bief list

More information

SPA7010U/SPA7010P: THE GALAXY. Solutions for Coursework 1. Questions distributed on: 25 January 2018.

SPA7010U/SPA7010P: THE GALAXY. Solutions for Coursework 1. Questions distributed on: 25 January 2018. SPA7U/SPA7P: THE GALAXY Solutions fo Cousewok Questions distibuted on: 25 Jnuy 28. Solution. Assessed question] We e told tht this is fint glxy, so essentilly we hve to ty to clssify it bsed on its spectl

More information

Week 8. Topic 2 Properties of Logarithms

Week 8. Topic 2 Properties of Logarithms Week 8 Topic 2 Popeties of Logithms 1 Week 8 Topic 2 Popeties of Logithms Intoduction Since the esult of ithm is n eponent, we hve mny popeties of ithms tht e elted to the popeties of eponents. They e

More information

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4

MATHEMATICS IV 2 MARKS. 5 2 = e 3, 4 MATHEMATICS IV MARKS. If + + 6 + c epesents cicle with dius 6, find the vlue of c. R 9 f c ; g, f 6 9 c 6 c c. Find the eccenticit of the hpeol Eqution of the hpeol Hee, nd + e + e 5 e 5 e. Find the distnce

More information

Topics for Review for Final Exam in Calculus 16A

Topics for Review for Final Exam in Calculus 16A Topics fo Review fo Finl Em in Clculus 16A Instucto: Zvezdelin Stnkov Contents 1. Definitions 1. Theoems nd Poblem Solving Techniques 1 3. Eecises to Review 5 4. Chet Sheet 5 1. Definitions Undestnd the

More information

The Area of a Triangle

The Area of a Triangle The e of Tingle tkhlid June 1, 015 1 Intodution In this tile we will e disussing the vious methods used fo detemining the e of tingle. Let [X] denote the e of X. Using se nd Height To stt off, the simplest

More information

Physics 11b Lecture #11

Physics 11b Lecture #11 Physics 11b Lectue #11 Mgnetic Fields Souces of the Mgnetic Field S&J Chpte 9, 3 Wht We Did Lst Time Mgnetic fields e simil to electic fields Only diffeence: no single mgnetic pole Loentz foce Moving chge

More information

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3 DEPATMENT OF CIVIL AND ENVIONMENTAL ENGINEEING FLID MECHANICS III Solutions to Poblem Sheet 3 1. An tmospheic vote is moelle s combintion of viscous coe otting s soli boy with ngul velocity Ω n n iottionl

More information

1 Using Integration to Find Arc Lengths and Surface Areas

1 Using Integration to Find Arc Lengths and Surface Areas Novembe 9, 8 MAT86 Week Justin Ko Using Integtion to Find Ac Lengths nd Sufce Aes. Ac Length Fomul: If f () is continuous on [, b], then the c length of the cuve = f() on the intevl [, b] is given b s

More information

Electricity & Magnetism Lecture 6: Electric Potential

Electricity & Magnetism Lecture 6: Electric Potential Electicity & Mgnetism Lectue 6: Electic Potentil Tody s Concept: Electic Potenl (Defined in tems of Pth Integl of Electic Field) Electicity & Mgnesm Lectue 6, Slide Stuff you sked bout:! Explin moe why

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

EECE 260 Electrical Circuits Prof. Mark Fowler

EECE 260 Electrical Circuits Prof. Mark Fowler EECE 60 Electicl Cicuits Pof. Mk Fowle Complex Numbe Review /6 Complex Numbes Complex numbes ise s oots of polynomils. Definition of imginy # nd some esulting popeties: ( ( )( ) )( ) Recll tht the solution

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

( ) D x ( s) if r s (3) ( ) (6) ( r) = d dr D x

( ) D x ( s) if r s (3) ( ) (6) ( r) = d dr D x SIO 22B, Rudnick dpted fom Dvis III. Single vile sttistics The next few lectues e intended s eview of fundmentl sttistics. The gol is to hve us ll speking the sme lnguge s we move to moe dvnced topics.

More information

dx was area under f ( x ) if ( ) 0

dx was area under f ( x ) if ( ) 0 13. Line Integls Line integls e simil to single integl, f ( x) dx ws e unde f ( x ) if ( ) 0 Insted of integting ove n intevl [, ] (, ) f xy ds f x., we integte ove cuve, (in the xy-plne). **Figue - get

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

Physics 1502: Lecture 2 Today s Agenda

Physics 1502: Lecture 2 Today s Agenda 1 Lectue 1 Phsics 1502: Lectue 2 Tod s Agend Announcements: Lectues posted on: www.phs.uconn.edu/~cote/ HW ssignments, solutions etc. Homewok #1: On Mstephsics this Fid Homewoks posted on Msteingphsics

More information

Chapter 7. Kleene s Theorem. 7.1 Kleene s Theorem. The following theorem is the most important and fundamental result in the theory of FA s:

Chapter 7. Kleene s Theorem. 7.1 Kleene s Theorem. The following theorem is the most important and fundamental result in the theory of FA s: Chpte 7 Kleene s Theoem 7.1 Kleene s Theoem The following theoem is the most impotnt nd fundmentl esult in the theoy of FA s: Theoem 6 Any lnguge tht cn e defined y eithe egul expession, o finite utomt,

More information

Chapter 6 Thermoelasticity

Chapter 6 Thermoelasticity Chpte 6 Themoelsticity Intoduction When theml enegy is dded to n elstic mteil it expnds. Fo the simple unidimensionl cse of b of length L, initilly t unifom tempetue T 0 which is then heted to nonunifom

More information

Mark Scheme (Results) January 2008

Mark Scheme (Results) January 2008 Mk Scheme (Results) Jnuy 00 GCE GCE Mthemtics (6679/0) Edecel Limited. Registeed in Englnd nd Wles No. 4496750 Registeed Office: One90 High Holbon, London WCV 7BH Jnuy 00 6679 Mechnics M Mk Scheme Question

More information

1 Spherical multipole moments

1 Spherical multipole moments Jackson notes 9 Spheical multipole moments Suppose we have a chage distibution ρ (x) wheeallofthechageiscontained within a spheical egion of adius R, as shown in the diagam. Then thee is no chage in the

More information

ELECTRO - MAGNETIC INDUCTION

ELECTRO - MAGNETIC INDUCTION NTRODUCTON LCTRO - MAGNTC NDUCTON Whenee mgnetic flu linked with cicuit chnges, n e.m.f. is induced in the cicuit. f the cicuit is closed, cuent is lso induced in it. The e.m.f. nd cuent poduced lsts s

More information

3.1 Magnetic Fields. Oersted and Ampere

3.1 Magnetic Fields. Oersted and Ampere 3.1 Mgnetic Fields Oested nd Ampee The definition of mgnetic induction, B Fields of smll loop (dipole) Mgnetic fields in mtte: ) feomgnetism ) mgnetiztion, (M ) c) mgnetic susceptiility, m d) mgnetic field,

More information

Two dimensional polar coordinate system in airy stress functions

Two dimensional polar coordinate system in airy stress functions I J C T A, 9(9), 6, pp. 433-44 Intentionl Science Pess Two dimensionl pol coodinte system in iy stess functions S. Senthil nd P. Sek ABSTRACT Stisfy the given equtions, boundy conditions nd bihmonic eqution.in

More information

Friedmannien equations

Friedmannien equations ..6 Fiedmnnien equtions FLRW metic is : ds c The metic intevl is: dt ( t) d ( ) hee f ( ) is function which detemines globl geometic l popety of D spce. f d sin d One cn put it in the Einstein equtions

More information

r a + r b a + ( r b + r c)

r a + r b a + ( r b + r c) AP Phsics C Unit 2 2.1 Nme Vectos Vectos e used to epesent quntities tht e chcteized b mgnitude ( numeicl vlue with ppopite units) nd diection. The usul emple is the displcement vecto. A quntit with onl

More information

Review of Gaussian Quadrature method

Review of Gaussian Quadrature method Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge

More information

Math 259 Winter Solutions to Homework #9

Math 259 Winter Solutions to Homework #9 Mth 59 Winter 9 Solutions to Homework #9 Prolems from Pges 658-659 (Section.8). Given f(, y, z) = + y + z nd the constrint g(, y, z) = + y + z =, the three equtions tht we get y setting up the Lgrnge multiplier

More information

STD: XI MATHEMATICS Total Marks: 90. I Choose the correct answer: ( 20 x 1 = 20 ) a) x = 1 b) x =2 c) x = 3 d) x = 0

STD: XI MATHEMATICS Total Marks: 90. I Choose the correct answer: ( 20 x 1 = 20 ) a) x = 1 b) x =2 c) x = 3 d) x = 0 STD: XI MATHEMATICS Totl Mks: 90 Time: ½ Hs I Choose the coect nswe: ( 0 = 0 ). The solution of is ) = b) = c) = d) = 0. Given tht the vlue of thid ode deteminnt is then the vlue of the deteminnt fomed

More information

RELATIVE KINEMATICS. q 2 R 12. u 1 O 2 S 2 S 1. r 1 O 1. Figure 1

RELATIVE KINEMATICS. q 2 R 12. u 1 O 2 S 2 S 1. r 1 O 1. Figure 1 RELAIVE KINEMAICS he equtions of motion fo point P will be nlyzed in two diffeent efeence systems. One efeence system is inetil, fixed to the gound, the second system is moving in the physicl spce nd the

More information

CHAPTER 2 ELECTROSTATIC POTENTIAL

CHAPTER 2 ELECTROSTATIC POTENTIAL 1 CHAPTER ELECTROSTATIC POTENTIAL 1 Intoduction Imgine tht some egion of spce, such s the oom you e sitting in, is pemeted by n electic field (Pehps thee e ll sots of electiclly chged bodies outside the

More information

Waveguide Guide: A and V. Ross L. Spencer

Waveguide Guide: A and V. Ross L. Spencer Wveguide Guide: A nd V Ross L. Spencer I relly think tht wveguide fields re esier to understnd using the potentils A nd V thn they re using the electric nd mgnetic fields. Since Griffiths doesn t do it

More information

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) ,

Prerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) , R Pen Towe Rod No Conttos Ae Bistupu Jmshedpu 8 Tel (67)89 www.penlsses.om IIT JEE themtis Ppe II PART III ATHEATICS SECTION I (Totl ks : ) (Single Coet Answe Type) This setion ontins 8 multiple hoie questions.

More information

Green function and Eigenfunctions

Green function and Eigenfunctions Green function nd Eigenfunctions Let L e regulr Sturm-Liouville opertor on n intervl (, ) together with regulr oundry conditions. We denote y, φ ( n, x ) the eigenvlues nd corresponding normlized eigenfunctions

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

332:221 Principles of Electrical Engineering I Fall Hourly Exam 2 November 6, 2006

332:221 Principles of Electrical Engineering I Fall Hourly Exam 2 November 6, 2006 2:221 Principles of Electricl Engineering I Fll 2006 Nme of the student nd ID numer: Hourly Exm 2 Novemer 6, 2006 This is closed-ook closed-notes exm. Do ll your work on these sheets. If more spce is required,

More information

Math 4318 : Real Analysis II Mid-Term Exam 1 14 February 2013

Math 4318 : Real Analysis II Mid-Term Exam 1 14 February 2013 Mth 4318 : Rel Anlysis II Mid-Tem Exm 1 14 Febuy 2013 Nme: Definitions: Tue/Flse: Poofs: 1. 2. 3. 4. 5. 6. Totl: Definitions nd Sttements of Theoems 1. (2 points) Fo function f(x) defined on (, b) nd fo

More information

PDE Notes. Paul Carnig. January ODE s vs PDE s 1

PDE Notes. Paul Carnig. January ODE s vs PDE s 1 PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................

More information

Section 35 SHM and Circular Motion

Section 35 SHM and Circular Motion Section 35 SHM nd Cicul Motion Phsics 204A Clss Notes Wht do objects do? nd Wh do the do it? Objects sometimes oscillte in simple hmonic motion. In the lst section we looed t mss ibting t the end of sping.

More information

9.4 The response of equilibrium to temperature (continued)

9.4 The response of equilibrium to temperature (continued) 9.4 The esponse of equilibium to tempetue (continued) In the lst lectue, we studied how the chemicl equilibium esponds to the vition of pessue nd tempetue. At the end, we deived the vn t off eqution: d

More information

B.A. (PROGRAMME) 1 YEAR MATHEMATICS

B.A. (PROGRAMME) 1 YEAR MATHEMATICS Gdute Couse B.A. (PROGRAMME) YEAR MATHEMATICS ALGEBRA & CALCULUS PART B : CALCULUS SM 4 CONTENTS Lesson Lesson Lesson Lesson Lesson Lesson Lesson : Tngents nd Nomls : Tngents nd Nomls (Pol Co-odintes)

More information

(a) Counter-Clockwise (b) Clockwise ()N (c) No rotation (d) Not enough information

(a) Counter-Clockwise (b) Clockwise ()N (c) No rotation (d) Not enough information m m m00 kg dult, m0 kg bby. he seesw stts fom est. Which diection will it ottes? ( Counte-Clockwise (b Clockwise ( (c o ottion ti (d ot enough infomtion Effect of Constnt et oque.3 A constnt non-zeo toque

More information

f(k) e p 2 (k) e iax 2 (k a) r 2 e a x a a 2 + k 2 e a2 x 1 2 H(x) ik p (k) 4 r 3 cos Y 2 = 4

f(k) e p 2 (k) e iax 2 (k a) r 2 e a x a a 2 + k 2 e a2 x 1 2 H(x) ik p (k) 4 r 3 cos Y 2 = 4 Fouie tansfom pais: f(x) 1 f(k) e p 2 (k) p e iax 2 (k a) 2 e a x a a 2 + k 2 e a2 x 1 2, a > 0 a p k2 /4a2 e 2 1 H(x) ik p 2 + 2 (k) The fist few Y m Y 0 0 = Y 0 1 = Y ±1 1 = l : 1 Y2 0 = 4 3 ±1 cos Y

More information

Previously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system

Previously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system 436-459 Advnced contol nd utomtion Extensions to bckstepping contolle designs Tcking Obseves (nonline dmping) Peviously Lst lectue we looked t designing nonline contolles using the bckstepping technique

More information

ELECTROSTATICS. 4πε0. E dr. The electric field is along the direction where the potential decreases at the maximum rate. 5. Electric Potential Energy:

ELECTROSTATICS. 4πε0. E dr. The electric field is along the direction where the potential decreases at the maximum rate. 5. Electric Potential Energy: LCTROSTATICS. Quntiztion of Chge: Any chged body, big o smll, hs totl chge which is n integl multile of e, i.e. = ± ne, whee n is n intege hving vlues,, etc, e is the chge of electon which is eul to.6

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

1B40 Practical Skills

1B40 Practical Skills B40 Prcticl Skills Comining uncertinties from severl quntities error propgtion We usully encounter situtions where the result of n experiment is given in terms of two (or more) quntities. We then need

More information

2 mv2 qv (0) = 0 v = 2qV (0)/m. Express q. . Substitute for V (0) and simplify to obtain: v = q

2 mv2 qv (0) = 0 v = 2qV (0)/m. Express q. . Substitute for V (0) and simplify to obtain: v = q Pof Anchodoui Polems set # Physics 69 Mch 3, 5 (i) Eight eul chges e locted t cones of cue of side s, s shown in Fig Find electic potentil t one cone, tking zeo potentil to e infinitely f wy (ii) Fou point

More information

PX3008 Problem Sheet 1

PX3008 Problem Sheet 1 PX38 Poblem Sheet 1 1) A sphee of dius (m) contins chge of unifom density ρ (Cm -3 ). Using Guss' theoem, obtin expessions fo the mgnitude of the electic field (t distnce fom the cente of the sphee) in

More information

Orthogonal functions

Orthogonal functions Orthogonl functions Given rel vrible over the intervl (, b nd set of rel or complex functions U n (ξ, n =, 2,..., which re squre integrble nd orthonorml b U n(ξu m (ξdξ = δ n,m ( if the set of of functions

More information

= ΔW a b. U 1 r m 1 + K 2

= ΔW a b. U 1 r m 1 + K 2 Chpite 3 Potentiel électiue [18 u 3 mi] DEVOIR : 31, 316, 354, 361, 35 Le potentiel électiue est le tvil p unité de chge (en J/C, ou volt) Ce concept est donc utile dns les polèmes de consevtion d énegie

More information

This chapter is about energy associated with electrical interactions. Every

This chapter is about energy associated with electrical interactions. Every 23 ELECTRIC PTENTIAL whee d l is n infinitesiml displcement long the pticle s pth nd f is the ngle etween F nd d l t ech point long the pth. econd, if the foce F is consevtive, s we defined the tem in

More information

DEFINITION The inner product of two functions f 1 and f 2 on an interval [a, b] is the number. ( f 1, f 2 ) b DEFINITION 11.1.

DEFINITION The inner product of two functions f 1 and f 2 on an interval [a, b] is the number. ( f 1, f 2 ) b DEFINITION 11.1. 398 CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 11.1 ORTHOGONAL FUNCTIONS REVIEW MATERIAL The notions of generlized vectors nd vector spces cn e found in ny liner lger text. INTRODUCTION The concepts

More information

AQA Maths M2. Topic Questions from Papers. Circular Motion. Answers

AQA Maths M2. Topic Questions from Papers. Circular Motion. Answers AQA Mths M Topic Questions fom Ppes Cicul Motion Answes PhysicsAndMthsTuto.com PhysicsAndMthsTuto.com Totl 6 () T cos30 = 9.8 Resolving veticlly with two tems Coect eqution 9.8 T = cos30 T =.6 N AG 3 Coect

More information

Question 1: The dipole

Question 1: The dipole Septembe, 08 Conell Univesity, Depatment of Physics PHYS 337, Advance E&M, HW #, due: 9/5/08, :5 AM Question : The dipole Conside a system as discussed in class and shown in Fig.. in Heald & Maion.. Wite

More information

Available online at ScienceDirect. Procedia Engineering 91 (2014 ) 32 36

Available online at   ScienceDirect. Procedia Engineering 91 (2014 ) 32 36 Aville online t wwwsciencediectcom ScienceDiect Pocedi Engineeing 91 (014 ) 3 36 XXIII R-S-P semin Theoeticl Foundtion of Civil Engineeing (3RSP) (TFoCE 014) Stess Stte of Rdil Inhomogeneous Semi Sphee

More information

On the Eötvös effect

On the Eötvös effect On the Eötvös effect Mugu B. Răuţ The im of this ppe is to popose new theoy bout the Eötvös effect. We develop mthemticl model which loud us bette undestnding of this effect. Fom the eqution of motion

More information

ρ θ φ δ δ θ δ φ δ φ π δ φ π δ φ π

ρ θ φ δ δ θ δ φ δ φ π δ φ π δ φ π Physics 6 Fin Ex Dec. 6, ( pts Fou point chges with chge ± q e nged s in Figue. (5 pts. Wht is the chge density function ρ (, θφ,? (,, q ( ( cos ( / + ( ( / / ρ θ φ δ δ θ δ φ δ φ π δ φ π δ φ π b (5 pts.

More information

Unit 6. Magnetic forces

Unit 6. Magnetic forces Unit 6 Mgnetic foces 6.1 ntoduction. Mgnetic field 6. Mgnetic foces on moving electic chges 6. oce on conducto with cuent. 6.4 Action of unifom mgnetic field on flt cuent-cying loop. Mgnetic moment. Electic

More information

Chapter 2: Electric Field

Chapter 2: Electric Field P 6 Genel Phsics II Lectue Outline. The Definition of lectic ield. lectic ield Lines 3. The lectic ield Due to Point Chges 4. The lectic ield Due to Continuous Chge Distibutions 5. The oce on Chges in

More information

Phys. 506 Electricity and Magnetism Winter 2004 Prof. G. Raithel Problem Set 1 Total 30 Points. 1. Jackson Points

Phys. 506 Electricity and Magnetism Winter 2004 Prof. G. Raithel Problem Set 1 Total 30 Points. 1. Jackson Points Phys. 56 Electricity nd Mgnetism Winter 4 Prof. G. Rithel Prolem Set Totl 3 Points. Jckson 8. Points : The electric field is the sme s in the -dimensionl electrosttic prolem of two concentric cylinders,

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

Differential Equations 2 Homework 5 Solutions to the Assigned Exercises

Differential Equations 2 Homework 5 Solutions to the Assigned Exercises Differentil Equtions Homework Solutions to the Assigned Exercises, # 3 Consider the dmped string prolem u tt + 3u t = u xx, < x , u, t = u, t =, t >, ux, = fx, u t x, = gx. In the exm you were supposed

More information

Math 426: Probability Final Exam Practice

Math 426: Probability Final Exam Practice Mth 46: Probbility Finl Exm Prctice. Computtionl problems 4. Let T k (n) denote the number of prtitions of the set {,..., n} into k nonempty subsets, where k n. Argue tht T k (n) kt k (n ) + T k (n ) by

More information

22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 20

22.615, MHD Theory of Fusion Systems Prof. Freidberg Lecture 20 .615, MHD Theoy of Fusion Systes Pof. Feideg Lectue Resistive Wll Mode 1. We hve seen tht pefectly conducting wll, plced in close poxiity to the pls cn hve stong stilizing effect on extenl kink odes..

More information

Linear Inequalities. Work Sheet 1

Linear Inequalities. Work Sheet 1 Work Sheet 1 Liner Inequlities Rent--Hep, cr rentl compny,chrges $ 15 per week plus $ 0.0 per mile to rent one of their crs. Suppose you re limited y how much money you cn spend for the week : You cn spend

More information

4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve

4 7x =250; 5 3x =500; Read section 3.3, 3.4 Announcements: Bell Ringer: Use your calculator to solve Dte: 3/14/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: Use your clcultor to solve 4 7x =250; 5 3x =500; HW Requests: Properties of Log Equtions

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

CHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD

CHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD ollege Physics Student s Mnul hpte 8 HAPTR 8: LTRI HARG AD LTRI ILD 8. STATI LTRIITY AD HARG: OSRVATIO O HARG. ommon sttic electicity involves chges nging fom nnocoulombs to micocoulombs. () How mny electons

More information

Michael Rotkowitz 1,2

Michael Rotkowitz 1,2 Novembe 23, 2006 edited Line Contolles e Unifomly Optiml fo the Witsenhusen Counteexmple Michel Rotkowitz 1,2 IEEE Confeence on Decision nd Contol, 2006 Abstct In 1968, Witsenhusen intoduced his celebted

More information

Prof. Anchordoqui Problems set # 12 Physics 169 May 12, 2015

Prof. Anchordoqui Problems set # 12 Physics 169 May 12, 2015 Pof. Anchodoqui Poblems set # 12 Physics 169 My 12, 2015 1. Two concentic conducting sphees of inne nd oute dii nd b, espectively, cy chges ±Q. The empty spce between the sphees is hlf-filled by hemispheicl

More information

defined on a domain can be expanded into the Taylor series around a point a except a singular point. Also, f( z)

defined on a domain can be expanded into the Taylor series around a point a except a singular point. Also, f( z) 08 Tylo eie nd Mcluin eie A holomophic function f( z) defined on domin cn be expnded into the Tylo eie ound point except ingul point. Alo, f( z) cn be expnded into the Mcluin eie in the open dik with diu

More information

4.1. Probability Density Functions

4.1. Probability Density Functions STT 1 4.1-4. 4.1. Proility Density Functions Ojectives. Continuous rndom vrile - vers - discrete rndom vrile. Proility density function. Uniform distriution nd its properties. Expected vlue nd vrince of

More information

Review of Mathematical Concepts

Review of Mathematical Concepts ENEE 322: Signls nd Systems view of Mthemticl Concepts This hndout contins ief eview of mthemticl concepts which e vitlly impotnt to ENEE 322: Signls nd Systems. Since this mteil is coveed in vious couses

More information

Section 4: Integration ECO4112F 2011

Section 4: Integration ECO4112F 2011 Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic

More information

Solutions to Homework Set 3

Solutions to Homework Set 3 Solutions to Homework Set 3 1 Bent brs: First recll some elementry clculus results: the curvture κ, nd the rdius R of the osculting circle, t point (x, y on curve y(x re given by κ = 1 R = dψ = 1 d tn

More information

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016

Algebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016 Newton's Lw of Univesl Gvittion Gvittionl Foce lick on the topic to go to tht section Gvittionl Field lgeb sed Physics Newton's Lw of Univesl Gvittion Sufce Gvity Gvittionl Field in Spce Keple's Thid Lw

More information