Homework Assignment 4 Solution Set

Size: px
Start display at page:

Download "Homework Assignment 4 Solution Set"

Transcription

1 Homework Assignment 4 Solution Set PHYCS February, 24 Problem (Griffiths 2.37 If the plates are sufficiently large the field near them does not depend on d. The field between the plates is zero (the fields from each plate are equal in strength but oppositely directed, and the field just outside either plate is the sum of the contributions from each plate. Therefore, the pressure on each plate is P 2 σ( E outside + E between 2 σ(2 σ 2ɛ + ˆd σ2 2ɛ ˆd Q2 2ɛ A 2 ˆd. Of course, this is the same as considering the force on one plate due to an electric field E σ 2ɛ ˆd due to the other plate. Problem 2 (Grifffiths optional Let s consider the sphere to be centered at the origin with its equator in the x y plane. The electrostatic pressure may be thought of as being due to the average field surrounding the charge on any given element of surface. The field is zero inside the sphere, and outside the field is E Q 4πɛ R ˆr. So, for any 2 given element of surface the pressure will be d P σ 2 ( E outside + E inside σ 2 ( Q ˆr + 4πɛ R2 4πɛ 8πR 4 ˆr 32π 2 ɛ R 4 ˆr.

2 Since the pressure is in the radial direction we know that only the components in the z direction will contribute to the overall force. Integrating the z components of the pressure over the upper hemisphere gives F P da z hemisphere π 2 2π dθ 32π 2 ɛ R 2 π 32πɛ R 2. dφ( 32π 2 ɛ R 4 ˆr (R2 sin θˆr cos θ Problem 3 (Griffiths 2.34(a The field due to this configuration is zero everywhere except for between the two shells. Thus, U ɛ E 2 dτ 2 allspace ɛ b ( 2 q 2 a 4πɛ r 2 4πr 2 dr ( q2 2 4πɛ a. b Problem 4 (Griffiths 3.26 To find the approximate potential (in the context of this chapter, anyway we should use the multipole expansion and keep only the first non-vanishing term. The monopole gives V ( r 4πɛ r 4πɛ r 2π 4πɛ r π 2 4πɛ r. V ρ( r dτ dr dθ dφ k R r 2 (R 2r sin θ r 2 sin θ π dr dθ kr(r 2r sin 2 θ dr kr(r 2r The total charge is zero, so the monopole term in the potential is zero. Thus, we should next check the dipole term. We have V ( r 4πɛ r 2 r cos Θρ( r dτ V 2

3 where Θ is the angle between r and r. In this case, we have been asked to find the potential at a point on the z axis, so Θ θ. Therefore V (z 4πɛ z 2 2π 4πɛ z 2. dr dθ dφ r cos θ k R r 2 (R 2r sin θ r 2 sin θ π dr dθ kr R(R 2r sin 2 θ cos θ This, too, is zero (this time because of the θ integral! So, we continue with V 2 (z 4πɛ z 3 4πɛ z 3 2πkR 5 4πɛ 6z 3 πkr5 92ɛ z 3. ( 3 dr dθ dφ r 2 2 cos2 θ 2 ( 3 dr dθ dφ 2 cos2 θ 2 π sin 2 θ dθ ( 3 2 cos2 θ 2 k R r 2 (R 2r sin θ r 2 sin θ kr(rr 2 2r 3 sin 2 θ This, then, is the approximate potential for points far from our charge distribution. Problem 5 (Griffiths 3.3 For a pure dipole the potential and field are given in the book, and they are exact (they are due to the dipole moment only, since all other terms in the multipole expansion vanish. We can easily calculate a F q (a,, q E(a,, qp 4πɛ a 3 (2 cos π 2 ˆr + sin π 2 ˆθ qp 4πɛ a 3 ˆθ. Note: at the point (a,,, ˆθ ẑ. b F q (,, a q E(,, a Note: at the point (,, a, ˆr ẑ. qp (2 cos ˆr + sin ˆθ 4πɛ a3 qp 2πɛ a 3 ˆr. 3

4 c The path one takes to move the charge from one point to another should not change the final answer. The fact that we are using an approximate potential does not change the fact that we still must satisfy Laplace s equation, and the field is still a conservative one. Therefore, W U U (,,a U (a,, qp cos 4πɛ a 2 qp cos π 2 4πɛ a 2 qp 4πɛ a 2. Problem 6 (Griffiths optional For such a distribution of point charges the integrals in the multipole expansion become sums over the point charges because ρ( r q δ( r r + q 2 δ( r r 2 + q 3 δ( r r 3. Therefore, the first term in the multipole expansion is The second term is V ( r 4πɛ r (q + q 2 + q 3 4πɛ q r. V ( r 4πɛ r 2 (q r cos Θ + q 2 r 2 cos Θ 2 + q 3 r 3 cos Θ 3. Here the only tricky thing is figuring out how to write Θ i. However, we can exploit the fact that r r i rr i cos Θ i to write V ( r 4πɛ r 2 ˆr (q r + q 2 r 2 + q 3 r 3 (sin θ cos φ, sin θ sin φ, cos θ (q(,, a q(, a, q(, a,. 4πɛ r2 Therefore, including only the two lowest terms in the multipole expansion, the potential from this three charge configuration is approximately V ( r V + V 2 [ q 4πɛ r + q ] r 2 (a cos θ a sin θ sin φ + a sin θ sin φ ( q 4πɛ r + q r 2 a cos θ. Apparently, the dipole contribution from the two charges on the y axis exactly cancel and we are left with only the dipole term from the charge on the z axis. 4

5 To find the field we simply take the gradient. Thus, E( r V ( r ( V r ˆr + r q 4πɛ V θ ˆθ + [( r 2 + a cos θ r 3 r sin θ V φ ˆφ ˆr + a sin θ ˆθ r 2 ] Problem 7 (Griffiths 3.6 We have seen that the potential (and, therefore, the field of such a configuration may be modeled by the placement of image charges that are the mirror-images (across the x y plane of the existing charges. Thus, the charge +q at z 3d experiences a force due to the field of three other point charges. It may be written as F ( q 2q 4πɛ (2d ( q 2 ẑ 72 4πɛ d 2q (4d 2 q (6d 2 ẑ Problem 8 (Griffiths 3. As seen in class, the image charges should be located as though they were the images in mirrors in the x and y planes. Thus, there are three images in all: an image q at the points (a, b and ( a, b, and an image q at the point ( a, b. The potential, then, is the potential of four point charges: V (x, y q ( 4πɛ (x a2 + (y b + 2 (x + a2 + (y + b 2 (x + a2 + (y b 2 (x a2 + (y + b 2 The force on the charge q at the point (a, b will be the same as if it were due to the three image charges only. So, F q (a, b [( q 2 ( 4πɛ (2a 2 + 2a ˆx + ((2a 2 + (2b (2b 2 + 2b ((2a 2 + (2b The potential due to the three images charges after q is brought to the point (a, b is [ ] V (a, b q 4πɛ 2a 2b +. (2a2 + (2b 2 So, the work required to bring the charge in from infinity is half the energy due to this potential (because the image charges accumulate for free. Therefore, W q 2 [ 4πɛ 4 a ] b +. a2 + b 2 5 ] ŷ.

6 This method does not work for any arbitrary angle of intersection of two conducting planes. In order for the reflections to terminate at some finite number of image charges, the set of the real charge and all of its reflections in the conducting planes must form an abelian group. This is fancy math-talk for saying that you should be able to get from the charge to any of it s images by successive application of a discreet set of operators. In our case, the operator that takes you from the real charge to one of its images is more than spacial. It should consist of a rotation about the point of intersection of the planes as well as a reversal of charge (the first image charge has the same sign as the real charge, but the next has opposite sign, etc.. This is a very formal proof, but your reasoning may be more straight forward and less rigorous. The short story is, only those situations in which the angle of intersection of the planes is an even divisor of 36 will be solvable via the image charge method. So, for the image method to work we require θ intersection 36 ; n {, 2, 3, }. 2n 6

Electromagnetism: Worked Examples. University of Oxford Second Year, Part A2

Electromagnetism: Worked Examples. University of Oxford Second Year, Part A2 Electromagnetism: Worked Examples University of Oxford Second Year, Part A2 Caroline Terquem Department of Physics caroline.terquem@physics.ox.ac.uk Michaelmas Term 2017 2 Contents 1 Potentials 5 1.1 Potential

More information

Department of Physics IIT Kanpur, Semester II,

Department of Physics IIT Kanpur, Semester II, Department of Physics IIT Kanpur, Semester II, 7-8 PHYA: Physics II Solution # 4 Instructors: AKJ & SC Solution 4.: Force with image charges (Griffiths rd ed. Prob.6 As far as force is concerned, this

More information

Indiana University Physics P331: Theory of Electromagnetism Review Problems #3

Indiana University Physics P331: Theory of Electromagnetism Review Problems #3 Indiana University Physics P331: Theory of Electromagnetism Review Problems #3 Note: The final exam (Friday 1/14 8:00-10:00 AM will be comprehensive, covering lecture and homework material pertaining to

More information

Expansion of 1/r potential in Legendre polynomials

Expansion of 1/r potential in Legendre polynomials Expansion of 1/r potential in Legendre polynomials In electrostatics and gravitation, we see scalar potentials of the form V = K d Take d = R r = R 2 2Rr cos θ + r 2 = R 1 2 r R cos θ + r R )2 Use h =

More information

Scattering. March 20, 2016

Scattering. March 20, 2016 Scattering March 0, 06 The scattering of waves of any kind, by a compact object, has applications on all scales, from the scattering of light from the early universe by intervening galaxies, to the scattering

More information

Problem Set #4: 4.1,4.7,4.9 (Due Monday, March 25th)

Problem Set #4: 4.1,4.7,4.9 (Due Monday, March 25th) Chapter 4 Multipoles, Dielectrics Problem Set #4: 4.,4.7,4.9 (Due Monday, March 5th 4. Multipole expansion Consider a localized distribution of charges described by ρ(x contained entirely in a sphere of

More information

Physics 3323, Fall 2016 Problem Set 2 due Sep 9, 2016

Physics 3323, Fall 2016 Problem Set 2 due Sep 9, 2016 Physics 3323, Fall 26 Problem Set 2 due Sep 9, 26. What s my charge? A spherical region of radius R is filled with a charge distribution that gives rise to an electric field inside of the form E E /R 2

More information

0.4. math. r r = r 2 + r 2 2r r cos(θ ) 0.5. Coordiates.

0.4. math. r r = r 2 + r 2 2r r cos(θ ) 0.5. Coordiates. .. Electrostatics... Multipole. F = E(r) = E(r) = S n qq i r r i 3 (r r i) i= n i= q i r r i 3 (r r i) d 3 r ρ(r ) 3 (r r ) ds E(r) = 4πQ enclosed E(r) = 4πρ(r ) E(r) = Φ(r) d 3 r ρ(r ) r dl E W = q(φ(r

More information

7 Curvilinear coordinates

7 Curvilinear coordinates 7 Curvilinear coordinates Read: Boas sec. 5.4, 0.8, 0.9. 7. Review of spherical and cylindrical coords. First I ll review spherical and cylindrical coordinate systems so you can have them in mind when

More information

Magnetostatics III Magnetic Vector Potential (Griffiths Chapter 5: Section 4)

Magnetostatics III Magnetic Vector Potential (Griffiths Chapter 5: Section 4) Dr. Alain Brizard Electromagnetic Theory I PY ) Magnetostatics III Magnetic Vector Potential Griffiths Chapter 5: Section ) Vector Potential The magnetic field B was written previously as Br) = Ar), 1)

More information

Errata Instructor s Solutions Manual Introduction to Electrodynamics, 3rd ed Author: David Griffiths Date: September 1, 2004

Errata Instructor s Solutions Manual Introduction to Electrodynamics, 3rd ed Author: David Griffiths Date: September 1, 2004 Errata Instructor s Solutions Manual Introduction to Electrodynamics, 3rd ed Author: David Griffiths Date: September, 004 Page 4, Prob..5 (b): last expression should read y +z +3x. Page 4, Prob..6: at

More information

(b) For the system in question, the electric field E, the displacement D, and the polarization P = D ɛ 0 E are as follows. r2 0 inside the sphere,

(b) For the system in question, the electric field E, the displacement D, and the polarization P = D ɛ 0 E are as follows. r2 0 inside the sphere, PHY 35 K. Solutions for the second midterm exam. Problem 1: a The boundary conditions at the oil-air interface are air side E oil side = E and D air side oil side = D = E air side oil side = ɛ = 1+χ E.

More information

Topic 7. Electric flux Gauss s Law Divergence of E Application of Gauss Law Curl of E

Topic 7. Electric flux Gauss s Law Divergence of E Application of Gauss Law Curl of E Topic 7 Electric flux Gauss s Law Divergence of E Application of Gauss Law Curl of E urface enclosing an electric dipole. urface enclosing charges 2q and q. Electric flux Flux density : The number of field

More information

Electric fields in matter

Electric fields in matter Electric fields in matter November 2, 25 Suppose we apply a constant electric field to a block of material. Then the charges that make up the matter are no longer in equilibrium: the electrons tend to

More information

University of Illinois at Chicago Department of Physics

University of Illinois at Chicago Department of Physics University of Illinois at Chicago Department of Physics Electromagnetism Qualifying Examination January 4, 2017 9.00 am - 12.00 pm Full credit can be achieved from completely correct answers to 4 questions.

More information

The Helmholtz theorem at last!

The Helmholtz theorem at last! Problem. The Helmholtz theorem at last! Recall in class the Helmholtz theorem that says that if if E =0 then E can be written as E = φ () if B =0 then B can be written as B = A (2) (a) Let n be a unit

More information

Multipole moments. November 9, 2015

Multipole moments. November 9, 2015 Multipole moments November 9, 5 The far field expansion Suppose we have a localized charge distribution, confined to a region near the origin with r < R. Then for values of r > R, the electric field must

More information

FORMULA SHEET FOR QUIZ 2 Exam Date: November 8, 2017

FORMULA SHEET FOR QUIZ 2 Exam Date: November 8, 2017 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II November 5, 207 Prof. Alan Guth FORMULA SHEET FOR QUIZ 2 Exam Date: November 8, 207 A few items below are marked

More information

PHY481 - Outline of solutions to Homework 3

PHY481 - Outline of solutions to Homework 3 1 PHY481 - Outline of solutions to Homework 3 Problem 3.8: We consider a charge outside a conducting sphere that is neutral. In order that the sphere be neutral, we have to introduce a new image charge

More information

EE 333 Electricity and Magnetism, Fall 2009 Homework #9 solution

EE 333 Electricity and Magnetism, Fall 2009 Homework #9 solution EE 333 Electricity and Magnetism, Fall 009 Homework #9 solution 4.10. The two infinite conducting cones θ = θ 1, and θ = θ are maintained at the two potentials Φ 1 = 100, and Φ = 0, respectively, as shown

More information

Summary: Applications of Gauss Law

Summary: Applications of Gauss Law Physics 2460 Electricity and Magnetism I, Fall 2006, Lecture 15 1 Summary: Applications of Gauss Law 1. Field outside of a uniformly charged sphere of radius a: 2. An infinite, uniformly charged plane

More information

Summary: Curvilinear Coordinates

Summary: Curvilinear Coordinates Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 10 1 Summary: Curvilinear Coordinates 1. Summary of Integral Theorems 2. Generalized Coordinates 3. Cartesian Coordinates: Surfaces of Constant

More information

Fall Lee - Midterm 2 solutions

Fall Lee - Midterm 2 solutions Fall 2009 - Lee - Midterm 2 solutions Problem 1 Solutions Part A Because the middle slab is a conductor, the electric field inside of the slab must be 0. Parts B and C Recall that to find the electric

More information

1. (3) Write Gauss Law in differential form. Explain the physical meaning.

1. (3) Write Gauss Law in differential form. Explain the physical meaning. Electrodynamics I Midterm Exam - Part A - Closed Book KSU 204/0/23 Name Electro Dynamic Instructions: Use SI units. Where appropriate, define all variables or symbols you use, in words. Try to tell about

More information

Physics 6C Review 1. Eric Reichwein Department of Physics University of California, Santa Cruz. July 16, Figure 1: Coulombs Law

Physics 6C Review 1. Eric Reichwein Department of Physics University of California, Santa Cruz. July 16, Figure 1: Coulombs Law Physics 6C Review 1 Eric Reichwein Department of Physics University of California, Santa Cruz July 16, 2012 1 Review 1.1 Coulombs Law Figure 1: Coulombs Law The steps for solving any problem of this type

More information

CHAPTER 3 POTENTIALS 10/13/2016. Outlines. 1. Laplace s equation. 2. The Method of Images. 3. Separation of Variables. 4. Multipole Expansion

CHAPTER 3 POTENTIALS 10/13/2016. Outlines. 1. Laplace s equation. 2. The Method of Images. 3. Separation of Variables. 4. Multipole Expansion CHAPTER 3 POTENTIALS Lee Chow Department of Physics University of Central Florida Orlando, FL 32816 Outlines 1. Laplace s equation 2. The Method of Images 3. Separation of Variables 4. Multipole Expansion

More information

xy 2 e 2z dx dy dz = 8 3 (1 e 4 ) = 2.62 mc. 12 x2 y 3 e 2z 2 m 2 m 2 m Figure P4.1: Cube of Problem 4.1.

xy 2 e 2z dx dy dz = 8 3 (1 e 4 ) = 2.62 mc. 12 x2 y 3 e 2z 2 m 2 m 2 m Figure P4.1: Cube of Problem 4.1. Problem 4.1 A cube m on a side is located in the first octant in a Cartesian coordinate system, with one of its corners at the origin. Find the total charge contained in the cube if the charge density

More information

Physics 505 Fall 2005 Homework Assignment #7 Solutions

Physics 505 Fall 2005 Homework Assignment #7 Solutions Physics 505 Fall 005 Homework Assignment #7 Solutions Textbook problems: Ch. 4: 4.10 Ch. 5: 5.3, 5.6, 5.7 4.10 Two concentric conducting spheres of inner and outer radii a and b, respectively, carry charges

More information

UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 110A. Homework #7. Benjamin Stahl. March 3, 2015

UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 110A. Homework #7. Benjamin Stahl. March 3, 2015 UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS A Homework #7 Benjamin Stahl March 3, 5 GRIFFITHS, 5.34 It will be shown that the magnetic field of a dipole can written in the following

More information

Homework Assignment 6 Solution Set

Homework Assignment 6 Solution Set Homework Assignment 6 Solution Set PHYCS 440 Mrch, 004 Prolem (Griffiths 4.6 One wy to find the energy is to find the E nd D fields everywhere nd then integrte the energy density for those fields. We know

More information

Physics 342 Lecture 4. Probability Density. Lecture 4. Physics 342 Quantum Mechanics I

Physics 342 Lecture 4. Probability Density. Lecture 4. Physics 342 Quantum Mechanics I Physics 342 Lecture 4 Probability Density Lecture 4 Physics 342 Quantum Mechanics I Monday, February 4th, 28 We review the basic notions of probability, in particular the role of probability density in

More information

Curvilinear Coordinates

Curvilinear Coordinates University of Alabama Department of Physics and Astronomy PH 106-4 / LeClair Fall 2008 Curvilinear Coordinates Note that we use the convention that the cartesian unit vectors are ˆx, ŷ, and ẑ, rather than

More information

Solutions to PS 2 Physics 201

Solutions to PS 2 Physics 201 Solutions to PS Physics 1 1. ke dq E = i (1) r = i = i k eλ = i k eλ = i k eλ k e λ xdx () (x x) (x x )dx (x x ) + x dx () (x x ) x ln + x x + x x (4) x + x ln + x (5) x + x To find the field for x, we

More information

B Field Creation Detecting B fields. Magnetic Fields. PHYS David Blasing. Wednesday June 26th 1 / 26

B Field Creation Detecting B fields. Magnetic Fields. PHYS David Blasing. Wednesday June 26th 1 / 26 Magnetic Fields PHYS 272 - David Blasing Wednesday June 26th 1 / 26 Magnetic ( B) Fields This is a significant change, until now we have discussed just E fields. Now we are talking about a totally different

More information

Classical Field Theory: Electrostatics-Magnetostatics

Classical Field Theory: Electrostatics-Magnetostatics Classical Field Theory: Electrostatics-Magnetostatics April 27, 2010 1 1 J.D.Jackson, Classical Electrodynamics, 2nd Edition, Section 1-5 Electrostatics The behavior of an electrostatic field can be described

More information

Electromagnetism Answers to Problem Set 9 Spring 2006

Electromagnetism Answers to Problem Set 9 Spring 2006 Electromagnetism 70006 Answers to Problem et 9 pring 006. Jackson Prob. 5.: Reformulate the Biot-avart law in terms of the solid angle subtended at the point of observation by the current-carrying circuit.

More information

PHYS 110B - HW #4 Spring 2004, Solutions by David Pace Any referenced equations are from Griffiths Problem statements are paraphrased

PHYS 110B - HW #4 Spring 2004, Solutions by David Pace Any referenced equations are from Griffiths Problem statements are paraphrased PHYS 11B - HW #4 Spring 4, Solutions by David Pace Any referenced equations are from Griffiths Problem statements are paraphrased [1.] Problem 8. from Griffiths Reference problem 7.31 figure 7.43. a Let

More information

Maxwell s equations for electrostatics

Maxwell s equations for electrostatics Maxwell s equations for electrostatics October 6, 5 The differential form of Gauss s law Starting from the integral form of Gauss s law, we treat the charge as a continuous distribution, ρ x. Then, letting

More information

Problem Set #3: 2.11, 2.15, 2.21, 2.26, 2.40, 2.42, 2.43, 2.46 (Due Thursday Feb. 27th)

Problem Set #3: 2.11, 2.15, 2.21, 2.26, 2.40, 2.42, 2.43, 2.46 (Due Thursday Feb. 27th) Chapter Electrostatics Problem Set #3:.,.5,.,.6,.40,.4,.43,.46 (Due Thursday Feb. 7th). Coulomb s Law Coulomb showed experimentally that for two point charges the force is - proportional to each of the

More information

Physics 342 Lecture 23. Radial Separation. Lecture 23. Physics 342 Quantum Mechanics I

Physics 342 Lecture 23. Radial Separation. Lecture 23. Physics 342 Quantum Mechanics I Physics 342 Lecture 23 Radial Separation Lecture 23 Physics 342 Quantum Mechanics I Friday, March 26th, 2010 We begin our spherical solutions with the simplest possible case zero potential. Aside from

More information

Solution to Homework 1. Vector Analysis (P201)

Solution to Homework 1. Vector Analysis (P201) Solution to Homework 1. Vector Analysis (P1) Q1. A = 3î + ĵ ˆk, B = î + 3ĵ + 4ˆk, C = 4î ĵ 6ˆk. The sides AB, BC and AC are, AB = 4î ĵ 6ˆk AB = 7.48 BC = 5î + 5ĵ + 1ˆk BC = 15 1.5 CA = î 3ĵ 4ˆk CA = 6

More information

Physics 7B Midterm 2 Solutions - Fall 2017 Professor R. Birgeneau

Physics 7B Midterm 2 Solutions - Fall 2017 Professor R. Birgeneau Problem 1 Physics 7B Midterm 2 Solutions - Fall 217 Professor R. Birgeneau (a) Since the wire is a conductor, the electric field on the inside is simply zero. To find the electric field in the exterior

More information

Electromagnetism Answers to Problem Set 8 Spring Jackson Prob. 4.1: Multipole expansion for various charge distributions

Electromagnetism Answers to Problem Set 8 Spring Jackson Prob. 4.1: Multipole expansion for various charge distributions Electromagnetism 76 Answers to Problem Set 8 Spring 6. Jackson Prob. 4.: Multipole expansion for various charge distributions (a) In the first case, we have 4 charges in the xy plane at distance a from

More information

3.4-7 First check to see if the loop is indeed electromagnetically small. Ie sinθ ˆφ H* = 2. ˆrr 2 sinθ dθ dφ =

3.4-7 First check to see if the loop is indeed electromagnetically small. Ie sinθ ˆφ H* = 2. ˆrr 2 sinθ dθ dφ = ECE 54/4 Spring 17 Assignment.4-7 First check to see if the loop is indeed electromagnetically small f 1 MHz c 1 8 m/s b.5 m λ = c f m b m Yup. (a) You are welcome to use equation (-5), but I don t like

More information

Physics 342 Lecture 22. The Hydrogen Atom. Lecture 22. Physics 342 Quantum Mechanics I

Physics 342 Lecture 22. The Hydrogen Atom. Lecture 22. Physics 342 Quantum Mechanics I Physics 342 Lecture 22 The Hydrogen Atom Lecture 22 Physics 342 Quantum Mechanics I Friday, March 28th, 2008 We now begin our discussion of the Hydrogen atom. Operationally, this is just another choice

More information

Written Examination. Antennas and Propagation (AA ) June 22, 2018.

Written Examination. Antennas and Propagation (AA ) June 22, 2018. Written Examination Antennas and Propagation (AA. 7-8 June, 8. Problem ( points A circular loop of radius a = cm is positioned at a height h over a perfectly electric conductive ground plane as in figure,

More information

UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 110A. Homework #6. Benjamin Stahl. February 17, 2015

UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 110A. Homework #6. Benjamin Stahl. February 17, 2015 UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS A Homework #6 Benjamin Stahl February 7, 5 GRIFFITHS, 5.9 The magnetic field at a point, P, will be found for each of the steady current

More information

Solution Set Eight. 1 Problem #1: Toroidal Electromagnet with Gap Problem #4: Self-Inductance of a Long Solenoid. 9

Solution Set Eight. 1 Problem #1: Toroidal Electromagnet with Gap Problem #4: Self-Inductance of a Long Solenoid. 9 : Solution Set Eight Northwestern University, Electrodynamics I Wednesday, March 9, 6 Contents Problem #: Toroidal Electromagnet with Gap. Problem #: Electromagnetic Momentum. 3 3 Problem #3: Type I Superconductor.

More information

Electrostatics. Chapter Maxwell s Equations

Electrostatics. Chapter Maxwell s Equations Chapter 1 Electrostatics 1.1 Maxwell s Equations Electromagnetic behavior can be described using a set of four fundamental relations known as Maxwell s Equations. Note that these equations are observed,

More information

NYU Physics Preliminary Examination in Electricity & Magnetism Fall 2011

NYU Physics Preliminary Examination in Electricity & Magnetism Fall 2011 This is a closed-book exam. No reference materials of any sort are permitted. Full credit will be given for complete solutions to the following five questions. 1. An impenetrable sphere of radius a carries

More information

Physics 342 Lecture 27. Spin. Lecture 27. Physics 342 Quantum Mechanics I

Physics 342 Lecture 27. Spin. Lecture 27. Physics 342 Quantum Mechanics I Physics 342 Lecture 27 Spin Lecture 27 Physics 342 Quantum Mechanics I Monday, April 5th, 2010 There is an intrinsic characteristic of point particles that has an analogue in but no direct derivation from

More information

Vector Integrals. Scott N. Walck. October 13, 2016

Vector Integrals. Scott N. Walck. October 13, 2016 Vector Integrals cott N. Walck October 13, 16 Contents 1 A Table of Vector Integrals Applications of the Integrals.1 calar Line Integral.........................1.1 Finding Total Charge of a Line Charge..........1.

More information

WKB Approximation in 3D

WKB Approximation in 3D 1 WKB Approximation in 3D We see solutions ψr of the stationary Schrodinger equations for a spinless particle of energy E: 2 2m 2 ψ + V rψ = Eψ At rst, we just rewrite the Schrodinger equation in the following

More information

Junior-level Electrostatics Content Review

Junior-level Electrostatics Content Review Junior-level Electrostatics Content Review Please fill out the following exam to the best of your ability. This will not count towards your final grade in the course. Do your best to get to all the questions

More information

Electrodynamics I Midterm - Part A - Closed Book KSU 2005/10/17 Electro Dynamic

Electrodynamics I Midterm - Part A - Closed Book KSU 2005/10/17 Electro Dynamic Electrodynamics I Midterm - Part A - Closed Book KSU 5//7 Name Electro Dynamic. () Write Gauss Law in differential form. E( r) =ρ( r)/ɛ, or D = ρ, E= electricfield,ρ=volume charge density, ɛ =permittivity

More information

PHY 2049: Introductory Electromagnetism

PHY 2049: Introductory Electromagnetism PHY 2049: Introductory Electromagnetism Notes From the Text 1 Introduction Chris Mueller Dept. of Physics, University of Florida 3 January, 2009 The subject of this course is the physics of electricity

More information

Problem Set #5: 5.7,5.9,5.13 (Due Monday, April 8th)

Problem Set #5: 5.7,5.9,5.13 (Due Monday, April 8th) Chapter 5 Magnetostatics Problem Set #5: 5.7,5.9,5.13 (Due Monday, April 8th) 5.1 Biot-Savart Law So far we were concerned with static configuration of charges known as electrostatics. We now switch to

More information

3: Gauss s Law July 7, 2008

3: Gauss s Law July 7, 2008 3: Gauss s Law July 7, 2008 3.1 Electric Flux In order to understand electric flux, it is helpful to take field lines very seriously. Think of them almost as real things that stream out from positive charges

More information

Supporting Information

Supporting Information Supporting Information A: Calculation of radial distribution functions To get an effective propagator in one dimension, we first transform 1) into spherical coordinates: x a = ρ sin θ cos φ, y = ρ sin

More information

G( x x, t) = 1 (2π) 3. d 3 k e k2 t/µσ e i k ( x x ) A( k, t)e k x d 3 k. A( x, t) = 1

G( x x, t) = 1 (2π) 3. d 3 k e k2 t/µσ e i k ( x x ) A( k, t)e k x d 3 k. A( x, t) = 1 Physics 505 Fall 2005 Homework Assignment #10 Solutions Textbook problems: Ch. 6: 6.3, 6.4, 6.14, 6.18 6.3 The homogeneous diffusion equation (5.160) for the vector potential for quasi-static fields in

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

3. Calculating Electrostatic Potential

3. Calculating Electrostatic Potential 3. Calculating Electrostatic Potential 3. Laplace s Equation 3. The Method of Images 3.3 Separation of Variables 3.4 Multipole Expansion 3.. Introduction The primary task of electrostatics is to study

More information

Tutorial 3 - Solutions Electromagnetic Waves

Tutorial 3 - Solutions Electromagnetic Waves Tutorial 3 - Solutions Electromagnetic Waves You can find formulas you require for vector calculus at the end of this tutorial. 1. Find the Divergence and Curl of the following functions - (a) k r ˆr f

More information

Transformed E&M I homework. Multipole Expansion (Griffiths Chapter 3)

Transformed E&M I homework. Multipole Expansion (Griffiths Chapter 3) Transformed E&M I homework Multipole Expansion (Griffiths Chapter 3) Multipole Expansion Question 1. Multipole moment of charged wire CALCULATION; EXPANSION (U. Nauenberg, HW3, solutions available) A charge

More information

Math 221 Examination 2 Several Variable Calculus

Math 221 Examination 2 Several Variable Calculus Math Examination Spring Instructions These problems should be viewed as essa questions. Before making a calculation, ou should explain in words what our strateg is. Please write our solutions on our own

More information

A Catalog of Hidden Momenta

A Catalog of Hidden Momenta A Catalog of Hidden Momenta David J. Griffiths Department of Physics, Reed College Portland, Oregon 9722 May 12, 218 Abstract Electromagnetic fields carry momentum: P em = ɛ (E B dτ. But if the center-of-energy

More information

ECE 3025: Electromagnetics TEST 2 (Fall 2006)

ECE 3025: Electromagnetics TEST 2 (Fall 2006) Name: GTID: ECE 3025: Electromagnetics TEST 2 (Fall 2006) Please read all instructions before continuing with the test. This is a closed notes, closed book, closed calculator, closed friend, open mind

More information

Read this cover page completely before you start.

Read this cover page completely before you start. I affirm that I have worked this exam independently, without texts, outside help, integral tables, calculator, solutions, or software. (Please sign legibly.) Read this cover page completely before you

More information

3 Chapter. Gauss s Law

3 Chapter. Gauss s Law 3 Chapter Gauss s Law 3.1 Electric Flux... 3-2 3.2 Gauss s Law (see also Gauss s Law Simulation in Section 3.10)... 3-4 Example 3.1: Infinitely Long Rod of Uniform Charge Density... 3-9 Example 3.2: Infinite

More information

Electrodynamics Exam Solutions

Electrodynamics Exam Solutions Electrodynamics Exam Solutions Name: FS 215 Prof. C. Anastasiou Student number: Exercise 1 2 3 4 Total Max. points 15 15 15 15 6 Points Visum 1 Visum 2 The exam lasts 18 minutes. Start every new exercise

More information

PHYS 281: Midterm Exam

PHYS 281: Midterm Exam PHYS 28: Midterm Exam October 28, 200, 8:00-9:20 Last name (print): Initials: No calculator or other aids allowed PHYS 28: Midterm Exam Instructor: B. R. Sutherland Date: October 28, 200 Time: 8:00-9:20am

More information

Quiz 4 (Discussion Session) Phys 1302W.400 Spring 2018

Quiz 4 (Discussion Session) Phys 1302W.400 Spring 2018 Quiz 4 (Discussion ession) Phys 1302W.400 pring 2018 This group quiz consists of one problem that, together with the individual problems on Friday, will determine your grade for quiz 4. For the group problem,

More information

lim = F F = F x x + F y y + F z

lim = F F = F x x + F y y + F z Physics 361 Summary of Results from Lecture Physics 361 Derivatives of Scalar and Vector Fields The gradient of a scalar field f( r) is given by g = f. coordinates f g = ê x x + ê f y y + ê f z z Expressed

More information

where the last equality follows from the divergence theorem. Now since we did this for an arbitrary volume τ, it must hold locally:

where the last equality follows from the divergence theorem. Now since we did this for an arbitrary volume τ, it must hold locally: 8 Electrodynamics Read: Boas Ch. 6, particularly sec. 10 and 11. 8.1 Maxwell equations Some of you may have seen Maxwell s equations on t-shirts or encountered them briefly in electromagnetism courses.

More information

Conductors and Insulators

Conductors and Insulators Conductors and Insulators Lecture 11: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay Self Energy of a Charge Distribution : In Lecture 1 we briefly discussed what we called

More information

Vector Potential for the Magnetic Field

Vector Potential for the Magnetic Field Vector Potential for the Magnetic Field Let me start with two two theorems of Vector Calculus: Theorem 1: If a vector field has zero curl everywhere in space, then that field is a gradient of some scalar

More information

l=0 The expansion coefficients can be determined, for example, by finding the potential on the z-axis and expanding that result in z.

l=0 The expansion coefficients can be determined, for example, by finding the potential on the z-axis and expanding that result in z. Electrodynamics I Exam - Part A - Closed Book KSU 15/11/6 Name Electrodynamic Score = 14 / 14 points Instructions: Use SI units. Where appropriate, define all variables or symbols you use, in words. Try

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

Chapter 21 Electric Charge and the Electric Field

Chapter 21 Electric Charge and the Electric Field Chapter 21 Electric Charge and the Electric Field 1 Electric Charge Electrostatics is the study of charges when they are stationery. Figure 1: This is Fig. 21.1 and it shows how negatively charged objects

More information

Homework Two. Abstract: Liu. Solutions for Homework Problems Two: (50 points total). Collected by Junyu

Homework Two. Abstract: Liu. Solutions for Homework Problems Two: (50 points total). Collected by Junyu Homework Two Abstract: Liu. Solutions for Homework Problems Two: (50 points total). Collected by Junyu Contents 1 BT Problem 13.15 (8 points) (by Nick Hunter-Jones) 1 2 BT Problem 14.2 (12 points: 3+3+3+3)

More information

ECE 3209 Electromagnetic Fields Final Exam Example. University of Virginia Solutions

ECE 3209 Electromagnetic Fields Final Exam Example. University of Virginia Solutions ECE 3209 Electromagnetic Fields Final Exam Example University of Virginia Solutions (print name above) This exam is closed book and closed notes. Please perform all work on the exam sheets in a neat and

More information

Applications of Ampere s Law

Applications of Ampere s Law Applications of Ampere s Law In electrostatics, the electric field due to any known charge distribution ρ(x, y, z) may alwaysbeobtainedfromthecoulomblaw it sauniversal tool buttheactualcalculation is often

More information

EECS 117 Lecture 7: Electrostatics Review

EECS 117 Lecture 7: Electrostatics Review EECS 117 Lecture 7: Electrostatics Review Prof. Niknejad University of California, Berkeley University of California, Berkeley EECS 117 Lecture 7 p. 1/19 Existence of Charge Charge, like mass, is an intrinsic

More information

Physics 525, Condensed Matter Homework 8 Due Thursday, 14 th December 2006

Physics 525, Condensed Matter Homework 8 Due Thursday, 14 th December 2006 Physics 525, Condensed Matter Homework 8 Due Thursday, 14 th December 2006 Jacob Lewis Bourjaily Problem 1: Little-Parks Experiment Consider a long, thin-walled, hollow cylinder of radius R and thickness

More information

Physics 7B, Speliotopoulos Final Exam, Spring 2014 Berkeley, CA

Physics 7B, Speliotopoulos Final Exam, Spring 2014 Berkeley, CA Physics 7B, Speliotopoulos Final Exam, Spring 4 Berkeley, CA Rules: This final exam is closed book and closed notes. In particular, calculators are not allowed during this exam. Cell phones must be turned

More information

University of Alabama Department of Physics and Astronomy. PH 125 / LeClair Spring A Short Math Guide. Cartesian (x, y) Polar (r, θ)

University of Alabama Department of Physics and Astronomy. PH 125 / LeClair Spring A Short Math Guide. Cartesian (x, y) Polar (r, θ) University of Alabama Department of Physics and Astronomy PH 125 / LeClair Spring 2009 A Short Math Guide 1 Definition of coordinates Relationship between 2D cartesian (, y) and polar (r, θ) coordinates.

More information

Electromagnetism Physics 15b

Electromagnetism Physics 15b Electromagnetism Physics 15b Lecture #5 Curl Conductors Purcell 2.13 3.3 What We Did Last Time Defined divergence: Defined the Laplacian: From Gauss s Law: Laplace s equation: F da divf = lim S V 0 V Guass

More information

Lecture 17 - The Secrets we have Swept Under the Rug

Lecture 17 - The Secrets we have Swept Under the Rug 1.0 0.5 0.0-0.5-0.5 0.0 0.5 1.0 Lecture 17 - The Secrets we have Swept Under the Rug A Puzzle... What makes 3D Special? Example (1D charge distribution) A stick with uniform charge density λ lies between

More information

Physics 7B, Speliotopoulos Final Exam, Fall 2014 Berkeley, CA

Physics 7B, Speliotopoulos Final Exam, Fall 2014 Berkeley, CA Physics 7B, Speliotopoulos Final Exam, Fall 4 Berkeley, CA Rules: This final exam is closed book and closed notes. In particular, calculators are not allowed during this exam. Cell phones must be turned

More information

Phys. 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel Problem Set 1

Phys. 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel Problem Set 1 Phys. 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel Problem Set Problem.3 a): By symmetry, the solution must be of the form ρ(x) = ρ(r) = Qδ(r R)f, with a constant f to be specified by the condition

More information

Physics 505 Fall Homework Assignment #9 Solutions

Physics 505 Fall Homework Assignment #9 Solutions Physics 55 Fall 25 Textbook problems: Ch. 5: 5.2, 5.22, 5.26 Ch. 6: 6.1 Homework Assignment #9 olutions 5.2 a) tarting from the force equation (5.12) and the fact that a magnetization M inside a volume

More information

Integrals in Electrostatic Problems

Integrals in Electrostatic Problems PHYS 119 Integrals in Electrostatic Problems Josh McKenney University of North Carolina at Chapel Hill (Dated: January 6, 2016) 1 FIG. 1. Three positive charges positioned at equal distances around an

More information

Lecture 13: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay. Poisson s and Laplace s Equations

Lecture 13: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay. Poisson s and Laplace s Equations Poisson s and Laplace s Equations Lecture 13: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay We will spend some time in looking at the mathematical foundations of electrostatics.

More information

Field Theory exam II Solutions

Field Theory exam II Solutions Field Theory exam II Solutions Problem 1 (a) Consider point charges, one with charge q located at x 1 = (1, 0, 1), and the other one with charge q at x = (1, 0, 1). Compute the multipole moments q lm in

More information

Problems in Magnetostatics

Problems in Magnetostatics Problems in Magnetostatics 8th February 27 Some of the later problems are quite challenging. This is characteristic of problems in magnetism. There are trivial problems and there are tough problems. Very

More information

Notes: Most of the material presented in this chapter is taken from Jackson, Chap. 2, 3, and 4, and Di Bartolo, Chap. 2. 2π nx i a. ( ) = G n.

Notes: Most of the material presented in this chapter is taken from Jackson, Chap. 2, 3, and 4, and Di Bartolo, Chap. 2. 2π nx i a. ( ) = G n. Chapter. Electrostatic II Notes: Most of the material presented in this chapter is taken from Jackson, Chap.,, and 4, and Di Bartolo, Chap... Mathematical Considerations.. The Fourier series and the Fourier

More information

Physics 506 Winter 2008 Homework Assignment #4 Solutions. Textbook problems: Ch. 9: 9.6, 9.11, 9.16, 9.17

Physics 506 Winter 2008 Homework Assignment #4 Solutions. Textbook problems: Ch. 9: 9.6, 9.11, 9.16, 9.17 Physics 56 Winter 28 Homework Assignment #4 Solutions Textbook problems: Ch. 9: 9.6, 9., 9.6, 9.7 9.6 a) Starting from the general expression (9.2) for A and the corresponding expression for Φ, expand

More information

The dielectrophoretic force on dielectric spherical objects

The dielectrophoretic force on dielectric spherical objects 10064 - Physics Project The dielectrophoretic force on dielectric spherical objects Andreas Ammitzbøll s042191 Janosch Michael Cristoph Rauba s042606 Marco Haller Schultz s042357 Supervisors: Niels Asger

More information

E & M TEST BANK. August 2016

E & M TEST BANK. August 2016 August 2016 E & M TEST BANK 1. Charge in an Atom [500 level] A certain charge distribution ρ(r) generates an electric potential Φ(r) = e 0a e 2r/a ( a r + 1), with constants e 0 and a. (a) Find the electric

More information

PHYSICS 7B, Section 1 Fall 2013 Midterm 2, C. Bordel Monday, November 4, pm-9pm. Make sure you show your work!

PHYSICS 7B, Section 1 Fall 2013 Midterm 2, C. Bordel Monday, November 4, pm-9pm. Make sure you show your work! PHYSICS 7B, Section 1 Fall 2013 Midterm 2, C. Bordel Monday, November 4, 2013 7pm-9pm Make sure you show your work! Problem 1 - Current and Resistivity (20 pts) a) A cable of diameter d carries a current

More information