Pearson Physics Level 30 Unit VI Forces and Fields: Chapter 10 Solutions



Similar documents
The force between electric charges. Comparing gravity and the interaction between charges. Coulomb s Law. Forces between two charges

Solution Derivations for Capa #8

Episode 401: Newton s law of universal gravitation

Voltage ( = Electric Potential )

Charges, Coulomb s Law, and Electric Fields

Voltage ( = Electric Potential )

FXA Candidates should be able to : Describe how a mass creates a gravitational field in the space around it.

Physics 235 Chapter 5. Chapter 5 Gravitation

9.5 Amortization. Objectives

12. Rolling, Torque, and Angular Momentum

Magnetic Field and Magnetic Forces. Young and Freedman Chapter 27

Mechanics 1: Motion in a Central Force Field

The Electric Potential, Electric Potential Energy and Energy Conservation. V = U/q 0. V = U/q 0 = -W/q 0 1V [Volt] =1 Nm/C

The Role of Gravity in Orbital Motion

(a) The centripetal acceleration of a point on the equator of the Earth is given by v2. The velocity of the earth can be found by taking the ratio of

Chapter 19: Electric Charges, Forces, and Fields ( ) ( 6 )( 6

PY1052 Problem Set 8 Autumn 2004 Solutions

Exam 3: Equation Summary

10. Collisions. Before During After

Gravitation. AP Physics C

Chapter 22. Outside a uniformly charged sphere, the field looks like that of a point charge at the center of the sphere.

2 r2 θ = r2 t. (3.59) The equal area law is the statement that the term in parentheses,

Vector Calculus: Are you ready? Vectors in 2D and 3D Space: Review

F G r. Don't confuse G with g: "Big G" and "little g" are totally different things.

Displacement, Velocity And Acceleration

Deflection of Electrons by Electric and Magnetic Fields

Determining solar characteristics using planetary data

Multiple choice questions [60 points]

PHYSICS 111 HOMEWORK SOLUTION #13. May 1, 2013

Chapter 30: Magnetic Fields Due to Currents

Experiment 6: Centripetal Force

Gauss Law. Physics 231 Lecture 2-1

4a 4ab b (count number of places from first non-zero digit to

Chapter 2. Electrostatics

Chapter 17 The Kepler Problem: Planetary Mechanics and the Bohr Atom

Gravity. A. Law of Gravity. Gravity. Physics: Mechanics. A. The Law of Gravity. Dr. Bill Pezzaglia. B. Gravitational Field. C.

AP Physics Electromagnetic Wrap Up

7 Circular Motion. 7-1 Centripetal Acceleration and Force. Period, Frequency, and Speed. Vocabulary

Experiment MF Magnetic Force

CHAPTER 5 GRAVITATIONAL FIELD AND POTENTIAL

Electrostatic properties of conductors and dielectrics

8.4. Motion of Charged Particles in Magnetic Fields

Figure 2. So it is very likely that the Babylonians attributed 60 units to each side of the hexagon. Its resulting perimeter would then be 360!

Lesson 7 Gauss s Law and Electric Fields

2. Orbital dynamics and tides

Physics HSC Course Stage 6. Space. Part 1: Earth s gravitational field

UNIT CIRCLE TRIGONOMETRY

A r. (Can you see that this just gives the formula we had above?)

Multiple choice questions [70 points]

Forces & Magnetic Dipoles. r r τ = μ B r

Lab #7: Energy Conservation


Problems of the 2 nd and 9 th International Physics Olympiads (Budapest, Hungary, 1968 and 1976)

9.4 Annuities. Objectives. 1. Calculate the future value of an ordinary annuity. 2. Perform calculations regarding sinking funds.

est using the formula I = Prt, where I is the interest earned, P is the principal, r is the interest rate, and t is the time in years.

An Introduction to Omega

1240 ev nm 2.5 ev. (4) r 2 or mv 2 = ke2

Gravitation and Kepler s Laws Newton s Law of Universal Gravitation in vectorial. Gm 1 m 2. r 2

NUCLEAR MAGNETIC RESONANCE

Solutions for Physics 1301 Course Review (Problems 10 through 18)

INVESTIGATION OF FLOW INSIDE AN AXIAL-FLOW PUMP OF GV IMP TYPE

Converting knowledge Into Practice

Worked Examples. v max =?

Moment and couple. In 3-D, because the determination of the distance can be tedious, a vector approach becomes advantageous. r r

Fluids Lecture 15 Notes

The Binomial Distribution

The Detection of Obstacles Using Features by the Horizon View Camera

Analytical Proof of Newton's Force Laws

Magnetic Bearing with Radial Magnetized Permanent Magnets

Lecture 16: Color and Intensity. and he made him a coat of many colours. Genesis 37:3

Coordinate Systems L. M. Kalnins, March 2009

STUDENT RESPONSE TO ANNUITY FORMULA DERIVATION

Carter-Penrose diagrams and black holes

Newton s Law of Universal Gravitation and the Scale Principle

Spirotechnics! September 7, Amanda Zeringue, Michael Spannuth and Amanda Zeringue Dierential Geometry Project

Chapter 2 Coulomb s Law

SELF-INDUCTANCE AND INDUCTORS

Mechanics 1: Work, Power and Kinetic Energy

Things to Remember. r Complete all of the sections on the Retirement Benefit Options form that apply to your request.

(Ch. 22.5) 2. What is the magnitude (in pc) of a point charge whose electric field 50 cm away has a magnitude of 2V/m?

Lab M4: The Torsional Pendulum and Moment of Inertia

Do Vibrations Make Sound?

TECHNICAL DATA. JIS (Japanese Industrial Standard) Screw Thread. Specifications

TORQUE AND ANGULAR MOMENTUM IN CIRCULAR MOTION

Introduction to Fluid Mechanics

Products of the Second Pillar Pension

4.1 - Trigonometric Functions of Acute Angles

Lesson 8 Ampère s Law and Differential Operators

CHAPTER 9 THE TWO BODY PROBLEM IN TWO DIMENSIONS

Comparing Availability of Various Rack Power Redundancy Configurations

Comparing Availability of Various Rack Power Redundancy Configurations

Skills Needed for Success in Calculus 1

A discus thrower spins around in a circle one and a half times, then releases the discus. The discus forms a path tangent to the circle.

CHAPTER 10 Aggregate Demand I

VISCOSITY OF BIO-DIESEL FUELS

2. TRIGONOMETRIC FUNCTIONS OF GENERAL ANGLES

GAUSS S LAW APPLIED TO CYLINDRICAL AND PLANAR CHARGE DISTRIBUTIONS ` E MISN CHARGE DISTRIBUTIONS by Peter Signell, Michigan State University

Graphs of Equations. A coordinate system is a way to graphically show the relationship between 2 quantities.

Chapter 4: Fluid Kinematics

Lecture L26-3D Rigid Body Dynamics: The Inertia Tensor

Transcription:

Peason Physics Level 30 Unit VI Foces and Fields: hapte 10 Solutions Student Book page 518 oncept heck 1. It is easie fo ebonite to eove electons fo fu than fo silk.. Ebonite acquies a negative chage when ubbed with fu. Student Book page 5 oncept heck If you bing the balloon nea a wall, atos on the suface of the wall expeience chage sepaation though the pocess of induction. As a esult, the suface of the wall closest to the balloon is oppositely chaged and attaction occus. Since the balloon and the wall ae insulatos, the tansfe of chage while they ae in contact is difficult and occus vey slowly. The balloon will stick to the wall fo a long tie until enough chage has leaked fo the balloon to the wall. Then epulsion will occu because the balloon and the wall ae now siilaly chaged. 10.1 heck and Reflect Student Book page 53 Knowledge 1. Electostatics is the study of electic chages at est o stationay chages.. A siple expeient would involve testing diffeent objects to see if they eithe attacted o epelled each othe. The fact that thee wee two possible actions was evidence that thee wee two diffeent types of effects. This led scientists to conclude that thee ust be two diffeent types of chages. 3. (a) Siilaities: Both display attaction o epulsion. Thee ust be two diffeent types of electic chages and two diffeent types of agnetic poles. They both affect objects at a distance. (b) Diffeences: Electic objects ust be ubbed befoe any effects ae obseved. Magnetic objects affect only cetain othe objects. Magnetic effects see to oiginate fo two poles while electic effects do not. 4. (a) Electical conductivity depends on how tightly the nucleus of an ato holds onto the ato s oute electons. (b) Insulato, seiconducto, conducto, supeconducto (c) Students answes ay vay. The following ae exaples of possible answes: Insulato glass Seiconducto seleniu Peason Physics Solutions Unit VI hapte 10 1 opyight 007 Peason Education anada

onducto coppe Supeconducto coppe oxide alloys (d) Seleniu is an insulato in the dak and a conducto when exposed to light. Applications 5. (a) The object neae the top of the chat will becoe negatively chaged since it has a geate affinity fo electons. (b) The geate the sepaation of two objects in the chat, the easie the tansfe of chage. 6. (a) To chage the glass sphee positively by fiction, ub the glass od with silk. (b) To chage the glass sphee positively by conduction, touch the glass sphee with a positively chaged object. (c) To chage the glass sphee positively by induction, bing a negatively chaged object nea the glass sphee. Gound the glass sphee, and then eove the negatively chaged object. The glass sphee will now have a positive chage. 7. To chage the sphee negatively by induction, bing a positively chaged object nea the sphee. Gound the sphee. Reove the gounding souce, and then eove the positively chaged object. The sphee will now have a negative chage. 8.(a) As a negatively chaged od appoaches a neutal pith ball, electons in the pith ball ae epelled and shift to the othe side of the atos, causing the atos to becoe polaized. Since the positive nuclei ae close to the negatively chaged od than the negative electons, thee will be a net foce of attaction and the pith ball will swing towad the negatively chaged od. (b) Afte they touch, the pith ball will becoe negatively chaged by conduction. Thee will be a net foce of epulsion between the two objects and the pith ball will swing away fo the negatively chaged od.. (a) Both objects will becoe negatively chaged by conduction. Howeve, since aluiniu is a conducto, the chages edistibute evenly thoughout the suface of the aluiniu and since glass is an insulato, the chages eain on the suface of the glass at the point of contact with the negatively chaged object. (b) When a negatively chaged sall etal sphee touches a lage etal sphee, electons tansfe fo the salle etal sphee to the lage etal sphee, and the lage etal sphee will becoe chaged by conduction. Since it is lage, it will now have a lage negative chage than the salle etal sphee. 10. (a) When a negatively chaged object is held nea the knob of an electoscope, electons in the knob ae epelled and ove to the leaves. The leaves divege, indicating that they ae now chaged with negative chages. (b) When the knob is gounded, the excess electons in the leaves ae eoved to the gound. The leaves will now etun to thei oiginal neutal position. (c) The chaging od holds the chages in the knob in place. If the chaging od is eoved befoe the gound, then the knob will becoe gounded, and thee will be no esidual chage on the electoscope. Extensions 11. Student answes ay vay. One exaple pocedue would be: Rub the ebonite od with fu. The ebonite is now chaged negatively. Touch the ebonite od to chage the electoscope negatively by conduction. Bing the electoscope nea the unknown chage. Peason Physics Solutions Unit VI hapte 10 opyight 007 Peason Education anada

If the unknown sphee is negatively chaged, the electoscope leaves should divege futhe. If the unknown sphee is positively chaged, the leaves on the electoscope would convege. 1. haging an object by fiction involves a tansfe of electons fo one object to anothe. No new chage has been poduced. The net positive chage on the glass od will equal the net negative chage on the silk, as would be pedicted by the law of consevation of chage. Student Book page 530 Exaple 10.1 Pactice Poble 1. Given 1 q1 1.60 10 1 q + 1.60 10 11 5. 10 Requied agnitude of the electostatic foce acting on the two chages ( F e ) Accoding to Newton s thid law, the electostatic foces acting on the two chages ae the sae in agnitude but opposite in diection. To calculate the agnitude of the electostatic foce acting on the two chages, use qq 1 Fe k The negative and the positive signs fo the chages ae only used in the final answe to indicate if the electostatic foce is attactive o epulsive. oulob s constant is k 8. 10 N /. The electostatic foce is attactive because the chages ae opposite to each othe. qq 1 Fe k 8. 10 N 1 1 (1.60 10 )(1.60 10 ) 11 (5. 10 ) 8 8. 10 N Paaphase The electostatic foce acting on the two chages is 8. 10 8 N. Since one chage is negative and the othe is positive, the electostatic foce is attactive. Peason Physics Solutions Unit VI hapte 10 3 opyight 007 Peason Education anada

Student Book page 531 Exaple 10. Pactice Poble 1. Given 6 q1 3.00 10 6 q +.00 10 1 1.0 c o 1.0 10 Requied agnitude of the electostatic foce acting on the two sphees afte they have touched one anothe and sepaated ( F e ) Accoding to Newton s thid law, the electostatic foces acting on the two sphees ae the sae in agnitude but opposite in diection. When the two sphees oentaily touch one anothe, the esulting net chage is divided equally between the. When a sphee with a negative chage of 3.00 10 6 oentaily touches a sphee with a positive chage of.00 10 6, then.00 10 6 is eoved fo the fist sphee to neutalize the +.00 10 6 of the second sphee. This leaves a negative chage of 1.00 10 6 on the fist sphee, divided equally between the two sphees. Each sphee now has a negative chage of 0.500 10 6. To calculate the agnitude of the electostatic foce acting on the two sphees, use: qq 1 Fe k oulob s constant is k 8. 10 N /. The new electostatic foce will be epulsive because both chages ae negative: qq 1 Fe k 8. 10 N 6 6 (0.500 10 )(0.500 10 ) 1 (1.0 10 ) 1 1.56 10 N Paaphase The electostatic foce acting on the two chages is 1.56 10 1 N. Since both sphees have a negative chage, the electostatic foce is epulsive. oncept heck Siilaities: Both foces vay with 1. Both foces ae action-at-a-distance foces. Both foces have siila equations. Student Book page 53 Peason Physics Solutions Unit VI hapte 10 4 opyight 007 Peason Education anada

Diffeences: Electostatic foces ae uch geate. Electostatic foces can be attactive o epulsive. Gavitational foces ae only attactive. Exaple 10.3 Pactice Pobles 1. The electostatic foce of chage A on chage B is attactive to the left, and the electostatic foce of chage on chage B is attactive to the ight. The poduct of the chages is the sae fo both foces. The only facto affecting the agnitudes of the foces of chages A and on B is 1 distance, whee Fe. Since the distance between chages A and B is one-half the distance between chages and B, then F 1 B is o 1 4 F AB. The vecto diaga is shown below. The vecto to the left epesents F AB. It is fou ties longe than the vecto to the ight, which epesents F. B. The two electostatic foces ae vectos along the sae line, so the net foce can be deteined by adding the vectos in the solution to Pactice Poble 1. The esultant vecto is shown below. Student Book page 533 Exaple 10.4 Pactice Pobles 1. Given q q q A B AB B.50 10 + 1.50 10 1.00 10 1.50 c o 1.50 10.00 c o.00 10 Peason Physics Solutions Unit VI hapte 10 5 opyight 007 Peason Education anada

Requied net electostatic foce on sphee B ( F net ) The electostatic foce of q A on q B is an attactive foce to the left. The electostatic foce of q on q B is an attactive foce to the ight. Since the two electostatic foces ae vectos along the sae line, the net foce can be deteined by adding the vectos. F net F AB + F B Use the following convention fo vecto addition: Vectos diected to the ight ae positive. Vectos diected to the left ae negative. F F + F net AB B 8. 10 N (.50 10 )(1.50 10 + (1.50 10 ) 4 5 1.48 10 N + 3.371 10 N 4 1.16 10 N ) 8. 10 N (1.50 10 )(1.00 10 ) (.00 10 ) Paaphase The net electostatic foce on the second chage is 1.16 10 4 N towad the.50 10 chage o left.. Given qa.50 10 qb + 1.50 10 q 1.00 10 3.50 c o 3.50 10 A Requied position of the second sphee so that the net electostatic foce on it is zeo ( AB) The distance of the second chage fo the fist chage is AB. The distance of the second chage fo the thid chage is ( 3.50 10 c AB ). The electostatic foce of q A on q B is an attactive foce to the left. The electostatic foce of q on q B is an attactive foce to the ight. Since the two electostatic foces ae foce vectos along the sae line, the net foce can be deteined by adding the vectos. F net F AB + F B Use the following convention fo vecto addition: Vectos diected to the ight ae positive. Vectos diected to the left ae negative. If the F net 0, then: Peason Physics Solutions Unit VI hapte 10 6 opyight 007 Peason Education anada

0 F + F 8. 10 N (.50 10 )(1.50 10 ) 8. 10 ( AB) 8. 10 8. 10 8. 10 AB B 8. 10 N (.50 10 )(1.50 10 ) 8. 10 + ( AB ) N (1.50 10 )(1.00 10 ) (3.50 10 AB) 8. 10 N (1.50 10 )(1.00 10 ) N (.50 10 )(1.50 10 ) ( AB ) (3.50 10 AB) N (.50 10 )(1.50 10 ) ( AB) N (3.50 10 AB) (1.50 10 )(1.00 10 ) AB.5 3.50 10 AB (3.50 10 AB ) (.5) AB (3.50 10 ) (.5) ( AB)(.5) AB (3.50 10 ) (.5) AB(1 +.5). 10 AB N (1.50 10 )(1.00 10 ) (3.50 10 AB) Paaphase The net electostatic foce on the second chage is 0 N when the second chage is placed. 10 to the ight of the fist chage. Student Book page 534 Exaple 10.5 Pactice Pobles 1. Given q.50 q q X Y Z XY ZY + 3.00 + 4.00 1.0 10 1.0 10 Requied net electostatic foce on chage Y ( F net ) The electostatic foce of q X on q Y is an attactive foce ( F XY ) on a line fo chage Y towad chage X. The electostatic foce of q Z on q Y is a epulsive foce ( F ZY ) diectly upwad fo chage Y. Peason Physics Solutions Unit VI hapte 10 7 opyight 007 Peason Education anada

X Y To calculate the electostatic foce of q X on q Y use: F q q XY k XY Daw a fee-body diaga of the electostatic foces on chage Y: F q q k X Y XY XY F N 8. 10 (.50 )(3.00 ) (1.0 10 ) 4.68 10 N qq k Z Y ZY ZY N 8. 10 (4.00 )(3.00 ) (1.0 10 ) 7.4 10 N Deteine the agnitude of the foce using the Pythagoean theoe: F (7.4 10 N) + (4.68 10 N) net 8.83 10 N Deteine the angle θ: 7.4 10 N tanθ 4.68 10 N θ 58 Fo the vecto diaga, this angle is [58 o N of W]. Using the atesian ethod, this angle is 180 o 58 o 1 o. Peason Physics Solutions Unit VI hapte 10 8 opyight 007 Peason Education anada

Paaphase The net electostatic foce on chage Y is 8.83 10 N [1 ].. Given qa.00 μ qb + 1.00 μ q.00 μ AB 1.00 10 B 1.00 10 Requied net electostatic foce on chage B ( F net ) Daw a fee-body diaga of the electostatic foces on chage B: Deteine the electostatic foce of chage A on chage B (attaction towad A): qaqb F k AB AB N 6 6 8. 10 (.00 10 )(1.00 10 ) (1.00 10 ) 1.78 10 N Deteine the electostatic foce of chage on chage B (attaction towad ): qq B F k B B N 6 6 8. 10 (.00 10 )(1.00 10 ) (1.00 10 ) 1.78 10 N Peason Physics Solutions Unit VI hapte 10 opyight 007 Peason Education anada

Deteine the agnitude of the net foce using the Pythagoean theoe: F (1.78 10 N) + (1.78 10 N) net.54 10 N Deteine the angle θ: 1.78 10 N tanθ 1.78 10 N θ 45.0 Fo the vecto diaga, the diection of the net foce is [45.0 S of W] o [5 ]. Paaphase The net electostatic foce on chage B is.54 10 N [45.0 S of W] o [5 ]. Student Book page 535 Exaple 10.6 Pactice Pobles 1. Given qx.50 qy + 3.00 qz + 4.00 XY 1.0 10 ZY 1.0 10 Requied net electostatic foce on chage Y ( F net ) Daw a fee-body diaga of the electostatic foces on chage Y (see below). Peason Physics Solutions Unit VI hapte 10 10 opyight 007 Peason Education anada

Deteine the electostatic foce of chage X on chage Y (attaction towad X): qxqy FXY k XY 8. 10 N (.50 )(3.00 ) (1.0 10 ) 4.68 10 N Deteine the electostatic foce of chage Z on chage Y (epulsion away fo Z): qq Z Y F k ZY ZY 8. 10 N (4.00 )(3.00 ) (1.0 10 ) 7.4 10 N Resolve F XY into x and y coponents as shown below: F (4.68 10 N)(cos 60.0 ) XY x.341 10 N F (4.68 10 N)(sin 60.0 ) XY y 4.055 10 N The electostatic foce of chage Z on chage Y has only an x coponent. F ZY 7.4 10 N x F ZY 0 N y Find the su of the x and y coponents. F net F XY + F ZY F net x F XY x + F ZY x.341 10 N + ( 7.4 10 N) 5.151 10 N Peason Physics Solutions Unit VI hapte 10 11 opyight 007 Peason Education anada

F net y F XY y + F ZY y 4.055 10 N + 0 N 4.055 10 N Use tigonoety to deteine the diection and agnitude of the net electostatic foce on chage Y, as shown below. Deteine the agnitude of the foce using the Pythagoean theoe: F (5.151 10 N) + (4.055 10 N) net 6.56 10 N Deteine the angle θ using the tangent function: 4.055 10 N tanθ 5.151 10 N θ 38. Fo the diaga, this angle is N of W. The diection of the net foce is [38. N of W] o [ ]. Paaphase The net electostatic foce on chage Y is 6.56 10 N [ ].. Given Identify the chages as A, B,, and D as shown in the diaga. q q q q A B D AB B +.0 +.0 +.0 +.0 1 4.00 10 1 3.00 10 Peason Physics Solutions Unit VI hapte 10 1 opyight 007 Peason Education anada

Requied net electostatic foce on chage B ( F net ) Daw a fee-body diaga of the electostatic foces on chage B like the one below: Deteine the distance between chages B and D by using the Pythagoean theoe (see diaga below). 1 1 DB (3.00 10 ) + (4.00 10 ) 1 5.00 10 Deteine the electostatic foce of chage A on chage B (epulsion away fo A): qaqb FAB k AB 8. 10 N (.0 )(.0 ) 1 (4.00 10 ) 11.71 10 N Deteine the electostatic foce of chage on chage B (epulsion away fo ): qq B F k B B 8. 10 N (.0 )(.0 ) 1 (3.00 10 ) 11 4.835 10 N Peason Physics Solutions Unit VI hapte 10 13 opyight 007 Peason Education anada

Deteine the electostatic foce of chage D on chage B: qdqb F k DB DB 8. 10 N (.0 )(.0 ) 1 (5.00 10 ) 11 1.740 10 N Use tigonoety to deteine the angle θ 1 of the diagonal line fo D to B. 1 3.00 10 tanθ1 1 4.00 10 θ1 36.87 Then esolve F DB into x and y coponents. F F DB x DB y 11 11 (1.740 10 N)(cos36.87 ) 1.3 10 N 11 (1.740 10 N)(sin 36.87 ) 11 1.044 10 N + F B F net F AB + F DB F net x F AB x + F B x + F DB x.71 10 11 N + 0 N + 1.3 10 11 N 4.111 10 11 N F net y F AB y + F B y + F DB y 0 N + 4.835 10 11 N + 1.044 10 11 N 5.87 10 11 N Use tigonoety to deteine the net electostatic foce on chage B, as shown in the following diaga. Peason Physics Solutions Unit VI hapte 10 opyight 007 Peason Education anada

Deteine the agnitude of the net foce using the Pythagoean theoe: 11 11 F (5.87 10 N) + (4.111 10 N) net 11 7.17 10 N Deteine the angle θ using the tangent function: 11 5.87 10 N tanθ 11 4.111 10 N θ 55.0 Fo the diaga, the diection of the net foce is [55.0 ]. Paaphase The net electostatic foce on chage B is 7.17 10 11 N [55.0 ] by the atesian ethod. Student Book page 538 10. heck and Reflect Knowledge 1. The two factos that affect the electostatic foce ae the aount of each chage and the distance between the centes of the chages. The atheatical elationships ae: 1 Fe and F e q1q.. oulob easoned that since thee appeaed to be syety in natue, the electostatic foces should vay accoding to the invese squae law that explained gavitational foces. 3. The sallest unit of chage is one electon. It has a chage of 1.60 10 1, which was deteined expeientally by Millikan. 4. oulob would touch the two objects togethe, easoning that chage would be divided equally each tie they wee in contact. Applications 5. (a) If Fe q1q, doubling one chage doubles the foce, which would be 0 N. (b) If Fe q1q, doubling both chages quaduples the foce, which would be 40 N. Peason Physics Solutions Unit VI hapte 10 15 opyight 007 Peason Education anada

6. (a) Attaining a lage chage on objects is difficult because of the lage foces of epulsion: the object loses this chage to the suoundings. 5.0 10 10 (b) 3.1 10 e 1 1.60 10 e 10 The object gained 3.1 10 electons. 7. Given 5 q1 5.00 10 5 q 6.00 10.00 Requied (a) electostatic foce ( F e ) (b) new electostatic foce ( F e ) (a) Accoding to Newton s thid law, the electostatic foces acting on the two sphees ae the sae in agnitude but opposite in diection. Thus, it is only necessay to calculate the agnitude of the foces. qq 1 Fe k 8. 10 N 5 5 (5.00 10 )(6.00 10 ) (.00 ) 6.74 N (b) When a sphee with a chage of 5.00 10 5 oentaily touches a sphee with a chage of 6.00 10 5, this leaves a chage of 11.0 10 5, divided equally between the two sphees. Each sphee now has a chage of 5.50 10 5. The new electostatic foce will be epulsive because both sphees have like chages: qq 1 Fe k 8. 10 N 5 5 (5.50 10 )(5.50 10 ) (.00 ) 6.80 N Paaphase (a) The electostatic foce acting on the two chages is 6.74 N. Since the chages ae like chages, the electostatic foce is epulsive. (b) The electostatic foce acting on the two chages is 6.80 N. Since the sphees have like chages, the electostatic foce is epulsive. Peason Physics Solutions Unit VI hapte 10 16 opyight 007 Peason Education anada

8. Given qa.00 qb + 3.00 q.00 AB.00 B 3.00 Requied (a) net electostatic foce on chage A ( F net ) (b) net electostatic foce on chage B ( F net ) (a) The distance between q A and q is: A.00 + 3.00 5.00 The electostatic foce of q B and q A is an attactive foce to the ight. The electostatic foce of q and q A is an epulsive foce to the left. Since the two electostatic foces ae vectos along the sae line, the net foce can be deteined by adding the vectos. F net F BA + F A onside ight to be positive. 8. 10 N (.00 )(3.00 ) 8. 10 N (.00 )(.00 ) Fnet + (.00 ) (5.00 ) 10 1.34 10 N + ( 1.438 10 N) 10 1.1 10 N (b) The electostatic foce of q A on q B is an attactive foce to the left. The electostatic foce of q on q B is an attactive foce to the ight. Since the two electostatic foces ae vectos along the sae line, the net foce on chage B can be deteined by adding the vectos. F net F AB + F B 8. 10 N (.00 )(3.00 ) 8. 10 N (.00 )(3.00 ) Fnet + (.00 ) (3.00 ) 10 1.34 10 N + 5.3 10 N 7.50 10 N Paaphase (a) The net electostatic foce on chage A is 1.1 10 10 N towad chage B. (b) The net electostatic foce on chage B is 7.50 10 N towad chage A. Peason Physics Solutions Unit VI hapte 10 17 opyight 007 Peason Education anada

Extensions. No. The electostatic foces ae caused by both chages, and accoding to Newton s thid law, each foce is equal in agnitude but opposite in diection. 10. Thee ae also foces of electostatic epulsion acting in you body that balance the attactive foces. Student Book pages 540 541 hapte 10 Review Knowledge 1. An electostatic chage is a chage at est.. They depend on how tightly the atos of these ateials hold onto thei outeost electons. 3. Seleniu becoes a good conducto when exposed to light and an insulato when kept in the dak. Mateials with this popety ae called seiconductos. 4. Objects ay becoe chaged by fiction, conduction, o induction that includes a gounding step. 5. The object ade of a ateial that holds onto electons oe tightly will becoe negatively chaged. 6. These pocesses ae alike in that thee is a tansfe of electons though contact. The diffeence is that fiction allows oe contact between the two objects. 7. The law of chages states that like chages epel and unlike chages attact. 8. Benjain Fanklin is given cedit fo naing the chages.. The law of consevation of chages states that the total aount of chage in a syste ust eain the sae. 10. Seleniu becoes a seiconducto when it is exposed to light, wheeas geaniu becoes a seiconducto when doped with galliu. Fo this eason, seleniu is used in photocopies. 11. Given 6 q1 3.00 10 6 q.50 10 0.00 Requied electostatic foce ( F e ) Accoding to Newton s thid law, the electostatic foces exeted by the two chages on each othe ae the sae in agnitude but opposite in diection. qq 1 Fe k 8. 10 N 6 6 (3.00 10 )(.50 10 ) (0.00 ) 1.6 N The electostatic foce is epulsive because the chages ae the sae. Peason Physics Solutions Unit VI hapte 10 18 opyight 007 Peason Education anada

Paaphase The electostatic foce acting on the two chages is 1.6 N. Since both chages ae negative, the electostatic foce is epulsive. 1. Given q1 5.00 q 5.00 3 F 5.00 10 N e Requied distance between the two chages () qq 1 Fe k kq q F 1 e N 8. 10 (5.00 )(5.00 ) 3 (5.00 10 N ) 3 6.70 10 Paaphase The distance between the two chages is 6.70 10 3. 13. Given qa.50 qb + 3.0 q 1.60 AB 1.50 B 1.70 3.0 A Requied (a) net electostatic foce on chage B ( F net ) (b) net electostatic foce on chage ( F net ) (a) The electostatic foce of q A on q B is an attactive foce to the left. The electostatic foce of q on q B is an attactive foce to the ight. Since the two electostatic foces ae vectos along the sae line, the net foce on q B can be deteined by adding the vectos. F net F AB + F B hoose ight to be positive. Peason Physics Solutions Unit VI hapte 10 1 opyight 007 Peason Education anada

8. 10 N (.50 )(3.0 ) 8. 10 N (1.60 )(3.0 ) Fnet + (1.50 ) (1.70 ) 10 10 3.16 10 N + 1.53 10 N 10 1.60 10 N (b) The electostatic foce of q A on q is a epulsive foce to the ight. The electostatic foce of q on q B is an attactive foce to the left. Since the two electostatic foces ae vectos along the sae line, the net foce can be deteined by adding the vectos. F net F A + F B hoose ight to be positive. 8. 10 N (.50 )(1.60 ) 8. 10 N (1.60 )(3.0 ) Fnet (3.0 ) (1.70 ) 10 3.51 10 N 1.53 10 N 10 1.4 10 N Paaphase (a) The net electostatic foce on chage B is 1.60 10 10 N [left]. (b) The net electostatic foce on chage is 1.4 10 10 N [left]. Applications. As electons fo the RT stike the sceen of the TV, the sceen becoes negatively chaged and attacts dust paticles in the ai by induction. 15. In ateials with low electical esistance, vey little enegy is lost to heat, so they ae bette conductos of electicity. 16. The flae ionizes the ai nea the chaged od, allowing electons fo the od to leak to the chaged ai olecules. 17. Attaction can occu between a chaged object and a neutal object; howeve, epulsion can occu only if two objects have the sae chage. 18. hages fo chaged objects will leak into the huid ai. 1. Initially, the chaged ball attacts the neutal ball though the pocess of induction. When they touch, the neutal ball acquies a like chage though conduction, and epulsion occus. 0. Any chage that the od accuulates though fiction with the fu is gounded by you hand, so the od eains neutal. 1. No. The peson will be able to dischage soe of the chage fo the object only by becoing chaged though conduction.. (a) The agnitude of the foce is popotional to the two chages and the distance 1 between the: Fe q1q and Fe Peason Physics Solutions Unit VI hapte 10 0 opyight 007 Peason Education anada

F 1 ( ) e 16 The new electostatic foce will be 16 ties geate if the chage on both sphees is doubled and the sepaation distance is halved. F e 16 10 N 160N The new foce is 160 N. qq 1 [You can also deteine the answe by using the equation Fe k.] (b) The agnitude of the foce is popotional to the two chages and the distance 1 between the: Fe q1q and Fe 3 Fe 3 6 The new electostatic foce will be 6 as uch. 6 F e 10N 6.7 N The new foce will be 6.7 N. qq 1 [You can also deteine the answe by using the equation Fe k 3. Given 1 q 1.60 10 q e + p e 11 1 + 1.60 10 5. 10 31.11 10 kg 7 + 1.67 10 kg p Requied F e atio of the electostatic foce to the gavitational foce F g To calculate the agnitude of the electostatic foce acting on the two chages, use: q q + e p Fe k To calculate the agnitude of the gavitational foce on the two asses, use: + e p F G g.] Peason Physics Solutions Unit VI hapte 10 1 opyight 007 Peason Education anada

oulob s constant is k 8. 10 N /. The univesal gavitation constant is G 6.67 10 11 N /kg. q q + e p k F e F + g e p G N 1 8. 10 (1.60 10 ) 11 N 6.67 10 31 7 (.11 10 kg )(1.67 10 kg ) kg 3.7 10 Paaphase The electostatic foce is.7 10 3 ties geate than the gavitational foce. 4. Given Daw and label the equilateal tiangle with X at the top vetex, Y at the botto left vetex, and Z at the botto ight. qx.50 qy.50 qz.50 XY 0.00 ZY 0.00 0.00 XZ Requied net electostatic foce on chage X ( F net ) net electostatic foce on chage Y ( F net ) net electostatic foce on chage Z ( F net ) Peason Physics Solutions Unit VI hapte 10 opyight 007 Peason Education anada

The angles in an equilateal tiangle ae all 60.0. Fist calculate the net electostatic foce on chage X. The electostatic foce of q Y on q X is a epulsive foce 60.0 to the x-axis. The electostatic foce of q Z on q X is a epulsive foce 10 to the x-axis. Since the two electostatic foces on chage X ae vectos that ae at an angle, use the coponent ethod to deteine the net foce. Daw a fee-body diaga of the electostatic foces on chage X: The agnitude of the electostatic foce of chage Y on chage X (epulsion away fo Y) is: qxqy F k YX XY 8. 10 N (.50 )(.50 ) (0.00) 1 1.405 10 N The electostatic foce of chage Y on chage X is 1.405 10 1 N [60.0 ]. The agnitude of the electostatic foce of chage Z on chage X (epulsion away fo Z) is: q q F Z k 8. 10 N (.50 )(.50 ) (0.00) 1 1.405 10 N X ZX XZ The electostatic foce of chage Z on chage X is 1.405 10 1 N [10 ]. Peason Physics Solutions Unit VI hapte 10 3 opyight 007 Peason Education anada

Resolve F YX into x and y coponents: F (1.405 10 N)(cos 60.0 ) YX x 7.05 10 N 1 F (1.405 10 N)(sin 60.0 ) YX y 1 11 1 1.17 10 N Resolve F ZX into x and y coponents: F F ZX ZX x y 1 (1.405 10 N)(cos 60.0 ) 11 7.05 10 N 1 (1.405 10 N)(sin 60.0 ) 1 1.17 10 N F net x 7.05 10 11 N + 7.05 10 11 N 0 F 1.17 10 1 N + 1.17 10 1 N net y.43 10 1 N The net electostatic foce on chage X is.43 10 1 N on a line staight upwad fo X, between chages Y and Z. Assuption This tiangle is an equilateal tiangle, and all the chages ae the sae. Accoding to Newton s thid law, the net foce on chage Y and the net foce on chage Z should have the sae agnitudes as the net foce on chage X. The diections of these foces ae away fo each chage and between the othe two chages, as shown in the following diaga. Peason Physics Solutions Unit VI hapte 10 4 opyight 007 Peason Education anada

Paaphase The net foces on the thee chages ae: X.43 10 1 N [0 o ] Y.43 10 1 N [10 ] Z.43 10 1 N [330 ] 5. (a) (b) The shape of the gaph suggests the following elationship: 1 1 Fe o (c) Recipocal of Sepaation Foce of Distance Squaed (1/ Repulsion ) ( F ) (N) ( 10 3 / e ) 10.0 360.0.50 8. 1.11 40.0 0.65 7.5 0.400.4 Peason Physics Solutions Unit VI hapte 10 5 opyight 007 Peason Education anada

(d) (e) ise slope un 360.0 N.4 N 10000 400 0.0360 N qq 1 (f) Since Fe k and Fe slope 1 F then slope kq1q (g) slope kq1q kq q e slope k 0.0360 N 8. 10 N 6.00 10 Each chage is.00 10 6. Extensions 6. No. A neutal object has equal nubes of opposite chages. 7. Yes. A single negative chage o electon and a single positive chage o poton can exist. 8. You cannot chage the coin by ubbing it between you finges because you ae gounded. Peason Physics Solutions Unit VI hapte 10 6 opyight 007 Peason Education anada

. Lightning on Eath is caused by dischages between negatively chaged clouds and the positively chaged suface of Eath. On Satun, lightning is caused by dischages between ings with diffeent chages. 30. (a) onnect the two sphees with the conducting wie. Rub the glass od with the silk to give it a positive chage by fiction. Hold the glass od nea one etal sphee. Repulsion of chages will cause chage igation, and, though induction, the sphee closest to the od will be negatively chaged. The othe etal sphee will have an equal positive chage. (b) With the sphees connected by the wie and the od close to one sphee, gound the two sphees. Then eove the gound and the chaged od. Each sphee will have an equal net negative chage due to the pocess of induction and gounding. 31. (a) A tansfe of electons has occued. One stip has acquied a net negative chage, while the othe stip has acquied an equal net positive chage, accoding to the law of consevation of chage. Since they have opposite chages, they attact. (b) When the two stips ae stuck to a table and peeled off, they both acquie the sae chage fo the table, and so they epel each othe. Peason Physics Solutions Unit VI hapte 10 7 opyight 007 Peason Education anada