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: HW2 CHAPTER 21, QUESTIONS 6, 14, 15, 17 CHAPTER 22, QUESTIONS 11 CHAPTER 21, PROBLEM 11 CHAPTER 22, PROBLEMS 1, 4, 13, 15
QUESTIONS : Q21-6) Contrast the net charge on a conductor to the "free charges". Q21-14) Explain why we use small test charges when measuring electric fields. Q21-15) When determining an electric field, must we use a positive test charge, or would a negative one do as well? Explain. Q21-17) Assume that the two opposite charges in Fig. 21-33a are 12.0 cm apart. Consider the magnitude of the electric field 2.5 cm from the positive charge. On which side of this charge top, bottom, left, or right is the electric field the strongest? The weakest? Q22-11) A point charge Q is surrounded by a spherical surface of radius r0, whose center is at Q. Later, the charge is moved to the right a distance r0, but the sphere remains where it was, Fig. 22-22. How is the electric flux FE through the sphere changed? Is the electric field at the surface of the sphere changed? For each "yes" answer, describe the changes. PROBLEMS : P21-11) Three positive particles of charges 11.0 mC are located at the corners of an equilateral triangle of side 15.0 cm (Fig. 21-47). Calculate the magnitude and direction the of net force on each particle. P22-1) A flat circle of radius 15 cm is placed in a uniform electric field of magnitude 5.8 x 102 N/C. What is the electric flux through the circle when its face is (a) perpendicular to the field lines (b) at 45 to the field lines (c) parallel to the field lines? P22-4) A uniform field E is parallel to the axis of a hollow hemisphere of radius R, Fig. 22-25. (a) What is the electric flux through the hemispherical surface? (b) What is the result if E is perpendicular to the axis? P22-13) A 15.0 cm diameter nonconducting sphere carries a total charge of 12.0 mC distributed uniformly throughout its volume. Graph the electric field as a function of the distance r from the center of the sphere from r = 0 to r = 30 cm. Additionally, write down the formulas you have used and explain why the field increases within the sphere and decays outside of it.
P22-15) A spherical cavity of radius 4.50 cm is at the center of a metal sphere of radius 18.0cm. A point charge Q = 5.50 mC rests at the very center of the cavity, whereas the metal conductor carries no net charge. Determine the electric field at a point (a) 3.0 cm from the center of the cavity and (b) 6.0 cm from the center of the cavity.

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Rochester - PHY - 113

SOLUTIONS HW2 Q21-6) The net charge on a conductor is the sum over all positive and negative charges. It is zero if positive and negative charges are equal. Free charges are charges that can move inside the conductor. The conductors itself, though,

Rochester - PHY - 113

HW3 : CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER 22, 23, 24, 25, 22, 23, 24, 25, QUESTION 14 QUESTIONS 6, 18 QUESTIONS 6, 10 QUESTIONS 8, 11 PROBLEMS PROBLEMS PROBLEMS PROBLEMS 6 24, 29, 46 15, 67 7, 15, 36 QUESTIONS: Q22-14) In

Rochester - PHY - 113

SOLUTIONS HW3:Q22-14) The total charge that is enclosed in case (a) is +1.0 mC 2.0 mC = -1mC, in case (b) it is zero (as object 3 is neutral). Thus, using Gauss's law:E =Qencl0r=- 1.0 10 -6 C0in case (a) and zero in case (b).Q23-

Rochester - PHY - 113

HW4 : CHAPTER 24, QUESTIONS 14, 15 CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER 21, 22, 23, 24, 25, PROBLEM 18 PROBLEM 9 PROBLEMS 23, 47 PROBLEM 68 PROBLEM 6QUESTIONS : Q24-14) A dielectric is pulled from between the plates of a capacitor which remains

Rochester - PHY - 113

SOLUTIONS HW4: Q24-14) The potential difference remains constant as the capacitor is connected to a battery. The capacitance changes back its value without dielectric, The charge on the plates decreases toC0 =C . KQ . K C 2 C 2 C 2 The energy

Rochester - PHY - 113

HW 1: CHAPTER 21, QUESTIONS 4, 5, 13 CHAPTER 21, PROBLEMS 2, 4, 7, 8, 10QUESTIONS : Q21-4) A positively charged rod is brought close to a neutral piece of paper, which it attracts. Draw a diagram showing the separation of charge and explain why att

Rochester - PHY - 113

WM 1 CHAPTER 21, QUESTIONS 9, 10, 18, 19 CHAPTER 21, PROBLEMS 3, 14, 33, 37QUESTIONS : Q21-9) The form of Coulomb's law is very similar to that for Newton's law of universal gravitation. What are the differences between these two laws? Compare also

Rochester - PHY - 113

SOLUTIONS WM1 : Q21-9) Gravitation is always attractive as masses are always >0. The Coulomb force can be attractive (for oppositely charged objects, + and -) or repulsive (for equal signs, + and + or and -). Compare the size of the two forces for t

Rochester - PHY - 113

WM 2 : CHAPTER 21, QUESTIONS 21 CHAPTER 22, QUESTIONS 1, 5, 12 CHAPTER 21, PROBLEM 40 CHAPTER 22, PROBLEMS 3, 7, 12, 21QUESTIONS : Q21-21) Given two point charges, Q and 2Q, a distance l apart, is there a point along the straight line that passes t

Rochester - PHY - 113

SOLUTIONS WM2Q21-21) Equal charges repel each other, opposite charges attract. Thus, if the charges have opposite sign they will attract each other and the electric field lines will point from the positive charge to the negative charge; there is no

Rochester - PHY - 113

WM3 : CHAPTER CHAPTER CHAPTER CHAPTER 22, 23, 24, 25, QUESTION 13 QUESTIONS 3, 8, 16 QUESTIONS 3, 12, 13 QUESTIONS 3, 6, 12CHAPTER 23, PROBLEMS 5, 30 CHAPTER 24, PROBLEMS 7, 42 CHAPTER 25, PROBLEMS 1, 10, 21 QUESTIONS: Q22-13) A point charge q is p

Rochester - PHY - 113

SOLUTIONS WM3 Q22-13) Yes. The charge inside the cavity will pull the electrons of the metal to the inside of the shell's surface, thus charging it with q. The outside of the shell will then be left with a charge +q. Q23-3) a) The electric potential

Rochester - PHY - 113

WM 4 : CHAPTER 26, QUESTIONS 8, 9, 11 CHAPTER 26, PROBLEMS 9, 11, 29, 31 Last workshop before the first midterm use it for questions! QUESTIONS: Q26-8) Explain why Kirchhoff's junction rule is based on conservation of electric charge. Q26-9) Explain

Rochester - PHY - 113

SOLUTIONS MW4: Q26-8) Kirchhoff's junction rule states that the sum of the currents which enter a node is equal to the sum of all the currents which exit the same node. Current is equal to the charge which flows over a certain time. Thus, the amount

Rochester - PHY - 113

Rochester - PHY - 113

P114 CLASS TEST #1APRINT NAME UR ID#Prof. Cline1. _ 2. _ 3. _ 4. _ TOTAL _1. 2. 3. 4. 5. 6. 7. 8. 9. 10.DO NOT SIT NEXT TO A STUDENT WITH THE SAME COLOR TEST. THIS IS A CLOSED BOOK EXAM Print your name on each sheet Do all problems - Problem

Rochester - PHY - 113

P114 PRACTICE TEST #2 1; Two concentric solenoids both have a length 50cm. The primary coil has 1000 turns and radius 5cm. The secondary coil has 10, 000 turns and radius 10 cm. Show all steps to get full credit. a; (6pts) Derive the self inductance

Rochester - PHY - 113

P114 PRACTICE TEST #2NAME UR ID#Prof. Cline1; Two concentric solenoids both have a length 50cm. The primary coil has 1000 turns and radius 5cm. The secondary coil has 10, 000 turns and radius 10 cm. Show all steps to get full credit. a; (6pts) D

Rochester - PHY - 113

P114 - PROBLEM SET 1Homework to be handed in by 1800 hours on Tuesday 30 January 2007. [1] Electric and gravitational forces. a) What would be the proton's mass if the gravitational force between two protons at rest were to equal the electric force?

Rochester - PHY - 113

P114PROBLEM SET 2Spring 2007Homework to be handed in by 1700 hours on Tuesday, 6 February 2007. [1] A uniform line charge extends from x = -2.5cm to x = 2.5 cm and has linear charge density = 4.5 nC/m. a: Find the total charge b: Find the elec

Rochester - PHY - 113

P114PROBLEM SET 3Spring 2007Homework to be handed in by 1800 hours on 13 February 2007. [1] If a conducting sphere is to be charged to a potential of 10,000 V, what is the smallest possible radius of the sphere such that the electric field will

Rochester - PHY - 113

P114PROBLEM SET 6Spring 2007Homework to be handed in by 1700 hours on 6 March 2007. [1] Consider a single turn circular loop of radius 20 cm. a) At what point along the axis is the magnetic field 1% of that at the center? b) Find the current in

Rochester - PHY - 113

P114PROBLEM SET 7Spring 2007Homework to be handed in by 1800 hours on Tuesday 20 March 2007. [1] A 2.0 cm by 1.5 cm rectangular coil has 300 turns and rotates in a magnetic field of 4000 G. a; What is the maximum emf generated when the coil rot

Rochester - PHY - 113

P114PROBLEM SET 8Spring 2007Homework to be handed in by 1700 hours on 27 March 2007. [1] A plane parallel plate capacitor with circular plates of radius 1 meter and separation 1 mm is being charged with a current of 1 amp. (a) Show that the tot

Rochester - PHY - 113

P114PROBLEM SET 9Spring 2007Homework to be handed in by 1700 hours on 3 April 2007. [1] Light of wavelength 550 nm is incident on a long, narrow slit. Find the angle of the first diffraction minimum if the width of the slit is: a) 1mm b) 0.1mm

Rochester - PHY - 113

P114 - Problem Set 10Homework to be handed in by 1700hrs on Tuesday 10 April 2007 . [1] Light conditions on a certain day for a certain film call for an aperture stop of f/8 at 1/250 sec. a; If you wish to take a picture of a humming bird at 1/1000

Rochester - PHY - 113

P114PROBLEM SET 11Spring 2007Homework to be handed in by 1700 hours on 17 April 2007. [1] A penny weighs 2.5 g. Calculate the rest energy for a penny. If you convert this energy completely to electrical energy and sell it at 10 cent/kW-hour, ho

Rochester - PHY - 113

Rochester - PHY - 113

Physics 113, Fall 2005 Homework assignment 6 Due Tuesday, Oct. 25, 2005, 4:00pm Read Giancoli chapters 8 and 9. 1. Giancoli, Chapter 8, problem 14 (page 200). 2. Giancoli Chapter 8, problem 55 (page 203). 3. Giancoli Chapter 8, problem 87 (page 205)

Rochester - PHY - 113

Physics 113, Fall 2005 Homework assignment 9 Due Tuesday, November 22, 2005, 4:00pm Read Giancoli chapters 13 and 14. 1. Giancoli, Chapter 12, problem 48 (page 326). 2. Giancoli, Chapter 12, problem 54 (page 327). 3. Giancoli, Chapter 13, problem 34

Rochester - PHY - 113

Physics 113, Fall 2005 Homework assignment 10 Due Tuesday, Dec.6 2005, 4:00pm Read Giancoli chapters 14, 15 and 16.(Exclude sections 14-7,14-8, and 15-10.) 1. Giancoli, Chapter 14, problem 28 (page 383). 2. Giancoli, Chapter 14, problem 42 (page 383

Rochester - PHY - 113

Rochester - PHY - 113

4-46. (a) See drawing (b) If we select all three blocks as the system, we have F m1 Fx = max: F = (m1 + m2 + m3)a, which gives a = F/(m1 + m2 + m3). x (c) For the three blocks individually, for Fx = max we have F Fnet1 = m1a = m1F/(m1 + m2 + m3); Fne

Rochester - PHY - 113

Rochester - PHY - 113

Solutions to problem set 10 P113 Fall 200612-71. (a) The cylinder will roll about the contact point A.We write = I about the point A: Fa(2R h) + FN1[R2 (R h)2]1/2 Mg[R2 (R h)2]1/2 = IA. When the cylinder does roll over the curb, contact

Rochester - PHY - 113

Rochester - PHY - 113

Solutions to problem set 11 P113 Fall 200613-5413-75. The liquid pressure is produced from the elevation of the bottle: P = gh. (a) (65 mm-Hg)(133 N/m2 mm-Hg) = (1.00 103 kg/m3)(9.80 m/s2)h, which gives h = 0.88 m. (b) (550 mm-H2O)(9.81 N/m2

Rochester - PHY - 113

Phys 113 Problem Set 1 Solutions 1)vc = 90 km h 1000 m 750 km h 1000 m = 25 m s , vt = = 20.83 m s h 3600 s km h 3600 s km xc = vc t = ( 25 m s ) t , xt = x0t + vt t = 100 m + ( 20.83 m s ) txc = ct , ( 25 m s ) t = 100 m + ( 20.83 m s ) t , ( 4.1

Rochester - PHY - 113

Physics 113, Fall 2005 Homework assignment 2 Due Monday, Sept. 19, 2005, 4:00pm Read Giancoli chapter 3. 1. Giancoli, Chapter 2, problem 61. (page 42). Assume the acceleration due to gravity is 9.80m/s2 . 2. Giancoli, Chapter 2, problem 70 (page 42)

Rochester - PHY - 113

Physics 113, Fall 2005 Homework assignment 3 Due Monday Sept. 26, 2005, 4:00pm Read Giancoli chapter 4. 1. Giancoli, Chapter 3, problem 30 (page 71). 2. Giancoli, Chapter 3, problem 45 (page 72). 3. Giancoli, Chapter 3, problem 55 (page 73). 4. Gian

Washington - ACCTG - 225

Washington - ACCTG - 225

To Succeed in Accounting 225 . Come to big classes . prepared Attend all quiz sections . prepared Read the class notes, chapters and do the problems without the solutions Explore the textbook website and do the end-ofchapter quizzes Be an activ

Washington - SCAND - 151

19th-SCAND 151 AND 20 -CENTURY FINNISH LITERATURE AND CULTURE Prof. Andrew Nestingen http:/faculty.washington.edu/akn/scand151.htm Winter 2008thCourse Information5 Credits Room: Sieg 224 T., Th: 1:30-3:20. Office Hours: M. 11 :30-1 :20Instr

Washington - ACCTG - 225

The standard cost card for the production of 1 unit of Zippy at Hanson's Manufacturing is as follows:AStandard Quantity or Hours1.5 lbs. 1.5 hours 1.5 hoursBStandard Price or RateAxBStandard Cost per Unit6.00 18.00 4.50 28.50InputsDirect

Virginia Tech - PHYS - 2306

Physics 2306 Spring 2008 First Exam FORM A1) The pressure fluctuation from ambient pressure due to a sound wave is given by p(x,t) = (2.010-5 N/m2) sin[(5 radians/meter) x + (4000 radians/second) t] Which one of the following statements is true? A.

Virginia Tech - PHYS - 2306

Physics 2306 Spring 2008 Second Exam FORM A1) The electric field lines of two charges, A and B, are shown in the figure. The magnitude of charge A is 12.0 C. What is the charge (in Coulombs) of B? A. -10 B. 10 C. -4.0 D. 4.0 E. -8.0 F. 8.0 G. 20 H.

UIllinois - CEE - 300

BName _FIRST HOUR EXAMINATION - TAM 324/CEE300Friday February 16, 2007 Work all problems in spaces provided One page of hand-written notes permitted.1. (10 points) Nitrogen gas can be used to improve the properties of steel. A sheet of steel

UIllinois - CEE - 300

AName _SECOND HOUR EXAMINATION - TAM 324/CEE300Wednesday March 7, 2007 Work all problems in spaces provided One page of hand-written notes permitted.1. (15 points) Assume a single-crystal BCC material. (a)(5pts) Sketch the BCC crystal unit c

Colorado - CVEN - 3227

8.1 (a) B = stress = y / 0.1; A = strain = x / 10^^Let, E(B A = a) = + a Nos. 1 2 3 4 5 6 ai 9x10-4 20x10-4 28x10-4 41x10-4 52x10-4 63x10-4 213x10-4 bi 10 20 30 40 50 60 210 aibi 9x10-3 40x10-3 84x10-3 164x10-3 260x10-3 378x10-3 935x10-3 ai2 8

Colorado - CVEN - 3227

8.2 (a)(b)Stopping Distance (m) Yi Xi2 Yi2 XiYi Yi'=a+bXi (Yi-Yi')2Vehicle No.Speed (kph) XiTotal:On the basis of calculations in the above table we obtain the respective sample means of X and Y as,X = 679/12 = 56.6 kph,and correspondin

Colorado - CVEN - 3227

8.3 (a)Plot of peak hour traffic vs daily traffic1.61.41.2Peak hour traffic (1000 vehicles)10.80.60.40.20 0 1 2 3 4 5 6 7Daily traffic (1000 vehicles)(b)Let X be the daily traffic volume and Y the peak hour traffic volume, b

Colorado - CVEN - 3227

8.4 (a)(b)Country # Per Capita GNP Per Capita Energy ConsumptionXiYiXi2Yi2XiYiYi'=a+bXi(Yi-Yi')2Total:On the basis of calculations in the above table we obtain the respective sample means of X and Y as,X = 37800/8 = 4725,and c

Colorado - CVEN - 3227

8.5 (a) Let X be the car weight in kips; X ~ N(3.33, 1.04). Hence X - 4.5 - 3.33 ) P(X > 4.5) = P( > 1.04 = P(Z > 1.125) = 1 (1.125) = 1 - 0.8697 0.130(b) Let Y be the gasoline mileage. For linear regression, we assume E(Y | X = x) = + x and s

Colorado - CVEN - 3227

8.6 (a)Let Y be the number of years of experience, and M be the measurement error in inches. We have the following data: i 1 2 3 4 5 yi 3 5 10 20 25 63 mi 1.5 0.8 1 0.8 0.5 4.6 yimi 4.5 4 10 16 12.5 47 yi2 9 25 100 400 625 1159From Eq. 8.4 & 8.3

Colorado - CVEN - 3227

8.9 (a) The formulas to use are=x yi =1 n ini- nx y - nx2xi =12 iand^ = y-xThe various quantities involved are calculated in the following table: n= 3 xi yi xiyi 66.15 168.36 640.56 875.07 xi2 1.1025 3.3489 9.8596 14.311 yi' (yi

Colorado - CVEN - 3227

2.3 (a)(b) Wind directionE190oE2 E3 E1Wind Speed60o 30o 0 (c) 15 35 45A and B are not mutually exclusive A and C are not mutually exclusive

Colorado - CVEN - 3227

2.4 Possible water level Outflow Inflow Outflow + 7' 5' 8' 6' 7' 7' 6' 5' 9' 6' 8' 7' 7' 5' 10' 6' 9' 7' 8'(a) Inflow 6' 6' 6' 7' 7' 7' 8' 8' 8'Hence possible combinations of inflow and outflow are (6',5'), (6',6'), (6',7'), (7',5'), (7',6'), (7

Colorado - CVEN - 3227

2.8 P(E1) = 0.8; P(E2) = 0.7; P(E3) = 0.95 P(E3 E 2 ) = 0.6; assume E2 and E3 are statistically independent of E1 (a) A = (E2 E3) E1 B = ( E 2 E 3 ) E1or E 2 E 3 E1(b) P(B) = P( E1 E 2 E 3 ) = P( E1 ) + P( E 2 E 3 ) P( E1 E 2 E 3 ) = 0.2 + P

Kansas State University - MATH - 240

Ohio State - CHEM - 121

Lecture 2Chapter 2 Atoms, Molecules, and IonsToday's reading: section 1 to section 7The Atomic Structure - the Electron Cloud Model1Properties of Protons, Neutrons and ElectronsParticle Proton Neutron Electron Charge +1 0 -1 Mass (g) 1.6727

Emory - CHEM - 141

SI Questions for Chapter 10 Sections 000 and 004Write the major Lewis structure(s) for the following molecules. Include any formal charges and resonance structures. If resonance is possible, determine the major resonance structure. H2CN2 (2 skeletal