Exam 1 Practice Problems Solutions

Save this PDF as:

Size: px
Start display at page:

Transcription

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8 Spring 13 Exam 1 Practice Problems Solutions Part I: Short Questions and Concept Questions Problem 1: Spark Plug Pictured at right is a typical spark plug (for scale, the thread diameter is about 1 mm) About what voltage does your car ignition system need to generate to make a spark, if the breakdown field in a gas/air mixture is about 1 times higher than in air? For those of you unfamiliar with spark plugs, the spark is generated in the gap at the left of the top picture (top of the bottom picture) The white on the right is a ceramic which acts as an insulator between the high voltage center and the grounded outer (threaded) part Answer From the picture, if the threads have a diameter of 1 mm, then the gap is about 5 mm The breakdown is 1 times higher than in air, so its 3 x 1 7 V/m So to get breakdown you need a potential difference across the gap of V Ed = (3"1 7 V #m \$1 )(5"1 \$4 m) = 15"1 4 V 1

2 Problem Consider the three charges + Q, + Q, and Q, and a mathematical spherical surface (it does not physically exist) as shown in the figure below What is the net electric flux on the spherical surface between the charges? Briefly explain your answer Answer Only the positive charge + Q enclosed in the surface contributes to the electric flux on that surface Since the charge enclosed is positive, the electric flux on the surface is also positive and equal to + Q / All three charges do contribute to the electric field However if a charge is outside a closed surface then the electric flux on that surface due only to that charge is zero The two charges + Q and Q lie outside the surface, so the electric flux on the surface due to those two charges is zero

3 Problem 3: A pyramid has a square base of side a, and four faces which are equilateral triangles A charge Q is placed on the center of the base of the pyramid What is the net flux of electric field emerging from one of the triangular faces of the pyramid? 1 Q 8 3 Qa 4 Q 5 Undetermined: we must know whether Q is infinitesimally above or below the plane? Answer : Explain your reasoning: Construct an eight faced closed surface consisting of two pyramids with the charge at the center The total flux by Gauss s law is justq / Since each face is identical, the flux through each face is one eight the total flux or Q / 8 3

4 Problem 4: The work done by an external agent in moving a positively charged object that starts from rest at infinity and ends at rest at the point P midway between two charges of magnitude +Q and Q, 1 is positive is negative 3 is zero 4 cannot be determined not enough info is given Answer negative The work done by the external agent in moving the positive charge q from infinity to the point P is equal to the negative of the work done by the electrostatic force Each charged object is the same distance d from the point P The work done by the electrostatic force is equal to the potential energy difference between infinity and the point P, which is given by # Q U = qv = q k e d " k e \$ % Q d & ' ( < 4

5 Problem 5 Four charges of equal magnitude (two positive and two negative) are placed on the corners of a square as pictured at left (positive charges at positions A & B, negative at positions C & D) You decide that you want to swap charges B & C so that the charges of like sign will be on the same diagonal rather than on the same side of the square In order to do this, you must do 1 positive work negative work 3 no net work 4 an indeterminate amount of work, as it depends on exactly how you choose to move the charges Answer By moving like sign charges farther apart and opposite sign charges closer together you reduce the energy of the system 5

6 Problem 6 Two opposite charges are placed on a line as shown in the figure below The charge on the right is three times the magnitude of the charge on the left Besides infinite, where else can electric field possibly be zero? 1 between the two charges to the right of the charge on the right 3 to the left of the charge on the left 4 the electric field is nowhere zero Answer: 3 The electric field from the positive charge E +3q and the electric field from the negative charge E q are shown for different regions with the arrow size indicative of the magnitude in the figure below The only place the fields can cancel is to the left of the negative charge For the region to the right of the charge of greater magnitude, the smaller charge is further away and so the magnitude of the field of the smaller charge can never equal the magnitude of the field of the larger charge 6

7 Part II Analytic Problems Problem 1 Non-uniformly charged sphere A sphere of radius R has a charge density = (r / R) where is a constant and r is the distance from the center of the sphere a) What is the total charge inside the sphere? Solution: The total charge inside the sphere is the integral r = R r = R Q = \$ 4"r dr = \$ (r / R)4"r dr = 4" r = R r 3 R \$ dr = 4" R r # = r # = b) Find the electric field everywhere (both inside and outside the sphere) Solution: r = R 4 4 = " R3 There are two regions of space: region I: r < R, and region II: r > R so we apply Gauss Law to each region to find the electric field For region I: r < R, we choose a sphere of radius r as our Gaussian surface Then, the electric flux through this closed surface is E I d A "" = E I 4#r Since the charge distribution is non-uniform, we will need to integrate the charge density to find the charge enclosed in our Gaussian surface In the integral below we use the integration variable r in order to distinguish it from the radius r of the Gaussian sphere Q enc = 1 r \$ =r "4# r\$ % d r \$ = 1 r \$ =r " ( r \$ / R)4# r \$ % d r \$ = " 4# r \$ =r r\$ 3 R % d r \$ = " 4#r4 = " #r4 4R R r \$ = r \$ = r \$ = Notice that the integration is primed and the radius of the Gaussian sphere appears as a limit of the integral 7

8 Recall that Gauss s Law equates electric flux to charge enclosed: "" E I d A = Q enc # So we substitute the two calculations above into Gauss s Law to arrive at: We can solve this equation for the electric field E I 4"r = # "r4 R\$ E I = E I ˆr = r 4R" ˆr, < r < R The electric field points radially outward and has magnitude E I = r For region II: r > R : we choose the same spherical Gaussian surface of radius r > R, and the electric flux has the same form E II d A "" = E II 4#r 4R", < r < R All the charge is now enclosed, Q enc = Q = " R 3, so the right hand side of Gauss s Law becomes Then Gauss s Law becomes Q enc = Q = " # R3 We can solve this equation for the electric field E II 4"r = # " R3 \$ 8

9 E II = E II ˆr = R3 4" r ˆr, r > R In this region of space, the electric field points radially outward and has magnitude E II = R3 4" r, r > R, so it falls off as 1 / r as we expect since outside the charge distribution, the sphere acts as if it all the charge were concentrated at the origin 9

10 Problem Electric field and force A positively charged wire is bent into a semicircle of radius R, as shown in the figure below The total charge on the semicircle is Q However, the charge per unit length along the semicircle is non-uniform and given by = cos" a) What is the relationship between, R and Q? b) If a particle with a charge q is placed at the origin, what is the total force on the particle? Show all your work including setting up and integrating any necessary integrals Answer (a) In order to find a relation between, R and Q it is necessary to integrate the charge density because the charge distribution is non-uniform # \$= " / # / / cos # " sin \$= & & \$ \$ \$ # \$=% " \$=% " / wire Q = ds = Rd = R = R (b) The force on the charged particle at the center P of the semicircle is given by F( P) = qe( P) The electric field at the center P of the semicircle is given by 1 E( P) = \$ 4"# wire ds rˆ r The unit vector, ˆr, located at the field point, is directed from the source to the field point and in Cartesian coordinates is given by rˆ = # sin " ˆi # cos " ˆj 1

11 Therefore the electric field at the center P of the semicircle is given by E # cos % Rd % ( P) = = (& sin & cos ) 4"\$ r 4"\$ i j 1 # ds 1 %=" / ˆ ˆ ' ˆ / r ' % % %=&" wire R There are two separate integrals for the x - and y -components The x -component of the electric field at the center P of the semicircle is given by %=" / %=" / # cos % sin % d % # cos % 1 Ex ( P) = & 4"\$ ' = = R 8"\$ R %=&" / %=&" / We expected this result by the symmetry of the charge distribution about the y-axis The y -component of the electric field at the center P of the semicircle is given by 1 % = " / # / cos % d % 1 % = " # (1 + cos %) d % Ey ( P) = & 4"\$ ' = & R 4"\$ ' R % =& " / % =& " / # # = & & 8"\$ 16"\$ # = & 8\$ R % = " / % = " / % sin % =& " / % =& " / R R Therefore the force on the charged particle at the point P is given by q ˆ 8" R F( P) = qe( P) = # j 11

12 Problem 3: Charged slab and sheets An infinite slab of charge carrying a uniform volume charge density has its boundaries located at x = - meters and x = + meters It is infinite in the y direction and in the z direction (out of the page) Two similarly infinite charge sheets (zero thickness) are located at x = -6 meters and x = +6 meters, with uniform surface charge densities 1 and respectively In the accessible regions you ve measured the electric field to be: E(x) = # ; x < 6 m %(1 N "C 1 )î \$ ; 6 m < x < m %(1 N "C 1 )î ; m < x < 6 m % & ; x > 6 m (a) (b) What is the charge density of the slab? Find the two surface charge densities 1 and of the left and right charged sheets Solution: (a) We begin by choosing a Gaussian surface shown in the figure below Because we are given the magnitude and direction of the electric field on both endcaps with 1

13 E E L = E R, we can calculate the flux on the closed surface (note it is negative because the electric fields point into the Gaussian surface) "" E d A = E L A L + E R A R = # E A S The charge enclosed is Q enc = Ad Therefore Gauss s Law becomes The charge density is then EA = "Ad # = " E d # = " (1 N \$C"1 ) (4 m) 1 (4% )(9 & 1 9 N \$ m \$C " ) = "44 & 1"11 C \$ m "3 (b) Find the two surface charge densities 1 and of the left and right charged sheets Using Gauss s law with the Gaussian pillboxes indicated in the figure above For the sheet on the left, the charge enclosed is Q enc = 1 A, the flux is "" E d A = EA, and so Gauss s Law becomes EA = 1 A " Therefore the surface charge density on the left is In a similar manner, 1 1 = E" = (1 N # C \$1 ) (4% )(9 & 1 9 N # m #C \$ ) = 89 & 1\$11 C # m \$ 1 = E" = (1 N #C \$1 ) (4% )(9 &1 9 N #m #C \$ ) = 89 &1\$11 C#m \$ S 13

14 Problem 4 Concentric charged cylinders A very long uniformly charged solid cylinder (length L and radius a ) carrying a positive charge +q is surrounded by a thin uniformly charged cylindrical shell (length L and radius b) with negative charge q, as shown in the figure You may ignore edge effects Find a vector expression for the electric field in each of the regions (i) r < a, (ii) a < r < b, and (iii) r > b Solution: (i) r < a : Choose a Gaussian surface shown in the figure below We use a Gaussian cylinder of length l and radius r < a Then, the flux is E d A "" = E#rl Because the charge density is uniform, the charge enclosed is given by So Gauss Law becomes Q enc = " dv = #r l volume enclosed Erl = "r l # \$ E(r) = "r # ˆr ; r < a where ˆr is an unit vector pointing perpendicular to the axis of symmetry, in the direction of increasing r 14

15 (ii) a < r < b : Choose a Gaussian surface shown in the figure below We use a Gaussian cylinder of length l and radius a < r < b Then, the flux is again E d A "" = E#rl Because the charge density is uniform, the charge enclosed is given by So Gauss Law becomes Q enc = " dv = #a l volume enclosed Erl = "r l # \$ E(r) = "a # 1 r ˆr ; a < r < b (iii) r > b : Choose a Gaussian surface shown in the figure below We use a Gaussian cylinder of length l and radius r > b Then, the flux is again E d A "" = E#rl

16 The inner cylinder has charge per unit length inner = q / L and the outer shell has charge per unit length outer = "q / L, thus inner = " outer Therefore the charge enclosed by the Gaussian surface is zero: Q enc = ( inner + outer )l = So by Gauss Law the electric field in the region r > b must also be zero Erl = " E = ; r > b Combining our results, we have that the electric field is given by E(r) = # r % ˆr ; r < a " % % a 1 \$ " r ˆr ; a < r < b % % ; r > b % &%

17 Problem 5 Electric field and potential of charged objects Two charges +q and q lie along the y-axis and are separated by a distance d as shown in the figure (a) Calculate the total electric field E at position A, a distance a from the y-axis Indicate its direction on the sketch (draw an arrow) Solution: See figure for arrow at point A E q a ˆ d ˆ q a ˆ d ˆ A = k \$ + " 1/ 1/ % + k \$ # + " 1/ 1/ % a + d & ( a + d ) i ( a + d ) j ' a + d & ( a + d ) i ( a + d ) j ' qd ˆ 1 qd E ˆ A = ke j = j ( a + d ) 4 " ( a + d ) 3/ 3/ (b) Calculate the total electric field E at position B, a distance b from the x-axis Indicate its direction on the sketch (draw an arrow)

18 Solution: See figure above for arrow at point B E # q q \$ ˆ 1 # q q \$ ˆ B = ke & % + ' = & % + ' ( ( b % d) ( b + d) ) j 4 " ( ( b % d) ( b + d) ) j (c) Find the electric potential V at position A with the electric potential zero at infinity q q V ( A) V (") = V ( A) = ke + k 1/ e = 1/ ( a + d ) ( a + d ) (d) Find the electric potential V at position B # q q \$ 1 # q q \$ V ( B) % V (&) = V ( B) = ke ' + % ( = ' + % ( ) b % d b + d * 4" ) b % d b + d * (e) A positively charged dust particle with mass m and charge + q is released from rest at point B In what direction will it accelerate (circle the correct answer)? LEFT RIGHT UP DOWN It Won t Accelerate Solution: The positively charged dust particle will accelerate down because the charge + q is closer to the dust particle and hence exerts a greater repulsive force than the negative charge q, which exerts a smaller attractive force because it is further away from the dust particle (f) Again, assuming the positively charged dust particle was released from rest, what will its speed be after it has traveled a distance s from its original position at point B? HINT: Don t try to simplify any fractions Solution: Assuming that there are no other forces acting on the dust particle, the change in potential energy of the dust particle as it moves from point B to a point a distance s from B down the y-axis, is given by The final electric potential is given by So the change in potential energy is U " U ( final) # U ( B) = # q( V ( final) # V ( B)) q q " V ( final) = ke \$ + # % & b + s # d b + s + d '

19 From conservation of energy q q " q q "" # U = + q % ke % + \$ & \$ ke % + \$ & b + s \$ d b + s + d b \$ d b + d & ' ' ( ' (( K + U = Since it was released from rest at point B, K( B ) =, so the change in kinetic energy is 1 K " K( final) = mv f From conservation of energy K + U = implies that K = " U, so 1 q q " q q "" mv f = #\$ U = # q ke ke % % + # & # % + # & b + s # d b + s + d b # d b + d & ' ' ( ' (( The final speed is thus v f qk e q q " q q "" = # m \$ \$ + # % # \$ + # % b + s # d b + s + d b # d b + d % && ' & ''

20 Problem 6: Electric field and electric potential of a non-uniformly charged rod A rod of length L lies along the x-axis with its left end at the origin The rod has a non-uniform charge density = "x, where is a positive constant (a) Express the total charge Q on the rod in terms of and L (b) Calculate the electric field at point P, shown in the Figure Take the limit d L What does the electric field look like in this limit? Is this what you expect? Explain Hint: the following integral may be useful: x dx = (x + a) ln(1+ x) = x " x a + ln(x + a) x + a + (for small x) (c) Calculate the electric potential at P Take the limit d L What does the electric potential look like in this limit? Is this what you expect? Explain Hint: the following integral may be useful: Solution x dx = x " a ln(x + a) x + a (a) Since = dq / dx = "x, the total charge may be obtained by integrating over the entire length of the rod: L Q = dq = "x dx = " L L (1) (b) Consider an infinitesimal charge element dq = dx' = "x'dx' located at a distance x' from the origin (see figure) The electric field at a point P due to this element is given by

21 d E dq = k e r (î) = k "x'dx' (î) () e ( x'+ d) where r = x + d is the distance between the charge element and P The electric field points in the negative direction Integrating over d E, we obtain E = k e # L x'dx' ( x'+ d) (" \$ d î) = k e & % x'+ d \$ * = k e ln 1+ L - +, d / " L ' & % L + d )("î) ( ( ) + ln x'+ d ' ) ( L ("î) (3) For d L, the ratio L / d is small, and Taylor-series expansion yields and ln 1+ L \$ " # d % & = L d ' 1 L + (4) d L L + d = L d(1+ L / d) = L d L d + (5) In this limit, the electric field becomes # E = k e L d " 1 L d " L d + L & % \$ d ( ' ("î) = k 1 # L& e \$ % d ' ( ("î) = k Q ("î) (6) e d which is the electric field due to a point charge Q at a distance d away from P This pointcharge limit is precisely what is expected when d L (c) The contribution to the electric potential at P by a differential charge element dq is dq dv = k e r = k x'dx' e x'+ d (7) Integrating over the entire length yields L x" d x" L % ) V = k e # = k x " + d e % & x " \$ d ln ( x " + d) ' ( = ke L \$ d ln 1+ L, ' / * + d - (8) & (

22 In the limit d L, we have ) # V = k e L " d L d " 1 &, + (L / d) \$ % ' ( = k e 1 * - L d = k Q e d (9) which is the potential due to a point charge Q at distance d, as expected

Chapter 3. Gauss s Law

3 3 3-0 Chapter 3 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

Exam 2 Practice Problems Part 1 Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Exam Practice Problems Part 1 Solutions Problem 1 Electric Field and Charge Distributions from Electric Potential An electric potential V ( z

Gauss's Law. Gauss's Law in 3, 2, and 1 Dimension

[ Assignment View ] [ Eðlisfræði 2, vor 2007 22. Gauss' Law Assignment is due at 2:00am on Wednesday, January 31, 2007 Credit for problems submitted late will decrease to 0% after the deadline has passed.

Exam 1 Solutions. PHY2054 Fall 2014. Prof. Paul Avery Prof. Andrey Korytov Sep. 26, 2014

Exam 1 Solutions Prof. Paul Avery Prof. Andrey Korytov Sep. 26, 2014 1. Charges are arranged on an equilateral triangle of side 5 cm as shown in the diagram. Given that q 1 = 5 µc and q 2 = q 3 = 2 µc

CHAPTER 24 GAUSS S LAW

CHAPTER 4 GAUSS S LAW 4. The net charge shown in Fig. 4-40 is Q. Identify each of the charges A, B, C shown. A B C FIGURE 4-40 4. From the direction of the lines of force (away from positive and toward

Exam No. 1 Solutions

Exam No. 1 Solutions I. (20 pts) Three positive charges q 1 = +2 μc, q 2 = +1 μc, and q 3 = +1 μc are arranged at the corners of an equilateral triangle of side 2 m as shown in the diagram. Calculate:

Chapter 22: Electric Flux and Gauss s Law

22.1 ntroduction We have seen in chapter 21 that determining the electric field of a continuous charge distribution can become very complicated for some charge distributions. t would be desirable if we

The Electric Field. Electric Charge, Electric Field and a Goofy Analogy

. The Electric Field Concepts and Principles Electric Charge, Electric Field and a Goofy Analogy We all know that electrons and protons have electric charge. But what is electric charge and what does it

Physics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings

1 of 11 9/7/2012 1:06 PM Logged in as Julie Alexander, Instructor Help Log Out Physics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings Course Home Assignments Roster Gradebook Item Library

Electric Forces & Fields, Gauss s Law, Potential

This test covers Coulomb s Law, electric fields, Gauss s Law, electric potential energy, and electric potential, with some problems requiring a knowledge of basic calculus. Part I. Multiple Choice +q +2q

Last Name: First Name: Physics 102 Spring 2007: Exam #1 Multiple-Choice Questions 1. Two small conducting spheres attract one another electrostatically. This can occur for a variety of reasons. Which of

Physics 210 Q ( PHYSICS210BRIDGE ) My Courses Course Settings

1 of 16 9/7/2012 1:10 PM Logged in as Julie Alexander, Instructor Help Log Out Physics 210 Q1 2012 ( PHYSICS210BRIDGE ) My Courses Course Settings Course Home Assignments Roster Gradebook Item Library

HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case.

HW6 Solutions Notice numbers may change randomly in your assignments and you may have to recalculate solutions for your specific case. Tipler 22.P.053 The figure below shows a portion of an infinitely

Chapter 22: The Electric Field. Read Chapter 22 Do Ch. 22 Questions 3, 5, 7, 9 Do Ch. 22 Problems 5, 19, 24

Chapter : The Electric Field Read Chapter Do Ch. Questions 3, 5, 7, 9 Do Ch. Problems 5, 19, 4 The Electric Field Replaces action-at-a-distance Instead of Q 1 exerting a force directly on Q at a distance,

Solution. Problem. Solution. Problem. Solution

4. A 2-g ping-pong ball rubbed against a wool jacket acquires a net positive charge of 1 µc. Estimate the fraction of the ball s electrons that have been removed. If half the ball s mass is protons, their

CHAPTER 26 ELECTROSTATIC ENERGY AND CAPACITORS

CHAPTER 6 ELECTROSTATIC ENERGY AND CAPACITORS. Three point charges, each of +q, are moved from infinity to the vertices of an equilateral triangle of side l. How much work is required? The sentence preceding

Exam 2 Practice Problems Part 2 Solutions

Problem 1: Short Questions MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8. Exam Practice Problems Part Solutions (a) Can a constant magnetic field set into motion an electron, which is initially

PSS 27.2 The Electric Field of a Continuous Distribution of Charge

Chapter 27 Solutions PSS 27.2 The Electric Field of a Continuous Distribution of Charge Description: Knight Problem-Solving Strategy 27.2 The Electric Field of a Continuous Distribution of Charge is illustrated.

( )( 10!12 ( 0.01) 2 2 = 624 ( ) Exam 1 Solutions. Phy 2049 Fall 2011

Phy 49 Fall 11 Solutions 1. Three charges form an equilateral triangle of side length d = 1 cm. The top charge is q = - 4 μc, while the bottom two are q1 = q = +1 μc. What is the magnitude of the net force

Electromagnetism Laws and Equations

Electromagnetism Laws and Equations Andrew McHutchon Michaelmas 203 Contents Electrostatics. Electric E- and D-fields............................................. Electrostatic Force............................................2

Coefficient of Potential and Capacitance

Coefficient of Potential and Capacitance Lecture 12: Electromagnetic Theory Professor D. K. Ghosh, Physics Department, I.I.T., Bombay We know that inside a conductor there is no electric field and that

Vector surface area Differentials in an OCS

Calculus and Coordinate systems EE 311 - Lecture 17 1. Calculus and coordinate systems 2. Cartesian system 3. Cylindrical system 4. Spherical system In electromagnetics, we will often need to perform integrals

Chapter 25: Capacitance

Chapter 25: Capacitance Most of the fundamental ideas of science are essentially simple, and may, as a rule, be expressed in a language comprehensible to everyone. Albert Einstein 25.1 Introduction Whenever

Chapter 16 Electric Forces and Fields

Chapter 16 Electric Forces and Fields 2. How many electrons does it take to make one coulomb of negative charge? A. 1.00 10 9 B. 6.25 10 18 C. 6.02 10 23 D. 1.66 10 18 E. 2.24 10 4 10. Two equal point

Physics 1653 Exam 3 - Review Questions

Physics 1653 Exam 3 - Review Questions 3.0 Two uncharged conducting spheres, A and B, are suspended from insulating threads so that they touch each other. While a negatively charged rod is held near, but

PHYS 155: Final Tutorial

Final Tutorial Saskatoon Engineering Students Society eric.peach@usask.ca April 13, 2015 Overview 1 2 3 4 5 6 7 Tutorial Slides These slides have been posted: sess.usask.ca homepage.usask.ca/esp991/ Section

Fall 97 Test 1, P. 2

2102 Fall 97 Test 1 Fall 97 Test 1, P. 2 Fall 97 Test 1, P. 3 Fall 97 Test 1, P. 4 Fall 97 Test 1, P. 5 5. (10 points) A spherical rubber balloon has a charge uniformly distributed over is surface. The

Objectives for the standardized exam

III. ELECTRICITY AND MAGNETISM A. Electrostatics 1. Charge and Coulomb s Law a) Students should understand the concept of electric charge, so they can: (1) Describe the types of charge and the attraction

ELECTROSTATICS. Ans: It is a fundamental property of matter which is responsible for all electrical effects

ELECTROSTATICS One Marks Questions with Answers: 1.What is an electric charge? Ans: It is a fundamental property of matter which is responsible for all electrical effects 2. Write the SI unit of charge?

1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 80 µ T at the loop center. What is the loop radius?

CHAPTER 3 SOURCES O THE MAGNETC ELD 1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 8 µ T at the loop center. What is the loop radius? Equation 3-3, with

Physics 2112 Unit 6: Electric Potential

Physics 2112 Unit 6: Electric Potential Today s Concept: Electric Potential (Defined in terms of Path Integral of Electric Field) Unit 6, Slide 1 Big Idea Last time we defined the electric potential energy

Physics 202, Lecture 3. The Electric Field

Physics 202, Lecture 3 Today s Topics Electric Field Quick Review Motion of Charged Particles in an Electric Field Gauss s Law (Ch. 24, Serway) Conductors in Electrostatic Equilibrium (Ch. 24) Homework

Gauss s Law for Gravity

Gauss s Law for Gravity D.G. impson, Ph.D. Department of Physical ciences and Engineering Prince George s Community College December 6, 2006 Newton s Law of Gravity Newton s law of gravity gives the force

Fall 12 PHY 122 Homework Solutions #8

Fall 12 PHY 122 Homework Solutions #8 Chapter 27 Problem 22 An electron moves with velocity v= (7.0i - 6.0j)10 4 m/s in a magnetic field B= (-0.80i + 0.60j)T. Determine the magnitude and direction of the

1. AREA BETWEEN the CURVES

1 The area between two curves The Volume of the Solid of revolution (by slicing) 1. AREA BETWEEN the CURVES da = {( outer function ) ( inner )} dx function b b A = da = [y 1 (x) y (x)]dx a a d d A = da

Chapter 4. Electrostatic Fields in Matter

Chapter 4. Electrostatic Fields in Matter 4.1. Polarization A neutral atom, placed in an external electric field, will experience no net force. However, even though the atom as a whole is neutral, the

PH 212 07-31-2015 Physics 212 Exam-3 Solution NAME: Write down your name also on the back of the package of sheets you turn in.

PH 1 7-31-15 Physics 1 Exam-3 Solution NAME: Write down your name also on the back of the package of sheets you turn in. SIGNATURE and ID: Return this hard copy exam together with your other answer sheets.

Problem 4.48 Solution:

Problem 4.48 With reference to Fig. 4-19, find E 1 if E 2 = ˆx3 ŷ2+ẑ2 (V/m), ε 1 = 2ε 0, ε 2 = 18ε 0, and the boundary has a surface charge density ρ s = 3.54 10 11 (C/m 2 ). What angle does E 2 make with

Physics 9 Summer 2010 Midterm Solutions

Physics 9 Summer 010 Midterm s For the midterm, you may use one sheet of notes with whatever you want to put on it, front and back Please sit every other seat, and please don t cheat! If something isn

VIII. Magnetic Fields - Worked Examples

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 003 VIII. Magnetic Fields - Worked Examples Example : Rolling rod A rod with a mass m and a radius R is mounted on two parallel rails

Physics 2212 GH Quiz #4 Solutions Spring 2015

Physics 1 GH Quiz #4 Solutions Spring 15 Fundamental Charge e = 1.6 1 19 C Mass of an Electron m e = 9.19 1 31 kg Coulomb constant K = 8.988 1 9 N m /C Vacuum Permittivity ϵ = 8.854 1 1 C /N m Earth s

As customary, choice (a) is the correct answer in all the following problems.

PHY2049 Summer 2012 Instructor: Francisco Rojas Exam 1 As customary, choice (a) is the correct answer in all the following problems. Problem 1 A uniformly charge (thin) non-conucting ro is locate on the

A2 Physics - Electric Fields Q&A Revision Sheet

Give the equation relating to the force between point charges in a vacuum If 'F' (the force) is negative what does that mean? If 'F' (the force) is positive what does that mean? State Coulomb's Law F is

Capacitance, Resistance, DC Circuits

This test covers capacitance, electrical current, resistance, emf, electrical power, Ohm s Law, Kirchhoff s Rules, and RC Circuits, with some problems requiring a knowledge of basic calculus. Part I. Multiple

Chapter 13. Gravitation

Chapter 13 Gravitation 13.2 Newton s Law of Gravitation In vector notation: Here m 1 and m 2 are the masses of the particles, r is the distance between them, and G is the gravitational constant. G = 6.67

AP2 Electrostatics. Three point charges are located at the corners of a right triangle as shown, where q 1. are each 1 cm from q 3.

Three point charges are located at the corners of a right triangle as shown, where q 1 = q 2 = 3 µc and q 3 = -4 µc. If q 1 and q 2 are each 1 cm from q 3, what is the magnitude of the net force on q 3?

Chapter 6 Circular Motion

Chapter 6 Circular Motion 6.1 Introduction... 1 6.2 Cylindrical Coordinate System... 2 6.2.1 Unit Vectors... 3 6.2.2 Infinitesimal Line, Area, and Volume Elements in Cylindrical Coordinates... 4 Example

Electric Fields and Gauss s Law law. January 21, 2014 Physics for Scientists & Engineers 2, Chapter 22 1

Electric Fields and Gauss s Law law January 21, 2014 Physics for Scientists & Engineers 2, Chapter 22 1 Announcements! First exam is next Tuesday, January 28 45 minute exam during lecture time You can

Lecture 5. Electric Flux and Flux Density, Gauss Law in Integral Form

Lecture 5 Electric Flux and Flux ensity, Gauss Law in Integral Form ections: 3.1, 3., 3.3 Homework: ee homework file LECTURE 5 slide 1 Faraday s Experiment (1837), Flux charge transfer from inner to outer

6. In cylindrical coordinate system, the differential normal area along a z is calculated as: a) ds = ρ d dz b) ds = dρ dz c) ds = ρ d dρ d) ds = dρ d

Electrical Engineering Department Electromagnetics I (802323) G1 Dr. Mouaaz Nahas First Term (1436-1437), Second Exam Tuesday 07/02/1437 H االسم: الرقم الجامعي: Start from here Part A CLO 1: Students will

CBE 6333, R. Levicky 1. Potential Flow

CBE 6333, R. Levicky Part I. Theoretical Background. Potential Flow Potential Flow. Potential flow is irrotational flow. Irrotational flows are often characterized by negligible viscosity effects. Viscous

Review B: Coordinate Systems

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of hysics 8.02 Review B: Coordinate Systems B.1 Cartesian Coordinates... B-2 B.1.1 Infinitesimal Line Element... B-4 B.1.2 Infinitesimal Area Element...

Problem 1 (25 points)

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2012 Exam Three Solutions Problem 1 (25 points) Question 1 (5 points) Consider two circular rings of radius R, each perpendicular

7.3 Volumes Calculus

7. VOLUMES Just like in the last section where we found the area of one arbitrary rectangular strip and used an integral to add up the areas of an infinite number of infinitely thin rectangles, we are

Applications of Integration Day 1

Applications of Integration Day 1 Area Under Curves and Between Curves Example 1 Find the area under the curve y = x2 from x = 1 to x = 5. (What does it mean to take a slice?) Example 2 Find the area under

University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 5 Solutions by P. Pebler

University of California, Berkeley Physics H7B Spring 1999 (Strovink) SOLUTION TO PROBLEM SET 5 Solutions by P. Pebler 1 Purcell 1.16 A sphere of radius a was filled with positive charge of uniform density

Electrostatics-E Field

1. Which diagram represents the electric field lines between two small electrically charged spheres? 2. Which graph best represents the relationship between the magnitude of the electric field strength,

Solution: (a) For a positively charged particle, the direction of the force is that predicted by the right hand rule. These are:

Problem 1. (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields as shown in the figure. (b) Repeat part (a), assuming the moving particle is

* Biot Savart s Law- Statement, Proof Applications of Biot Savart s Law * Magnetic Field Intensity H * Divergence of B * Curl of B. PPT No.

* Biot Savart s Law- Statement, Proof Applications of Biot Savart s Law * Magnetic Field Intensity H * Divergence of B * Curl of B PPT No. 17 Biot Savart s Law A straight infinitely long wire is carrying

Magnetic Fields; Sources of Magnetic Field

This test covers magnetic fields, magnetic forces on charged particles and current-carrying wires, the Hall effect, the Biot-Savart Law, Ampère s Law, and the magnetic fields of current-carrying loops

Physics 9 Fall 2009 Homework 2 - Solutions

Physics 9 Fall 009 Homework - s 1. Chapter 7 - Exercise 5. An electric dipole is formed from ±1.0 nc charges spread.0 mm apart. The dipole is at the origin, oriented along the y axis. What is the electric

If two identical balls each of mass m and having charge q are suspended by silk thread of length l from the same point o,then the distance between

If two identical balls each of mass m and having charge q are suspended by silk thread of length l from the same point o,then the distance between the balls is given by : = X = 2 ( ) 1 Two pith balls each

You bring a charge of -3C from infinity to a point P on the perpendicular bisector of a dipole as shown. Is the work that you do:

Physics 2102 Gabriela González You bring a charge of -3C from infinity to a point P on the perpendicular bisector of a dipole as shown. Is the work that you do: a) Positive? b) Negative? c) Zero? a -Q

Exercises on Voltage, Capacitance and Circuits. A d = (8.85 10 12 ) π(0.05)2 = 6.95 10 11 F

Exercises on Voltage, Capacitance and Circuits Exercise 1.1 Instead of buying a capacitor, you decide to make one. Your capacitor consists of two circular metal plates, each with a radius of 5 cm. The

Electrostatics Problems

Name AP Physics B Electrostatics Problems Date Mrs. Kelly 1. How many excess electrons are contained in a charge of 30 C? 2. Calculate and compare the gravitational and electrostatic force between an electron

Chapter 2 Coulomb s Law

- Chapter Coulomb s Law... - Chapter... -1 Coulomb s Law... -3.1 Electric Charge... -3. Coulomb's Law... -3.3 Principle of Superposition... -4 Example.1: Three Charges... -5.4 Electric Field... -6.4.1

Electrostatic Fields: Coulomb s Law & the Electric Field Intensity

Electrostatic Fields: Coulomb s Law & the Electric Field Intensity EE 141 Lecture Notes Topic 1 Professor K. E. Oughstun School of Engineering College of Engineering & Mathematical Sciences University

9 ROTATIONAL DYNAMICS

CHAPTER 9 ROTATIONAL DYNAMICS CONCEPTUAL QUESTIONS 1. REASONING AND SOLUTION The magnitude of the torque produced by a force F is given by τ = Fl, where l is the lever arm. When a long pipe is slipped

Copyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass

Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of

The Electric Force. From mechanics, the relationship for the gravitational force on an object is: m is the mass of the particle of interest,

. The Electric Force Concepts and Principles The Gravitational Analogy In introducing the concept of the electric field, I tried to illustrate it by drawing an analogy with the gravitational field, g.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. 8.02 Spring 2013 Conflict Exam Two Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 802 Spring 2013 Conflict Exam Two Solutions Problem 1 (25 points): answers without work shown will not be given any credit A uniformly charged

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Spring 2009 OBJECTIVES Experiment 2: Faraday Ice Pail 1. To explore the charging of objects by friction and by contact. 2. To explore the

Physics 2220 Module 09 Homework

Physics 2220 Module 09 Homework 01. A potential difference of 0.050 V is developed across the 10-cm-long wire of the figure as it moves though a magnetic field perpendicular to the page. What are the strength

Handout 7: Magnetic force. Magnetic force on moving charge

1 Handout 7: Magnetic force Magnetic force on moving charge A particle of charge q moving at velocity v in magnetic field B experiences a magnetic force F = qv B. The direction of the magnetic force is

Definition: Let S be a solid with cross-sectional area A(x) perpendicular to the x-axis at each point x [a, b]. The volume of S is V = A(x)dx.

Section 7.: Volume Let S be a solid and suppose that the area of the cross-section of S in the plane P x perpendicular to the x-axis passing through x is A(x) for a x b. Consider slicing the solid into

AP Physics C: Electricity and Magnetism 2011 Scoring Guidelines

AP Physics C: Electricity and Magnetism 11 Scoring Guidelines The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and

Physics 1B: E-Field from a Point-Source Integral E-Field of a Plane of Charge

1 Problem Physics 1B: E-Field from a Point-Source Integral E-Field of a Plane of Charge Using the remote-point-source expression for the electric field (rather than Gauss s law), solve for the electric

Magnetic Fields. Goals. Introduction. Mapping magnetic fields with iron filings

Lab 4. Magnetic Fields Goals To visualize the magnetic fields produced by several different configurations of simple bar magnets using iron filings. To use small magnetic compasses to trace out the magnetic

ELECTRIC FIELD LINES AND EQUIPOTENTIAL SURFACES

ELECTRIC FIELD LINES AND EQUIPOTENTIAL SURFACES The purpose of this lab session is to experimentally investigate the relation between electric field lines of force and equipotential surfaces in two dimensions.

Module 1 : A Crash Course in Vectors Lecture 2 : Coordinate Systems

Module 1 : A Crash Course in Vectors Lecture 2 : Coordinate Systems Objectives In this lecture you will learn the following Define different coordinate systems like spherical polar and cylindrical coordinates

Name Calculus AP Chapter 7 Outline M. C.

Name Calculus AP Chapter 7 Outline M. C. A. AREA UNDER A CURVE: a. If y = f (x) is continuous and non-negative on [a, b], then the area under the curve of f from a to b is: A = f (x) dx b. If y = f (x)

Engineering Math II Spring 2015 Solutions for Class Activity #2

Engineering Math II Spring 15 Solutions for Class Activity # Problem 1. Find the area of the region bounded by the parabola y = x, the tangent line to this parabola at 1, 1), and the x-axis. Then find

Transformed E&M I homework. Divergence and Curl of B (Ampereʼs Law) (Griffiths Chapter 5)

Transformed E&M I homework Divergence and Curl of B (Ampereʼs Law) (Griffiths Chapter 5) Divergence and curl of B (Connections between E and B, Ampere s Law) Question 1. B of cylinder with hole Pollack

Solution Derivations for Capa #10

Solution Derivations for Capa #10 1) The flywheel of a steam engine runs with a constant angular speed of 172 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel

Physics 121 Sample Common Exam 3 NOTE: ANSWERS ARE ON PAGE 6. Instructions: 1. In the formula F = qvxb:

Physics 121 Sample Common Exam 3 NOTE: ANSWERS ARE ON PAGE 6 Signature Name (Print): 4 Digit ID: Section: Instructions: Answer all questions 24 multiple choice questions. You may need to do some calculation.

Eðlisfræði 2, vor 2007

[ Assignment View ] [ Pri Eðlisfræði 2, vor 2007 29a. Electromagnetic Induction Assignment is due at 2:00am on Wednesday, March 7, 2007 Credit for problems submitted late will decrease to 0% after the

Rotation. Moment of inertia of a rotating body: w I = r 2 dm

Rotation Moment of inertia of a rotating body: w I = r 2 dm Usually reasonably easy to calculate when Body has symmetries Rotation axis goes through Center of mass Exams: All moment of inertia will be

Physics II Exam 2 Review

Physics II Exam 2 Review Christopher Lane 1,2 Justin Lucas 1,3 Julia Bielaski 1,2 1 Department Physics, Clarkson University 2 Department Mathematics, Clarkson University 3 Department Electrical and Computer

Physics 122 (Sonnenfeld), Spring 2013 ( MPSONNENFELDS2013 ) My Courses Course Settings

Signed in as Richard Sonnenfeld, Instructor Help Sign Out Physics 122 (Sonnenfeld), Spring 2013 ( MPSONNENFELDS2013 ) My Courses Course Settings Course Home Assignments Roster Gradebook Item Library Essential

1 of 7 4/13/2010 8:05 PM

Chapter 33 Homework Due: 8:00am on Wednesday, April 7, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy [Return to Standard Assignment View] Canceling a Magnetic Field

NAME. and 2I o. (1) Two long wires carry magnetic fields I o. , where I o

(1) Two long wires carry magnetic fields I o and 2I o, where I o is a constant. The two wires cross at the origin (but without making any electrical connection), and lie in the x-y plane. (a) Find the

Week 8 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

Physics 202 Midterm Exam 1 September 27 th, 2010

ID CODE: A Physics 202 Midterm Exam 1 September 27 th, 2010 Name:...Yibin Pan... Student ID:... Section:... TA (please circle): Matt Elbert Josh Isaacs Jianjia Fei Hyungjun Lim Zhen Liu Frank McNally Yongyan

PH202-5D Final Comprehensive Exam (August 10, 2007)

NAME SCORE PH202-5D Final Comprehensive Exam (August 0, 2007) You may not open the textbook nor notebook. A letter size information may be used. A calculator may be used. However, mathematics or physics

Physics 227: Lecture 3 Electric Field Calculations, Symmetry, Charged Particle Motion, Flux, Gauss s Law

Physics 227: Lecture 3 Electric Field Calculations, Symmetry, Charged Particle Motion, Flux, Gauss s Law Lecture 2 review: F 1 on 2 = k q 1 q 2 / r 2 Superposition: Forces are vectors, Coulomb forces add

3 rd International Physics Olympiad 1969, Brno, Czechoslovakia

3 rd International Physics Olympiad 1969, Brno, Czechoslovakia Problem 1. Figure 1 shows a mechanical system consisting of three carts A, B and C of masses m 1 = 0.3 kg, m 2 = 0.2 kg and m 3 = 1.5 kg respectively.

Chapter Biography of J. C. Maxwell Derivation of the Maxwell Speed Distribution Function

Chapter 10 10.1 Biography of J. C. Maxwell 10.2 Derivation of the Maxwell Speed Distribution Function The distribution of molecular speeds was first worked out by Maxwell before the development of statistical

Magnetism. d. gives the direction of the force on a charge moving in a magnetic field. b. results in negative charges moving. clockwise.

Magnetism 1. An electron which moves with a speed of 3.0 10 4 m/s parallel to a uniform magnetic field of 0.40 T experiences a force of what magnitude? (e = 1.6 10 19 C) a. 4.8 10 14 N c. 2.2 10 24 N b.