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Assignment MasteringPhysics: Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Physics 220 Assignment 3 Assignment is due at 11:59pm on Tuesday, September 26, 2006 Credit for problems submitted late will decrease to 0% over the course of 240 hour(s) after the deadline has passed. The wrong answer penalty is 1% per part. Multiple choice questions are penalized as described in the online help. There is no opened hint penalty or unopened hint bonus. You are allowed unlimited attempts per answer. This assignment is now complete. Section 3.1. Visualizing Electric Fields Description: Select the correct drawing of electric field lines for several situations and answer questions about why other choices are incorrect. Then, these ideas are demonstrated with an applet. Learning Goal: To understand the nature of electric fields and how to draw field lines. Electric field lines are a tool used to visualize electric fields. A field line is drawn beginning at a positive charge and ending at a negative charge. Field lines may also appear from the edge of a picture or disappear at the edge of the picture. Such lines are said to begin or end at infinity. The field lines are directed so that the electric field at any point is tangent to the field line at that point. The figure shows two different ways to visualize an electric field. On the left, vectors are drawn at various points to show the direction and magnitude of the electric field. On the right, electric field lines depict the same situation. Notice that, as stated above, the electric field lines are drawn such that their tangents point in the same direction as the electric field vectors on the left. Because of the nature of electric fields, field lines never cross. Also, the vectors shrink as you move away from the charge, and the electric field lines spread out as you move away from the charge. The spacing between electric field lines indicates the strength of the electric field, just as the length of vectors indicates the strength of the electric field. The greater the spacing between field lines, the weaker the electric field. Although the advantage of field lines over field vectors may not be apparent in the case of a single charge, electric field lines present a much less cluttered and more intuitive picture of more complicated charge arrangements. Part A 1 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Which of the following figures correctly depicts the field lines from an infinite uniformly negatively charged sheet? Note that the sheet is being viewed edge-on in all pictures. Hint A.1 Description of the field Recall that the field around an infinite charged sheet is always perpendicular to the sheet and that the field strength does not change, regardless of distance from the sheet. View Full Document

B C D Part B In the diagram from part A , what is wrong with figure B? (Pick only those statements that apply to figure B.) A. Field lines cannot cross each other. B. The field lines should be parallel because of the sheet's symmetry. C. The field lines should spread apart as they leave the sheet to indicate the weakening of the field with distance. D. The field lines should always end on negative charges or at infinity. Enter the letters corresponding to all choices that apply. Use no spaces or commas, and enter the letters in alphabetical order. For instance, if you think that the correct choices are C and D, enter CD. ANSWER: Part C 2 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Which of the following figures shows the correct electric field lines for an electric dipole? B C D Part D In the diagram from part C , what is wrong with figure D? (Pick only those statements that apply to figure D.) A. Field lines cannot cross each other. B. The field lines should turn sharply as you move from one charge to the other. C. The field lines should be smooth curves. D. The field lines should always end on negative charges or at infinity. Enter the letters corresponding to all choices that apply. Use no spaces or commas, and enter the letters in alphabetical order. For instance, if you think that the correct choices are C and D, enter CD. ANSWER: In even relatively simple setups as in the figure, electric field lines are quite helpful for understanding the field qualitatively (understanding the general direction in which a certain charge will move from a specific position, identifying locations where the field is roughly zero or where the field points a specific direction, etc.). A good figure with electric field lines can help you to organize your thoughts as well as check your calculations to see whether they make sense. 3 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Part E In the figure , the electric field lines are shown for a system of two point charges, could represent the magnitudes and signs of In the following, take and ? and . Which of the following to be a positive quantity. ANSWER: , , , , , Electric Fields and Forces 4 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Description: Electric forces and electric fields. Learning Goal: To understand Coulomb's law, electric fields, and the connection between the electric field and the electric force. Coulomb's law gives the electrostatic force charges and acting between two charges. The magnitude of the force between two depends on the product of the charges and the square of the distance , between the charges: where . The direction of the force is along the line connecting the two charges. If the charges have the same sign, the force will be repulsive. If the charges have opposite signs, the force will be attractive. In other words, opposite charges attract and like charges repel. Because the charges are not in contact with each other, there must be an intermediate mechanism to cause the force. This mechanism is the electric field. The electric field at any location is equal to the force per unit charge experienced by a charge placed at that location. In other words, if a charge experiences a force , the electric field at that point is . The electric field vector has the same direction as the force vector on a positive charge and the opposite direction to that of the force vector on a negative charge. An electric field can be created by a single charge or a distribution of charges. The electric field a distance point charge has magnitude from a . The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by , will not significantly affect the electric field if it is small compared to . (many times larger than the charge on a single electron). Imagine an isolated positive point charge with a charge Part A 5 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend? A. B. C. D. E. F. G. the distance between the positive charge and the electron the charge on the electron the mass of the electron the charge of the positive charge the mass of the positive charge the radius of the positive charge the radius of the electron Enter the letters corresponding to the correct choices in alphabetical order. For example, if you think that only B, D, and E are correct, enter BDE. ANSWER: Part B For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend? A. B. C. D. E. F. G. the distance between the positive charge and the electron the charge on the electron the mass of the electron the charge of the positive charge the mass of the positive charge the radius of the positive charge the radius of the electron Enter the letters corresponding to the correct choices in alphabetical order. For example, if you think that only B, D, and E are correct, enter BDE. ANSWER: Part C Recall that the positive charge is much greater than the charge on the electron. ANSWER: Part D 6 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... If the total positive charge is point P, a distance = 1.53 = 1.62106 , what is the magntidue of the electric field caused by this charge at from the charge? Enter your answer numerically in newtons per coulomb. ANSWER: Part E What is the direction of the electric field at point P? Enter the letter of the vector that represents the direction of . ANSWER: Part F Now find the magnitude of the force on an electron placed at point P. Recall that the charge on an electron has magnitude Part F.1 . Determine to how approach the problem What strategy can you use to calculate the force between the positive charge and the electron? 7 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... ANSWER: Use Coulomb's law. Multiply the electric field due to the positive charge by the charge on Do either of the above. Do neither of the above. the electron. Enter your answer numerically in newtons. ANSWER: Part G What is the direction of the force on an electron placed at point P? Enter the letter of the vector that represents the direction of . ANSWER: HW3_1 qE Part A In a region of space there is a uniform electric field of unit vectors to express your answer as a vector. ANSWER: 132 . What is the force on a proton in this field? Use Part B What is the force on an electron in the same field. Use unit vectors to express your answer as a vector. ANSWER: 8 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Problem 26.2 What are the strength and direction of the electric field at the position indicated by the dot in the figure ? Part A Specify the strength of the electric field. ANSWER: Part B Specify the direction as an angle above or below horizontal. ANSWER: Section. 3.2. HW3_2 BfromE At time t=0 a thread is emitted from a source particle which is located at the origin of a coordinate system (point S). The source is traveling in the +x direction with . The source has a charge of ,1.05 =1.00 . The thread arrives at a point P (x,y)=(2 ) a short time later. Part A Find the head line, . Write the x and y components of the vector as two numbers separated by a comma. 9 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Hint A.1 Lost? Look at the illustration to the left. Recall that the point S is at the origin. ANSWER: Part B Find , the magnitude of the head line. Lost? Hint B.1 Just use the pythagorean theorem. ANSWER: Part C Find the unit vector Hint C.1 . Enter it as components, as in Part A. Did you forget what a unit vector is? The unit vector is a vector divided by its magnitude. It is a vector of length 1 that points in the direction of the original vector. ANSWER: Part D Find the time at which the thread arrives at point P. We will call this time Hint D.1 Lost? . Remember that the thread travels at the speed of light. You can easily find the distance from S to P. ANSWER: Part E At time Hint E.1 the source has moved to point T along the x axis. What is the coordinate of point T? We'll call this Lost? You know how fast the source is moving and how long it travels. ANSWER: Part F 10 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Find the ray line, . Enter the x-component then the y component separated by a comma. The components will be in units of meters, but don't put m in your answer. Hint F.1 Lost? Remember that the ray line is the line that goes from T to P. Since you already know where those points are, the answer is easy. ANSWER: = m Part G Find the magnitude of the ray line, ANSWER: . Part H Find the value of Hint H.1 associated with ? . Did you forget ANSWER: Part I Find the value of Hint I.1 Lost? . Look at the figure -- remember that sine is opposite/hypotenuse. ANSWER: Part J Find the electric field using Eq. 2.14. Recall that the electric field is given by the force when the field particle is at rest divided by the charge of the field particle. Express your answer as the x component then the y component separated by a comma. ANSWER: Part K 11 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Use Eq. 3.4, separated by commas. Hint K.1 , to calculate the magnetic field. Enter, the x, y, and z components of as three numbers How do I do the cross product? , , , etc. Remember that ANSWER: Problem 34.2 Part A What is the magnitude of the force on the proton in the figure? ANSWER: Problem 34.4 Part A What electric field strength will allow the electron in the figure to pass through this region of space without being deflected? 12 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... ANSWER: Problem 34.6 An electron travels with electron at this point is Part A What is the electric field? Express vector ANSWER: in the form of , , , where the x, y, and z components are separated by commas. 5.40 8.90 through a point in space where 8.90 . 0.110 . The force on the Section 3.3. Problem 29.28 Part A What is the electric potential at the point indicated with the dot in the figure? ANSWER: Problem 29.26 In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A 13 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... What is the electric potential of the proton at the position of the electron? ANSWER: Part B What is the electron's potential energy? ANSWER: Problem 29.44 A proton's speed as it passes point A is . It follows the trajectory shown in the figure. Part A What is the proton's speed at point B? ANSWER: HW3_4 V Part A An electric potential is given by the formula electric field as a function of x. Hint A.1 How is E related to V? where is a positive constant. Find the magnitude of the ANSWER: 14 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Part B In what direction is the force on a positive field particle? Hint B.1 Need Help? Think of the problem like a potential energy problem. The potential well gets larger as x gets larger. If the field particle were negative, how would that affect your answer? ANSWER: toward positive x no force. toward negative x toward positive y toward negative y There is Section 3.5. HW3_5 FL Part A A few electric field lines are shown in the figure. Arrows have been intentionally omitted. What is the direction of the electric field at point P? ANSWER: left or right up or down (up is to the top of the screen) none of the above in the screen or out of the screen Part B Where is the field the strongest? B C D HW3_6 FC Part A 15 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... A magnetic field contour is shown in the figure. Arrows have been intentionally omitted. (The lines, of course, represent perpendicular surfaces.) What is the direction of the magnetic field at point P? ANSWER: left or right up or down (up is to the top of the screen) none of the above in the screen or out of the screen Part B Where is the field the strongest? B C D Section 3.6. Relativity is not needed in Problem 29.12. Problem 29.12 Part A What is the speed of a proton that has been accelerated from rest through a potential difference of ANSWER: Section 3.7. Relativity is not needed in Problem 32.34. Problem 32.34 The aurora is caused when electrons and protons, moving in the earth's magnetic field of molecules of the atmosphere and cause them to glow. Part A , collide with 16 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... What is the radius of the cyclotron orbit for an electron with speed ? ANSWER: Part B What is the radius of the cyclotron orbit for a proton with speed ANSWER: HW3_3 Linac You wish to build a linear accelerator that will accelerate electrons to 90% of the speed of light. You want adjacent drift tubes to be at +5000V and -5000V potential. In order to work this problem, you will need the relativistic formula for kinetic energy. ( doesn't work for really fast particles.) Kinetic energy is where is the rest energy of an object. The rest energy of an electron is 0.511 MeV. [Remember that if we want to convert eV to SI units, all we have to do is put everything in eV.] Part A Find the kinetic energy of an electron going 90% of the speed of light. Note that we want the answer in units of MeV. Hint A.1 Need help to get started? and . into the equation; however, for this problem, it's easiest just to leave Don't let the funny expression for kinetic energy bother you; just plug in the numbers. ANSWER: Part B How much kinetic energy does the electron gain each time it moves from one drift tube into the next? Hint B.1 Need help? The change in kinetic energy between drift tubes equals the change in potential energy. The potential energy is just . ANSWER: Part C How many drift tubes would you need for your accelerator? Hint C.1 Does it seem that you only need a fraction of a drift tube? 17 of 18 9/21/2006 9:58 AM MasteringPhysics: Assignment Print View file:///c:/Documents%20and%20Settings/lbr/Desktop/HW3_files/ass... Remember that the kinetic energy is in MeV ( ). ANSWER: 18 of 18 9/21/2006 9:58 AM ...