# PHYS 212 Final Exam (Old Material) Solutions - Practice Test

Size: px
Start display at page:

Transcription

1 PHYS 212 Final Exam (Old Material) Solutions - Practice Test 1E If the ball is attracted to the rod, it must be made of a conductive material, otherwise it would not have been influenced by the nearby positive charge. The reason it is then attracted to the rod is due to induction, where the electrons rearrange themselves to be as close to the positive charge as possible (or as far away from the negative charge if the rod was negatively charged). Thus we pick answer choice E. 2A Remember that according to Coulombs Law the force acting on each sphere will be equal in size but acting in opposite directions. This is why they repel. Since the forces acting on each sphere have the same magnitude, and since the spheres have the same mass as well, the acceleration acting on each sphere will also be the same (this is comes from Newton s second law: F = ma). Thus if they each experience the same acceleration, they will move the same distance apart from each other, and answer choice A is the correct choice depicting that. 3E We use Coulomb s law to determine the force on the negative charge due to the other two charges located on either side: The force due to the +50µC charge is: F 1 = k(50µc)(40µc) = 4.495N, in the - x direction 2 2 where the distance is 2m based on the location of each charge along the x-axis. The force due to the +30µC charge is: F 2 = k(30µc)(40µc) = 2.697N, in the + x direction 2 2 where the distance is 2m based on the location of each charge along the x-axis. Thus the net force is F1 + F2 = ( ) = N along the x-axis.

2 4D The general equation for electric field is E = F/Q, where represents a test charge placed in the field. So if we solve this equation for force, we get: F = EQ = (200)(-1.6 x ) = -3.2 x The charge is based on the charge of an electron (given on your test s data sheet) and since it is negative, it must travel in the opposite direction of the field, thus we choose answer choice D. 5A To answer this, we must use Coulomb s law to determine the magnitude and direction of each force acting on q1 from the other charges: F 12 = k(20µc)(20µc) 2 2 = 0.899N, in the + y axis F 13 = k(20µc)(40µc) 2 2 =1.798N, 30 o above the - x axis Note that we know the force from q3 is 30 0 above the x axis since all the charges form an equilateral triangle 60 degrees apart. In order to get the correct magnitude, we must first take the forces and place them in the respective x- and y-components: F12 is already along the y-axis, so need not be altered. F13,x axis = F13cos30 = 1.56N, F13,y axis = F13sin30 = 0.899N Thus using vector analysis we see that the total force along each axis is: y-axis: = N x-axis: 1.56 N The magnitude using Pythagorean theory: F = (1.798) 2 + (1.56) 2 = 2.38N Thus we choose answer choice A.

4 11E The most effective use of Gauss Law is that the total flux through an enclosed surface is given by total internal charge, q, divided by the constant e0. Since we are not given the total charge, we can substitute for it using q = ll, where the charge density, l, is q/l. Thus we can pick answer choice E to represent the total flux. 12C For any conducting spherical shell, the inside surface will have the same charge as the point charge inside the sphere but will have the opposite sign. Thus the inner surface must be -5µC. However, since the +5µC inside the sphere is not located at the center, the charge will not be uniformly distributed along the inner surface. Another rule for conducting spherical shells, is that the sum of the inner and outer surface charges must equal the excess charge of the sphere. So we get: Inner + Outer = Excess -5µC + Outer = +3µC Outer = +8µC The outer surface charge will be uniformly distributed, unlike the inner surface charge, and thus we pick answer choice C. 13B Outside of the cylinder, the field decreases inversely proportional to the distance form the center. Thus if the distance doubles, the field will be halved. 14A The uniformly distributed charge can be thought of as a point charge located at the center of the sphere. According the equation for electric potential of a point charge, the potential decreases as the distance form the point increases. Thus the closer to the center you are, the greater the potential will be, and so we pick answer choice A.

5 15E The only equation relating potential energy and electric potential is DV = DU/q. This means to calculate the change in potential energy, we need to multiply DV with q. Note that in this particular problem, q represents the negative charge that is moved from the x-axis to the y-axis. The change in electric potential, DV, is a scalar quantity that is essentially only interested in the fixed point charge and its distance, r, where we measure the potential. In this particular problem, the initial and final locations of the moving charge are the same distance from the fixed charge, thus the potential at each location is identical. This means that the change in potential as the negative charge moves from its initial to its final location is zero, and thus the change in potential energy is also zero. 16D This scenario is indicative of a positive point charge, which decreases in field strength the further you are from it. The potential of such a point charge will also decrease as you move away from the charge, and thus answer D is the appropriate choice. 17B The answer choice differ in the way they express the distance form the point Y, to any place on the rod, given by the position, x. Essentially, we can think of the dashed line as the distance we are trying to express, which we can calculate using Pythagorean theory: where the distance along the x-axis is given by L-x, the distance along the y-axis is given by y, and the vector distance will then be: (L x) 2 + (y) 2 Thus we choose answer choice B as it is the only option that expresses the distance from the rod to point Y in this correct format. 18B The energy used to pull apart the plates is transferred into stored energy in the capacitors, so we expect the energy stored to increase. Based on our equations for capacitors, the distance between plates is inversely proportional to the capacitance, and so doubling the distance halves the capacitance, which doubles the energy stored within them. Thus we choose answer choice B.

6 19E The key to such problems is to find the equivalent capacitance for each row and then to combine the rows and find their equivalent capacitance. The top row has 3 identical capacitors in series, and so we can solve for their equivalent capacitance using: 1 = 1 C eq, top C + 1 C + 1 C, C eq, top = C 3 = 2µF The middle row has 2 identical capacitors in series, and so we can solve for their equivalent capacitance using: 1 = 1 C eq, middle C + 1 C, C eq, middle = C 2 = 3µF The bottom row has just 1 capacitor, and so we leave its capacitance alone, Cbottom = 6µF We are now left with an equivalent circuit of just 3 capacitors, one on each row, that are in parallel. To get one final equivalent capacitance, CEQ, for capacitors in parallel, we use the equation: C EQ = C Top + C Middle + C Bottom =11µF

7 20D As usual with circuits such as these, we need to simplify the circuit to determine the equivalent capacitance, CEQ. Capacitors 2 and 3 are parallel to each other and can be combined simply be adding the capacitance values together to get C23 = 15 µf. The circuit is now equivalent to two capacitors in series, capacitor 1 with capacitance 30µF, and the combined parallel capacitors with equivalent capacitance 15µF. The last step in simplifying the circuit is to determine the equivalent capacitance of the 30µF and 15µF capacitors in series, which is: 1 = 1 C eq , C eq =10µF Now that we have our equivalent capacitance of 10 µf, we can use the equation below to determine total charge for this circuit: Q Total = C EQ V = (10µF)(120V ) =1200µC If we then return back to the simplified version of the circuit that had capacitor 1 in series with the equivalent capacitance of 15µF (from capacitors 2 and 3 in parallel), we can then see that since the capacitors are in series, the rule is that the charge going across each capacitor is the same and equal to the total charge. Thus the charge across capacitor is the same as the total charge and equal to 1200µC. 21E In order to use the equation for discharging, we should consider the amount of charge remaining after three time constants, which would have to be q(t)/q0. Also we can substitute time, t, with 3t, which is three time constants. We can also substitute RC, which is also equal to t, and we get: q(t) q 0 = e t RC 3τ τ = e = e 3 = = 4.98% This means that after three time constants, we have 4.98% left, which means that charge has been reduced by 95.02%. Note that the fact that charge had been reduced by 63% after one time constant was useless information if you solve it this way.

8 22D Using the appropriate equations for resistors in parallel, the two resistors at the top left of the circuit can be combined into equivalent resistance of 15 ohms. Also, the two resistors in series at the bottom are the equivalent of 60 ohms. Thus the circuit can be redrawn as shown velow: At this stage the current can get from a to b by travelling through either the 15 ohm resistor or the 60 ohm resistor. It does not need to go via the 30 ohm resistor at the top right. Which means that for all intents and purposes we can ignore it. So now we really have a simple parallel circuit with resistors of 15 and 60 ohms, and an equivalent resistance of 12 ohms. 23D Looking at the currents entering and leaving any of the junctions where the wires meet gives the following equation: I2 + I3 = -I1. This can be rewritten to get answer D. 24D We treat the resistors in series on the top half of the circuit as one resistor of 12 ohms, and the resistors at the bottom as one resistor of 596 ohms. So the circuit is really a parallel circuit with equivalent resistance of 11.8 ohms. 25D The voltage drops from 9 V to 8.5 V due to the internal resistance. This 0.5 V lost within the battery can be set equal to V = IR, allowing us to solve for the current, I, to get: I = V/R = 0.5/0.1 = 5 A

9 26A Remember that going across a battery from + to makes the potential drop negative, and going against the direction of current when crossing a resistor makes the drop positive. Only answer A obeys these rules. 27D Initially, the capacitor will not affect the resistance of the circuit, and you can pretend that it is not even there. Thus the current in the circuit will be I = V/R, regardless of the capacitance, and is thus 18/2 = 9 Amps. 28D For any RC circuit, you can use the loop law to include the capacitor (assuming it has charge) so that it looks as follows: ε Ri q C = 0 This can be rewritten to solve for current: i = ε q C R = 4 29D Since the current density, J, is not uniform along the cross-section of the conductor, we must use the equation: I = J da We must replace J with the function given in the question, and da represents the derivative of the area of the cylinder s cross-section, 2pr, giving us: I = 0 R " \$ # Kr 2 2R % '( 2πr) dr = Kπ & R R r 3 0 = KπR4 4R = KπR3 4

10 30A The appropriate equation we need to get resistivity based on the information given is: R = ρl A We already have the area, A, and the length, L, so we just need resistance, R, which we easily get from the equation R = V/I = 1/0.5 = 2.0 ohms. Then we plug in our resistance, R, and solve to get: ρ = RA L = (2.0)(10 6 ) = A The most basic definition of current is: I = C/s Thus, the time in seconds is given by s = C/A = (5)/(10) = 0.5 s 32C Remember that resistivity is a property of a material, and so both wires have the same resistivity (this rules out A). The graph shows that at any given voltage the current in wire A is greater than that of wire B. This can only happen if wire A has a smaller resistance, since current and resistance are inversely proportional. That means that statement ii is correct. We should also remember that if everything else is held constant, the shorter the conductor is, the smaller the resistance will be. So since wire A has the smaller resistance, we can assume it has a shorter length, and thus statement v is also correct. 33D Energy dissipation is the same thing as power. So let s see the equations for power: P = V 0 I = I 2 R = V 2 We can analyze each answer choice as follows: (a) False if you half V (with R constant) the power decreases by a factor of one fourth. (b) False if you half I (with R constant) the power decreases by a factor of one fourth. (c) False if you half R (with V constant) the power will double. (d) True according to P = I 2 R, if you half R (with I constant) the power will half. (e) False if you half both V and I the power decreases by a factor of one fourth.. R

11 34D The Biot-Savart Law for magnetic fields from a current is used to determine the direction of the field. By pointing your thumb in the direction of the current, your fingers curl in the direction of the magnetic field. From this we determine that the two wires on the right going into the page will cancel each other out at point P (the top right wire produces a field to the left, and the bottom right wire produces an equal field to the right). The other wires both produce magnetic fields pointing right at point P. 35B The equation for any circular loop is given by: B = µ i 0 2R = T However, since we are only dealing with half a loop, we will get half the amount of magnetic field, B/2, which equals 1.0 x 10-5 T. 36E For a charge/current moving in a circular path, the Biot-Savart Law says to point your thumb in the direction of motion and your fingers will curl in the direction of the magnetic field. Doing this with the particles causes your fingers to curl into the page at the center of the circle. However, all rules are reversed when the charge is negative, and so the field will point out of the page instead.

12 37B For each wire the direction of the magnetic field can be determined by pointing your thumb along the wire and curling your fingers in the direction of the field. Since both currents are parallel (and assuming the current moves to the right), the top wire produces a field going into the page, while the bottom wire makes a field going out of the page, as shown below: This means that at the midway point, which is 0.2 m from each wire, the fields will oppose each other, and the net magnetic field will be the difference between the two. For a straight wire carrying current, the magnetic field at a distance, R, from the wire is: Thus the net field will be B1 B2: B = µ i 0 2πR B 1 B \$ 2 = µ 0(4) ' \$ & ) µ 0(5) ' & ) = T % 2π(0.1) ( % 2π(0.1) ( 38C Remember that the capacitance is decided once the capacitor is built, and cannot be changed after that point. Thus, plate separation, plate area, and the insertion of a dielectric material, are the only factors that influence capacitance. 39B Ampere s Law tells us that the line integral for the magnetic field within the closed loop is given by µ0ienc, which in this instance is equal to µ0(5), as the two wires going into the page cancel out two of the three wires going out of the page, leaving only one 5A wire remaining. Since µ0 = 4p x 10-7, µ0ienc = 20p x 10-7 = 2p x 10-6 Tm. Lastly, remember that the sign of the line integral is determined by comparing the direction of the path with the direction of the magnetic field created by the currentcarrying wire. The path is clockwise, and since the current is moving out of the page, the right-hand-rule shows the magnetic field is directed counterclockwise. Thus, since the path and the magnetic field are moving in opposite directions, the line integral needs to be negative.

13 40A The RC circuit has an uncharged capacitor initially connected to a battery, thus the appropriate equation we need to use is for charging a capacitor: q(t) = q 0 (1 e t RC ) Since we are looking to solve for time, t, when the charge, q(t) is 75% of its maximum value, this is equivalent to substituting q(t) with 0.75q0 to get: 0.75q 0 = q 0 (1 e t RC ) The maximum charge cancels on both sides, and we simplify to get: 0.25 = e t RC The negatives cancel on each side, and we can replace 0.25 with 1/4: 1/ 4 = e t RC ln(1/ 4) = ln(e t RC ) ln(1/ 4) = t RC Lastly we solve for time, t, and end up with: t = RC ln(1/ 4) Note that although this answer looks very similar to B, it is not the same, as our solution has a negative sign in front of it. Given that none of the answer choices have negative signs, we have to try and be creative, and recognize that if we take the reciprocal of the value within the natural log function it flips the sign,; meaning that ln(1/4) is equal to ln(4). Thus the answer ends up being: t = RC ln(4)

### = e = e 3 = = 4.98%

PHYS 212 Exam 2 - Practice Test - Solutions 1E In order to use the equation for discharging, we should consider the amount of charge remaining after three time constants, which would have to be q(t)/q0.

### A) I B) II C) III D) IV E) V

1. A square loop of wire moves with a constant speed v from a field-free region into a region of uniform B field, as shown. Which of the five graphs correctly shows the induced current i in the loop as

### Physics 126 Fall 2004 Practice Exam 1. Answer will be posted about Oct. 5.

Physics 126 Fall 2004 Practice Exam 1. Answer will be posted about Oct. 5. 1. Which one of the following statements best explains why tiny bits of paper are attracted to a charged rubber rod? A) Paper

### Review. Spring Semester /21/14. Physics for Scientists & Engineers 2 1

Review Spring Semester 2014 Physics for Scientists & Engineers 2 1 Notes! Homework set 13 extended to Tuesday, 4/22! Remember to fill out SIRS form: https://sirsonline.msu.edu Physics for Scientists &

### AP Physics C. Electricity - Term 3

AP Physics C Electricity - Term 3 Interest Packet Term Introduction: AP Physics has been specifically designed to build on physics knowledge previously acquired for a more in depth understanding of the

### AP Physics C Electricity & Magnetism Mid Term Review

AP Physics C Electricity & Magnetism Mid Term Review 1984 37. When lighted, a 100-watt light bulb operating on a 110-volt household circuit has a resistance closest to (A) 10-2 Ω (B) 10-1 Ω (C) 1 Ω (D)

### Questions A hair dryer is rated as 1200 W, 120 V. Its effective internal resistance is (A) 0.1 Ω (B) 10 Ω (C) 12Ω (D) 120 Ω (E) 1440 Ω

Questions 4-41 36. Three 1/ µf capacitors are connected in series as shown in the diagram above. The capacitance of the combination is (A).1 µf (B) 1 µf (C) /3 µf (D) ½ µf (E) 1/6 µf 37. A hair dryer is

### AP Physics C. Magnetism - Term 4

AP Physics C Magnetism - Term 4 Interest Packet Term Introduction: AP Physics has been specifically designed to build on physics knowledge previously acquired for a more in depth understanding of the world

### Mansfield Independent School District AP Physics C: Electricity and Magnetism Year at a Glance

Mansfield Independent School District AP Physics C: Electricity and Magnetism Year at a Glance First Six-Weeks Second Six-Weeks Third Six-Weeks Lab safety Lab practices and ethical practices Math and Calculus

### PHY 131 Review Session Fall 2015 PART 1:

PHY 131 Review Session Fall 2015 PART 1: 1. Consider the electric field from a point charge. As you move farther away from the point charge, the electric field decreases at a rate of 1/r 2 with r being

### LAST Name (print) ALL WORK MUST BE SHOWN FOR THE FREE RESPONSE QUESTION IN ORDER TO RECEIVE FULL CREDIT.

Physics 107 LAST Name (print) First Mid-Term Exam FIRST Name (print) Summer 2013 Signature: July 5 UIN #: Textbooks, cell phones, or any other forms of wireless communication are strictly prohibited in

### A) 1, 2, 3, 4 B) 4, 3, 2, 1 C) 2, 3, 1, 4 D) 2, 4, 1, 3 E) 3, 2, 4, 1. Page 2

1. Two parallel-plate capacitors with different plate separation but the same capacitance are connected in series to a battery. Both capacitors are filled with air. The quantity that is NOT the same for

### Physics 212 Midterm 2 Form A

1. A wire contains a steady current of 2 A. The charge that passes a cross section in 2 s is: A. 3.2 10-19 C B. 6.4 10-19 C C. 1 C D. 2 C E. 4 C 2. In a Physics 212 lab, Jane measures the current versus

### 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

### Physics 420 Fall 2004 Quiz 1 Wednesday This quiz is worth 6 points. Be sure to show your work and label your final answers.

Quiz 1 Wednesday This quiz is worth 6 points. Be sure to show your work and label your final answers. 1. A charge q 1 = +5.0 nc is located on the y-axis, 15 µm above the origin, while another charge q

### Physics Jonathan Dowling. Final Exam Review

Physics 2102 Jonathan Dowling Physics 2102 Final Exam Review A few concepts: electric force, field and potential Electric force: What is the force on a charge produced by other charges? What is the force

### Chapter 10. Electrostatics

Chapter 10 Electrostatics 3 4 AP Physics Multiple Choice Practice Electrostatics 1. The electron volt is a measure of (A) charge (B) energy (C) impulse (D) momentum (E) velocity. A solid conducting sphere

### Physics 55 Final Exam Fall 2012 Dr. Alward Page 1

Physics 55 Final Exam Fall 2012 Dr. Alward Page 1 1. The specific heat of lead is 0.030 cal/g C. 300 g of lead shot at 100 C is mixed with 100 g of water at 70 C in an insulated container. The final temperature

### Chapter 1 The Electric Force

Chapter 1 The Electric Force 1. Properties of the Electric Charges 1- There are two kinds of the electric charges in the nature, which are positive and negative charges. - The charges of opposite sign

### Section 1: Electric Fields

PHY 132 Outline of Lecture Notes i Section 1: Electric Fields A property called charge is part of the basic nature of protons and electrons. Large scale objects become charged by gaining or losing electrons.

### a. Clockwise. b. Counterclockwise. c. Out of the board. d. Into the board. e. There will be no current induced in the wire

Physics 1B Winter 2012: Final Exam For Practice Version A 1 Closed book. No work needs to be shown for multiple-choice questions. The first 10 questions are the makeup Quiz. The remaining questions are

### cancel each other out. Thus, we only need to consider magnetic field produced by wire carrying current 2.

PC1143 2011/2012 Exam Solutions Question 1 a) Assumption: shells are conductors. Notes: the system given is a capacitor. Make use of spherical symmetry. Energy density, =. in this case means electric field

### Physics 6B. Practice Final Solutions

Physics 6B Practice Final Solutions . Two speakers placed 4m apart produce sound waves with frequency 45Hz. A listener is standing m in front of the left speaker. Describe the sound that he hears. Assume

### Exam 1 Solutions. The ratio of forces is 1.0, as can be seen from Coulomb s law or Newton s third law.

Prof. Eugene Dunnam Prof. Paul Avery Feb. 6, 007 Exam 1 Solutions 1. A charge Q 1 and a charge Q = 1000Q 1 are located 5 cm apart. The ratio of the electrostatic force on Q 1 to that on Q is: (1) none

### PHYS1212 Exam#2 Spring 2014

PHYS Exam# Spring 4 NAME There are 9 different pages in this quiz. Check now to see that you have all of them. CEDIT PAT A 6% PAT B 4% TOTAL % GADE All work and answers must be given in the spaces provided

### Coulomb s constant k = 9x10 9 N m 2 /C 2

1 Part 2: Electric Potential 2.1: Potential (Voltage) & Potential Energy q 2 Potential Energy of Point Charges Symbol U mks units [Joules = J] q 1 r Two point charges share an electric potential energy

### Quick Questions. 1. Two charges of +1 µc each are separated by 1 cm. What is the force between them?

92 3.10 Quick Questions 3.10 Quick Questions 1. Two charges of +1 µc each are separated by 1 cm. What is the force between them? 0.89 N 90 N 173 N 15 N 2. The electric field inside an isolated conductor

### Phys222 W16 Exam 2: Chapters Key. Name:

Name: Please mark your answer here and in the scantron. A positively charged particle is moving in the +y-direction when it enters a region with a uniform electric field pointing in the +y-direction. Which

### Final on December Physics 106 R. Schad. 3e 4e 5c 6d 7c 8d 9b 10e 11d 12e 13d 14d 15b 16d 17b 18b 19c 20a

Final on December11. 2007 - Physics 106 R. Schad YOUR NAME STUDENT NUMBER 3e 4e 5c 6d 7c 8d 9b 10e 11d 12e 13d 14d 15b 16d 17b 18b 19c 20a 1. 2. 3. 4. This is to identify the exam version you have IMPORTANT

### Phys 2025, First Test. September 20, minutes Name:

Phys 05, First Test. September 0, 011 50 minutes Name: Show all work for maximum credit. Each problem is worth 10 points. Work 10 of the 11 problems. k = 9.0 x 10 9 N m / C ε 0 = 8.85 x 10-1 C / N m e

### Louisiana State University Physics 2102, Exam 2, March 5th, 2009.

PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 2, March 5th, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),

### Louisiana State University Physics 2102, Exam 3 April 2nd, 2009.

PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 3 April 2nd, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),

### Where k = 1. The electric field produced by a point charge is given by

Ch 21 review: 1. Electric charge: Electric charge is a property of a matter. There are two kinds of charges, positive and negative. Charges of the same sign repel each other. Charges of opposite sign attract.

### AP Physics C Mechanics Objectives

AP Physics C Mechanics Objectives I. KINEMATICS A. Motion in One Dimension 1. The relationships among position, velocity and acceleration a. Given a graph of position vs. time, identify or sketch a graph

### Circuits Capacitance of a parallel-plate capacitor : C = κ ε o A / d. (ρ = resistivity, L = length, A = cross-sectional area) Resistance : R = ρ L / A

k = 9.0 x 109 N m2 / C2 e = 1.60 x 10-19 C ε o = 8.85 x 10-12 C2 / N m2 Coulomb s law: F = k q Q / r2 (unlike charges attract, like charges repel) Electric field from a point charge : E = k q / r2 ( towards

### PHYS 2212 (Modern) Review. Electric Force and Fields

PHYS 2212 (Modern) Review Electric Force and Fields A permanent dipole and a charged particle lie on the x-axis and are separated by a distance d as indicated in the figure. The dipole consists of positive

### Two point charges, A and B, lie along a line separated by a distance L. The point x is the midpoint of their separation.

Use the following to answer question 1. Two point charges, A and B, lie along a line separated by a distance L. The point x is the midpoint of their separation. 1. Which combination of charges would yield

### Exam 2 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses.

Exam 2 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 Part of a long, straight insulated wire carrying current i is bent into a circular

### AMPERE'S LAW. B dl = 0

AMPERE'S LAW The figure below shows a basic result of an experiment done by Hans Christian Oersted in 1820. It shows the magnetic field produced by a current in a long, straight length of current-carrying

### Physics 42 Exam 2 PRACTICE Name: Lab

Physics 42 Exam 2 PRACTICE Name: Lab 1 2 3 4 Conceptual Multiple Choice (2 points each) Circle the best answer. 1.Rank in order, from brightest to dimmest, the identical bulbs A to D. A. C = D > B > A

### Physics 2220 Fall 2010 George Williams SECOND MIDTERM - REVIEW PROBLEMS

Physics 0 Fall 010 George Williams SECOND MIDTERM - REVIEW PROBLEMS The last four problems are from last years second midterm. Solutions are available on the class web site.. There are no solutions for,

### Physics 196 Final Test Point

Physics 196 Final Test - 120 Point Name You need to complete six 5-point problems and six 10-point problems. Cross off one 5-point problem and one 10-point problem. 1. Two small silver spheres, each with

### The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d:

PHYS 102 Exams Exam 2 PRINT (A) The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d: It is connected to a battery with constant emf V.

### AP Physics C. Electricity and Magne4sm Review

AP Physics C Electricity and Magne4sm Review Electrosta4cs 30% Chap 22-25 Charge and Coulomb s Law Electric Field and Electric Poten4al (including point charges) Gauss Law Fields and poten4als of other

### Practice Exam 1. Necessary Constants and Equations: Electric force (Coulomb s Law): Electric field due to a point charge:

Practice Exam 1 Necessary Constants and Equations: Electric force (Coulomb s Law): Electric field due to a point charge: Electric potential due to a point charge: Electric potential energy: Capacitor energy:

### Calculus Relationships in AP Physics C: Electricity and Magnetism

C: Electricity This chapter focuses on some of the quantitative skills that are important in your C: Mechanics course. These are not all of the skills that you will learn, practice, and apply during the

### 104 Practice Exam 1-2/21/02

104 Practice Exam 1-2/21/02 1. One mole of a substance contains 6.02 > 10 23 protons and an equal number of electrons. If the protons could somehow be separated from the electrons and placed in separate

### the electrical nature of matter is inherent in its atomic structure E & M atoms are made up of p+, n, and e- the nucleus has p+ and n

Electric Forces and Fields E & M the electrical nature of matter is inherent in its atomic structure atoms are made up of p+, n, and e- a.k.a Electricity and Magnetism the nucleus has p+ and n surrounding

### Physics 102 Spring 2006: Final Exam Multiple-Choice Questions

Last Name: First Name: Physics 102 Spring 2006: Final Exam Multiple-Choice Questions For questions 1 and 2, refer to the graph below, depicting the potential on the x-axis as a function of x V x 60 40

### 2014 F 2014 AI. 1. Why must electrostatic field at the surface of a charged conductor be normal to the surface at every point? Give reason.

2014 F 1. Why must electrostatic field at the surface of a charged conductor be normal to the surface at every point? Give reason. 2. Figure shows the field lines on a positive charge. Is the work done

### Physics 2212 K Quiz #1 Solutions Summer q in = ρv = ρah = ρa 4

Physics 2212 K Quiz #1 Solutions Summer 2016 I. (18 points A uniform infinite insulating slab of charge has a positive volume charge density ρ, and a thickness 2t, extending from t to +t in the z direction.

### PH 102 Exam I N N N N. 3. Which of the following is true for the electric force and not true for the gravitational force?

Name Date INSTRUCTIONS PH 102 Exam I 1. nswer all questions below. ll problems have equal weight. 2. Clearly mark the answer you choose by filling in the adjacent circle. 3. There will be no partial credit

### AP Physics C: Electricity and Magnetism

18 AP Physics C: Electricity and Magnetism Scoring Guidelines College Board, Advanced Placement Program, AP, AP Central, and the acorn logo are registered trademarks of the College Board. AP Central is

### ConcepTest PowerPoints

ConcepTest PowerPoints Chapter 16 Physics: Principles with Applications, 7 th edition Giancoli 2014 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely

### Exam 2 Solutions. ε 3. ε 1. Problem 1

Exam 2 Solutions Problem 1 In the circuit shown, R1=100 Ω, R2=25 Ω, and the ideal batteries have EMFs of ε1 = 6.0 V, ε2 = 3.0 V, and ε3 = 1.5 V. What is the magnitude of the current flowing through resistor

### P202 Practice Exam 2 Spring 2004 Instructor: Prof. Sinova

P202 Practice Exam 2 Spring 2004 Instructor: Prof. Sinova Name: Date: (5)1. How many electrons flow through a battery that delivers a current of 3.0 A for 12 s? A) 4 B) 36 C) 4.8 10 15 D) 6.4 10 18 E)

### c. They have electric charges that move freely d. Electrons are added to the rod a. charges are of unlike signs b. charges are of like signs

Physics Review Chapter 17 & 18 Name: Date: Period: 1. What sentence best characterizes electron conductors? a. They have low mass density b. They have high tensile strength c. They have electric charges

### PHYS 241 EXAM #2 November 9, 2006

1. ( 5 points) A resistance R and a 3.9 H inductance are in series across a 60 Hz AC voltage. The voltage across the resistor is 23 V and the voltage across the inductor is 35 V. Assume that all voltages

### Q1. Ans: (1.725) =5.0 = Q2.

Coordinator: Dr. A. Naqvi Wednesday, January 11, 2017 Page: 1 Q1. Two strings, string 1 with a linear mass density of 1.75 g/m and string 2 with a linear mass density of 3.34 g/m are tied together, as

NSWRS - P Physics Multiple hoice Practice lectrostatics Solution nswer 1. y definition. Since charge is free to move around on/in a conductor, excess charges will repel each other to the outer surface

### Physics 2B Winter 2012 Final Exam Practice

Physics 2B Winter 2012 Final Exam Practice 1) When the distance between two charges is increased, the force between the charges A) increases directly with the square of the distance. B) increases directly

### , where the sum is over all pairs of charges (q 1, q 2 ) that are

February 12, 2010 PHY2054 Solutions Exam I 1-3. Particle 1 has a mass of 1.0 g and charge 6.0 µc, and particle 2 has a mass of 2.0 g and charge (3.0 5.0 8) µc. While particle 2 is held in place, particle

### Physics 2212 GH Quiz #4 Solutions Spring 2016

Physics 2212 GH Quiz #4 Solutions Spring 2016 I. (18 points) A bar (mass m, length L) is connected to two frictionless vertical conducting rails with loops of wire, in the presence of a uniform magnetic

### AP Physics C. Electric Circuits III.C

AP Physics C Electric Circuits III.C III.C.1 Current, Resistance and Power The direction of conventional current Suppose the cross-sectional area of the conductor changes. If a conductor has no current,

### Solutions to practice problems for PHYS117B.02 Exam I

Solutions to practice problems for PHYS117B.02 Exam I 1. Shown in the figure below are two point charges located along the x axis. Determine all the following: y +4 µ C 2 µ C x 6m 1m (a) Find the magnitude

### Exam 2, Phy 2049, Spring Solutions:

Exam 2, Phy 2049, Spring 2017. Solutions: 1. A battery, which has an emf of EMF = 10V and an internal resistance of R 0 = 50Ω, is connected to three resistors, as shown in the figure. The resistors have

### Physics 9 Spring 2012 Midterm 1 Solutions

Physics 9 Spring 22 NAME: TA: Physics 9 Spring 22 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

### 2 Coulomb s Law and Electric Field 23.13, 23.17, 23.23, 23.25, 23.26, 23.27, 23.62, 23.77, 23.78

College of Engineering and Technology Department of Basic and Applied Sciences PHYSICS I Sheet Suggested Problems 1 Vectors 2 Coulomb s Law and Electric Field 23.13, 23.17, 23.23, 23.25, 23.26, 23.27,

### AP Physics Electromagnetic Wrap Up

AP Physics Electromagnetic Wrap Up Here are the glorious equations for this wonderful section. This is the equation for the magnetic force acting on a moving charged particle in a magnetic field. The angle

### Physics 2212 G Quiz #4 Solutions Spring 2018 = E

Physics 2212 G Quiz #4 Solutions Spring 2018 I. (16 points) The circuit shown has an emf E, three resistors with resistance, and one resistor with resistance 3. What is the current through the resistor

### PHYS 272 (Spring 2018): Introductory Physics: Fields Homeworks

PHYS 272 (Spring 2018): Introductory Physics: Fields Homeworks Note: the 1st homework is simply signing the honor pledge (but still it is compulsory); the actual homework starts with #2. And, please sign

### 7. A capacitor has been charged by a D C source. What are the magnitude of conduction and displacement current, when it is fully charged?

1. In which Orientation, a dipole placed in uniform electric field is in (a) stable (b) unstable equilibrium. 2. Two point charges having equal charges separated by 1 m in distance experience a force of

### Q1. Three point charges are arranged as shown in FIGURE 1. Find the magnitude of the net electrostatic force on the point charge at the origin.

Coordinator: Saleem Rao Monday, May 01, 2017 Page: 1 Q1. Three point charges are arranged as shown in FIGURE 1. Find the magnitude of the net electrostatic force on the point charge at the origin. A) 1.38

### Exam 2 Fall 2014

1 95.144 Exam 2 Fall 2014 Section instructor Section number Last/First name Last 3 Digits of Student ID Number: Show all work. Show all formulas used for each problem prior to substitution of numbers.

### 4 pt. (in J) 3.A

Mark Reeves - Physics 22, Fall 2011 1 A point charge of mass 0.0699 kg and charge q = +6.87 µc is suspended by a thread between the vertical parallel plates of a parallel-plate capacitor, as shown in the

### Good Luck! Mlanie LaRoche-Boisvert - Electromagnetism Electromagnetism and Optics - Winter PH. Electromagnetism and Optics - Winter PH

1 Notes: 1. To submit a problem, just click the Submit button under it. The Submit All button is not necessary. 2. A problem accepted as correct by CAPA will be highlighted in green. Once you see this,

### AC vs. DC Circuits. Constant voltage circuits. The voltage from an outlet is alternating voltage

Circuits AC vs. DC Circuits Constant voltage circuits Typically referred to as direct current or DC Computers, logic circuits, and battery operated devices are examples of DC circuits The voltage from

### AP Physics Study Guide Chapter 17 Electric Potential and Energy Name. Circle the vector quantities below and underline the scalar quantities below

AP Physics Study Guide Chapter 17 Electric Potential and Energy Name Circle the vector quantities below and underline the scalar quantities below electric potential electric field electric potential energy

### Physics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe. Useful Information. Your name sticker. with exam code

Your name sticker with exam code Physics 227 Final Exam December 18, 2007 Prof. Coleman and Prof. Rabe SIGNATURE: 1. The exam will last from 4:00 p.m. to 7:00 p.m. Use a #2 pencil to make entries on the

### Tactics Box 23.1 Using Kirchhoff's Loop Law

PH203 Chapter 23 solutions Tactics Box 231 Using Kirchhoff's Loop Law Description: Knight/Jones/Field Tactics Box 231 Using Kirchhoff s loop law is illustrated Learning Goal: To practice Tactics Box 231

### melectron= 9.1x10-31 kg e = 1.6x10-19 C MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Exam #1, PHYS 102 Name Chapters 16, 17, & 18 8 February 2006 Constants k=9x109 Nm2/C2 e o =8.85x10-12 F/m mproton=1.673x10-27 kg melectron= 9.1x10-31 kg e = 1.6x10-19 C MULTIPLE CHOICE. Choose the one

### AP Physics C 1998 Multiple Choice Questions Electricity and Magnetism

AP Physics C 1998 Multiple Choice Questions Electricity and Magnetism The materials included in these files are intended for use by AP teachers for course and exam preparation in the classroom; permission

### PHYS 1102 EXAM - II. SECTION: (Circle one) 001 (TH 9:30 AM to 10:45AM) 002 (TH 3:30 PM to 4:45 PM) You have 1 hr 45 minutes to complete the test

PHYS 1102 EXAM - II SECTION: (Circle one) 001 (TH 9:30 AM to 10:45AM) 002 (TH 3:30 PM to 4:45 PM) Your Name: Student ID: You have 1 hr 45 minutes to complete the test PLEASE DO NOT START TILL YOU ARE INSTRUCTED

### Exam 1 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1

Exam 1 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 A rod of charge per unit length λ is surrounded by a conducting, concentric cylinder

### PHYSICS 2B FINAL EXAM ANSWERS WINTER QUARTER 2010 PROF. HIRSCH MARCH 18, 2010 Problems 1, 2 P 1 P 2

Problems 1, 2 P 1 P 1 P 2 The figure shows a non-conducting spherical shell of inner radius and outer radius 2 (i.e. radial thickness ) with charge uniformly distributed throughout its volume. Prob 1:

### (D) Blv/R Counterclockwise

1. There is a counterclockwise current I in a circular loop of wire situated in an external magnetic field directed out of the page as shown above. The effect of the forces that act on this current is

### PHYSICS - CLUTCH CH 22: ELECTRIC FORCE & FIELD; GAUSS' LAW

!! www.clutchprep.com CONCEPT: ELECTRIC CHARGE e Atoms are built up of protons, neutrons and electrons p, n e ELECTRIC CHARGE is a property of matter, similar to MASS: MASS (m) ELECTRIC CHARGE (Q) - Mass

### Objects usually are charged up through the transfer of electrons from one object to the other.

1 Part 1: Electric Force Review of Vectors Review your vectors! You should know how to convert from polar form to component form and vice versa add and subtract vectors multiply vectors by scalars Find

### Physics 2220 Fall 2010 George Williams THIRD MIDTERM - REVIEW PROBLEMS

Physics 2220 Fall 2010 George Williams THIRD MIDTERM - REVIEW PROBLEMS Solution sets are available on the course web site. A data sheet is provided. Problems marked by "*" do not have solutions. 1. An

### Be careful with your time. Be careful with your time. Be careful with your time a( R) Vf = V i + at R = R o

PHY 262 Exam 1 7/23/12 This exam is closed book and closed notes, but open calculator. You have the first 75 minutes of classtime on Monday to complete the exam (8:30 am 9:45 am). In addition to a couple

### Electricity. Revision Notes. R.D.Pilkington

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

### General Physics II (PHYS 104) Exam 2: March 21, 2002

General Physics II (PHYS 104) Exam 2: March 21, 2002 Name: Multiple Choice (3 points each): Answer the following multiple choice questions. Clearly circle the response (or responses) that provides the

### Class 6. Capacitance and Capacitors. Physics 106. Winter Press CTRL-L to view as a slide show. Class 6. Physics 106.

and in and Energy Winter 2018 Press CTRL-L to view as a slide show. From last time: The field lines are related to the field as follows: What is the electric potential? How are the electric field and the

### Chapter 12. Magnetism and Electromagnetism

Chapter 12 Magnetism and Electromagnetism 167 168 AP Physics Multiple Choice Practice Magnetism and Electromagnetism SECTION A Magnetostatics 1. Four infinitely long wires are arranged as shown in the

### PH2200 Practice Final Exam Summer 2003

INSTRUCTIONS 1. Write your name and student identification number on the answer sheet. 2. Please cover your answer sheet at all times. 3. This is a closed book exam. You may use the PH2200 formula sheet

### Turn in scantron You keep these question sheets

Exam 2 on OCT. 15. 2018 - Physics 106 R. Schad YOUR NAME ¼À Turn in scantron You keep these question sheets 1) This is to identify the exam version you have IMPORTANT Mark the A 2) This is to identify

### Physics 4. Magnetic Induction. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Physics 4 Magnetic Induction Before we can talk about induction we need to understand magnetic flux. You can think of flux as the number of field lines passing through an area. Here is the formula: flux

### Notes and Solved Problems for Common Exam 2

Notes and Solved Problems for Common Exam 2 3. GAUSS LAW Key concepts: Gaussian Surface, Flux, Enclosed Charge Gauss Law is equivalent to the Coulomb law but sometimes more useful. Gauss law considers