As.rFall 12 PHY 122 Homework Solutions #1

Similar documents
Chapter 21. = C and point charge

Version 001 Electrostatics I tubman (12125) 1

Sample Questions for the AP Physics 1 Exam

Chapter 18. Electric Forces and Electric Fields

circular motion & gravitation physics 111N

Forces between charges

Version A Page The diagram shows two bowling balls, A and B, each having a mass of 7.00 kilograms, placed 2.00 meters apart.

v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

If you put the same book on a tilted surface the normal force will be less. The magnitude of the normal force will equal: N = W cos θ

Solution. Problem. Solution. Problem. Solution

Static Electricity Page 1. Static Electricity. Introduction: Structure of Atoms 2 Sample Curriculum, Materials Needed

Chapter 6. Work and Energy

C B A T 3 T 2 T What is the magnitude of the force T 1? A) 37.5 N B) 75.0 N C) 113 N D) 157 N E) 192 N

Physics 11 Assignment KEY Dynamics Chapters 4 & 5

Lab 1. Charges and Electrostatics

Electromagnetism Extra Study Questions Short Answer

Physics 53. Gravity. Nature and Nature's law lay hid in night: God said, "Let Newton be!" and all was light. Alexander Pope

Physics 112 Homework 5 (solutions) (2004 Fall) Solutions to Homework Questions 5

Newton s Law of Motion

Work, Energy and Power Practice Test 1

Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion

Physics 111: Lecture 4: Chapter 4 - Forces and Newton s Laws of Motion. Physics is about forces and how the world around us reacts to these forces.

39 kg of water at 10 C is mixed with 360 kg of ice at -7 C.

Two-Body System: Two Hanging Masses

Use the following information to deduce that the gravitational field strength at the surface of the Earth is approximately 10 N kg 1.

Serway_ISM_V1 1 Chapter 4

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

Video Killed the Radio Star! Watch a video of me explaining the difference between static and kinetic friction by clicking here.

Physics 30 Worksheet #10 : Magnetism From Electricity

Static Electricity. Section 4.2. Explaining Static Electricity

LAB 6: GRAVITATIONAL AND PASSIVE FORCES

Unit 8A: Systems in Action (Pg. 2 85) Chapter 2: Getting to Work (pg )

Chapter 20 Electrostatics and Coulomb s Law 20.1 Introduction electrostatics Separation of Electric Charge by Rubbing

LAB 6 - GRAVITATIONAL AND PASSIVE FORCES

PS-6.2 Explain the factors that determine potential and kinetic energy and the transformation of one to the other.

Cambridge International Examinations Cambridge International General Certificate of Secondary Education

GRAVITATIONAL FIELDS PHYSICS 20 GRAVITATIONAL FORCES. Gravitational Fields (or Acceleration Due to Gravity) Symbol: Definition: Units:

Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m

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

Physics 2A, Sec B00: Mechanics -- Winter 2011 Instructor: B. Grinstein Final Exam

Exploring Magnetism. DataQuest

XX. Introductory Physics, High School

4 Gravity: A Force of Attraction

Atoms and Elements. Outline Atoms Orbitals and Energy Levels Periodic Properties Homework

B) 286 m C) 325 m D) 367 m Answer: B

Newton s Law of Universal Gravitation

Solution Derivations for Capa #11

Newton s Second Law. ΣF = m a. (1) In this equation, ΣF is the sum of the forces acting on an object, m is the mass of

Interaction at a Distance

Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.

1. Mass, Force and Gravity

Chapter 9. particle is increased.

HW Set II page 1 of 9 PHYSICS 1401 (1) homework solutions

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

Lesson 29: Newton's Law of Universal Gravitation

AP Physics Applying Forces

Newton s Law of Universal Gravitation describes the attractive gravitational force that exists between any two bodies with the following equation:

SURFACE TENSION. Definition

Mechanics 1: Conservation of Energy and Momentum

Conceptual Questions: Forces and Newton s Laws

AP Physics - Chapter 8 Practice Test

W02D2-2 Table Problem Newton s Laws of Motion: Solution

Time allowed: 1 hour 45 minutes

Exam 2 is at 7 pm tomorrow Conflict is at 5:15 pm in 151 Loomis

The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION PHYSICAL SETTING PHYSICS. Friday, June 20, :15 to 4:15 p.m.

ACTIVITY 6: Falling Objects

AP1 Oscillations. 1. Which of the following statements about a spring-block oscillator in simple harmonic motion about its equilibrium point is false?

Name Partners Date. Energy Diagrams I

TEACHER ANSWER KEY November 12, Phys - Vectors

CHAPTER 6 WORK AND ENERGY

Unit 3 Work and Energy Suggested Time: 25 Hours

CLASS TEST GRADE 11. PHYSICAL SCIENCES: PHYSICS Test 3: Electricity and magnetism

WORK DONE BY A CONSTANT FORCE

Name Period WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS. 1. Stored energy or energy due to position is known as energy.

Lesson 3 - Understanding Energy (with a Pendulum)

F N A) 330 N 0.31 B) 310 N 0.33 C) 250 N 0.27 D) 290 N 0.30 E) 370 N 0.26

CHAPTER 26 ELECTROSTATIC ENERGY AND CAPACITORS

SPEED, VELOCITY, AND ACCELERATION

Instructors Guide: Atoms and Their Isotopes

HW Set VI page 1 of 9 PHYSICS 1401 (1) homework solutions

WORKSHEET: KINETIC AND POTENTIAL ENERGY PROBLEMS

Q3.2.a The gravitational force exerted by a planet on one of its moons is 3e23 newtons when the moon is at a particular location.

AP1 Electricity. 1. A student wearing shoes stands on a tile floor. The students shoes do not fall into the tile floor due to

Physics 9e/Cutnell. correlated to the. College Board AP Physics 1 Course Objectives

Rotational Motion: Moment of Inertia

E X P E R I M E N T 8

A = 6561 times greater. B. 81 times greater. C. equally strong. D. 1/81 as great. E. (1/81) 2 = 1/6561 as great.

Lecture-IV. Contact forces & Newton s laws of motion

Physics 125 Practice Exam #3 Chapters 6-7 Professor Siegel

Experiment: Static and Kinetic Friction

Representing Vector Fields Using Field Line Diagrams

5. Forces and Motion-I. Force is an interaction that causes the acceleration of a body. A vector quantity.

H 2O gas: molecules are very far apart

PHYSICS 111 HOMEWORK SOLUTION, week 4, chapter 5, sec 1-7. February 13, 2013

A pure covalent bond is an equal sharing of shared electron pair(s) in a bond. A polar covalent bond is an unequal sharing.

This topic explores the key concepts of magnetism as they relate to: the phenomenon of magnetism magnetic forces and fields a theory of magnetism.

TEACHER S CLUB EXAMS GRADE 11. PHYSICAL SCIENCES: PHYSICS Paper 1

Name Class Period. F = G m 1 m 2 d 2. G =6.67 x Nm 2 /kg 2

PHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013

3 5 Sample Science Task

Transcription:

As.rFall 12 PHY 122 Homework Solutions #1 Chapter 21 Question 4 A positively charged rod is brought close to a neutral piece of paper, which it attracts. Draw a diagram showing the separation of charge in the paper, and explain why attraction occurs. Solution Fig 21-9. The neutral piece of paper is composed of atoms and molecules which consist of positively charged nuclei surrounded by negatively charged electrons. The introduction of a positively charged rod will attract the electrons while maintaining their bond within the atom or molecule. Since the negatively charged electrons are closer than the positively charged nuclei the paper will be feel a stronger Coulombic attractive force than a Coulombic repulsive force. Chapter 21 Question 6 Contrast the net charge on a conductor to the free charges in the conductor. Solution The net charge is difference between the total positive and negative charge on the conductor. The free charges are the sum of all the total electrons that free to move within the conductor. Thus it is possible for a conductor to have free charges of electrons but a net charge of zero.

Chapter 21 Question 7 Figures 21 7 and 21 8 show how a charged rod placed near an uncharged metal object can attract (or repel) electrons. There are a great many electrons in the metal, yet only some of them move as shown. Why not all of them? Solution Fig 21-7 Fig 21-8 In Figure 21-7, electrons will move until the attractive force on the remaining conduction electrons due to the incoming charged rod is balanced by the repulsive force from electrons that have already gathered at the left end of the neutral rod. In Figure 21-8, conduction electrons will be repelled by the incoming rod and will leave the stationary rod through the ground connection until the repulsive force on the remaining conduction electrons due to the incoming charged rod is balanced by the attractive force from the net positive charge on the stationary rod. Chapter 21 Question 10 We are not normally aware of the gravitational or electric force between two ordinary objects. What is the reason in each case? Give an example where we are aware of each one and why. Solution The gravitational force between everyday objects on the surface of the Earth is extremely small. Consider two objects sitting on the floor near each other. They are attracted to each other, but the force of static friction for each is much greater than the gravitational force each experiences from the other. Even in an absolutely frictionless environment, the acceleration resulting from the gravitational force would be so small that it would not be noticeable in a short time frame. We are aware of the gravitational force between objects if at least one of them is very massive, as in the case of the Earth and satellites or the Earth and you. The electric force between two objects is typically zero or close to zero because ordinary objects are typically neutral or close to neutral. We are aware of electric forces between objects when the objects are charged. An example is the electrostatic force (static cling) between pieces of clothing when you pull the clothes out of the dryer.

Chapter 21 Question 18 Consider the electric field at the three points indicated by the letters A, B, and C in Fig. 21 51. First draw an arrow at each point indicating the direction of the net force that a positive test charge would experience if placed at that point, then list the letters in order of decreasing field strength (strongest first). Solution At point C, the positive test charge would experience zero net force. At points A and B, the direction of the force on the positive test charge would be the same as the direction of the field. This direction is indicated by the arrows on the field lines. The strongest field is at point A, followed (in order of decreasing field strength) by B and then C.

Chapter 21 Problem 3 (I) What is the magnitude of the force a +25μC charge exerts on a +2.5mC charge 28 cm away? Chapter 21 Problem 8 A person scuffing her feet on a wool rug on a dry day accumulates a net charge of -46μC. How many excess electrons does she get, and by how much does her mass increase? Chapter 21 Problem 9 (II) What is the total charge of all the electrons in a 15-kg bar of gold? What is the net charge of the bar? (Gold has 79 electrons per atom and an atomic mass of 197 u.) Chapter 21 Problem 11 (II) Two positive point charges are a fixed distance apart. The sum of their charges is Q T? What charge must each have in order to (a) maximize the electric force between them, and (b) minimize it? Chapter 21 Problem 20 (III) Two small charged spheres hang from cords of equal length, as shown in Fig. 21 55 and make small angles θ 1 and θ 2 with the vertical. (a) If Q 1 =Q, Q 2 =2Q and m 1 = m 2 =m determine the ratio θ 1 /θ 2. (b) If Q 1 =Q, Q 2 =2Q, m 1 =m, and m 2 =2m determine the ratio θ 1 /θ 2. (c) Estimate the distance between the spheres for each case. Chapter 21 Problem 26 (I) What is the electric field at a point when the force on a 1.25μC charge placed at that point is F = (3.0i 3.9j) x 10-3 N?