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- Arizona
- Arizona State University - Tempe
- Physics
- Physics 131
- Adams/menendez
- [SOLUTIONS] Mastering Physics HW10.pdf

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HW10 Due: 11:59pm on Thursday, September 24, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] In HW10, there is one problem that must be written up and turned in at recitation. You will find that problem, with due date, at the Homework Schedule site on our class web page. GBA Electric Potential Energy versus Electric Potential Learning Goal: To understand the relationship and differences between electric potential and electric potential energy. In this problem we will learn about the relationships between electric force , electric field , potential energy , and electric potential . To understand these concepts, we will first study a system with which you are already familiar: the uniform gravitational field. Gravitational Force and Potential Energy First we review the force and potential energy of an object of mass in a uniform gravitational field that points downward (in the direction), like the gravitational field near the earth's surface. Part A Find the force on an object of mass in the uniform gravitational field when it is at height . Express in terms of , , , and . ANSWER: = Correct Because we are in a uniform field, the force does not depend on the object's location. Therefore, the variable does not appear in the correct answer. Part B Now find the gravitational potential energy of the object when it is at an arbitrary height . Take zero potential to be at position . Express in terms of , , and . Note that because potential energy is a scalar, and not a vector, there will be no unit vector in the answer. ANSWER: = Correct Part C In what direction does the object accelerate when released with initial velocity upward? ANSWER: upward downward upward or downward depending on its mass downward only if the ratio of to initial velocity is large enough Correct Electric Force and Potential Energy Now consider the analogous case of a particle with charge placed in a uniform electric field of strength , pointing downward (in the direction) Part D Find , the electric force on the charged particle at height . Hint D.1 Relationship between force and electric field Hint not displayed Express in terms of , , , and . ANSWER: = Correct Part E Now find the potential energy of this charged particle when it is at height . Take zero potential to be at position . Express (a scalar quantity) in terms of , , and . ANSWER: = Correct Part F In what direction does the charged particle accelerate when released with upward initial velocity? ANSWER: upward downward upward or downward depending on its charge downward only if the ratio of to initial velocity is large enough Correct Electric Field and Electric Potential The electric potential is defined by the relationship , where is the electric potential energy of a particle with charge . Part G Find the electric potential of the uniform electric field . Note that this field is not pointing in the same direction as the field in the previous section of this problem. Take zero potential to be at position . Express in terms of , , and . ANSWER: = Correct Part H The SI unit for electric potential is the volt. The volt is a derived unit, which means that it can be written in terms of other SI units. What are the dimensions of the volt in terms of the fundamental SI units? Express your answer in terms of the standard abbreviations for the fundamental SI units: (meters), (kilograms), (seconds), and (coulombs) ANSWER: volts = Correct Part I The electric field can be derived from the electric potential, just as the electrostatic force can be determined from the electric potential energy. The relationship between electric field and electric potential is , where is the gradient operator: . The partial derivative means the derivative of with respect to , holding all other variables constant. Consider again the electric potential corresponding to the field . This potential depends on the z coordinate only, so and . Find an expression for the electric field in terms of the derivative of . Express your answer as a vector in terms of the unit vectors , , and/or . Use for the derivative of with respect to . ANSWER: = Correct Part J A positive test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy. Hint J.1 Direction of the electric field Hint not displayed Hint J.2 Formula for the force on a charge in an electric field Hint not displayed Hint J.3 Formula for electric potential energy Hint not displayed Choose the appropriate answer combination to fill in the blanks correctly. ANSWER: higher; higher higher; lower lower; higher lower; lower Correct Part K A negative test charge will accelerate toward regions of ________ electric potential and ________ electric potential energy. Hint K.1 Direction of the electric field Hint not displayed Hint K.2 Formula for the force on a charge in an electric field Hint not displayed Hint K.3 Formula for electric potential energy Hint not displayed Choose the appropriate answer combination to fill in the blanks correctly. ANSWER: higher; higher higher; lower lower; higher lower; lower Correct A charge in an electric field will experience a force in the direction of decreasing potential energy. Since the electric potential energy of a negative charge is equal to the charge times the electric potential ( ), the direction of decreasing electric potential energy is the direction of increasing electric potential. In "Potential Difference and Potential near a Charged Sheet", the sheet is located at y=0. Also, the coordinates (x1,y1) and (x2,y2) are intended to refer to GENERAL points on the same side of the sheet, not the particular points shown in the figure. GBA Potential Difference and Potential near a Charged Sheet Let and be two points near and on the same side of a charged sheet with surface charge density . The electric field due to such a charged sheet has magnitude everywhere, and the field points away from the sheet, as shown in the diagram. Part A What is the potential difference between points A and B? Hint A.1 Formula for potential difference Hint not displayed Hint A.2 Calculating the line integral Hint not displayed Express your answer in terms of some or all of , , , , and . ANSWER: Correct Note that the expression will not yield the correct potential if you apply it to two points on opposite sides of the sheet. For example, the expression does not indicate that two points on opposite sides of the sheet and the same distance from it are at the same potential ( ), which is clear from the symmetry of the situation. If you take care in carrying out the integration to observe the change in the direction of the electric field as you pass from one side of the sheet to the other, you will find that the potential difference between A and B is actually given by . Recall that the potential difference, the quantity asked for in Part A, is a well-defined quantity for any situation. The potential, however, is only defined once you pick a point as the zero-potential point. Different choices simply change the potential by an additive constant, so the potential difference will stay the same, regardless of what point you designate as having zero potential. Part B If the potential at is taken to be zero, what is the value of the potential at a point at some positive distance from the surface of the sheet? Hint B.1How to approach the problem Hint not displayed ANSWER: 0 Correct Part C Now take the potential to be zero at instead of at infinity. What is the value of at point A some positive distance from the sheet? Hint C.1 How to approach the problem Hint not displayed ANSWER: 0 Correct Note that the potential is zero everywhere on the sheet, that is, at every point whose y coordinate is zero. You always have the freedom to choose a convenient location and reference potential with respect to which other potentials are measured, since it is potential differences and not absolute potentials that actually matter when one is doing something with charges in the real world. Potentials, however, are a useful calculational and bookkeeping tool. For example, if there were four points of interest in an electrical unit, there would be six possible potential differences, so it would be easier to keep track of the four potentials corresponding to the four points instead of working with potential differences. For the case of a charged sheet, it is clear that choosing the potential at the sheet to be zero is a more convenient choice than choosing the potential to be zero far away from the sheet. In this way, the potentials of points near the sheet remain finite. The opposite is true for a point charge. Work on a Charge Moving in a Constant Field A charge of 32.0 is placed in a uniform electric field that is directed vertically upward and has a magnitude of 3.60×10 4 . Part A What work is done by the electric force when the charge moves a distance of 0.460 in an easterly direction? ANSWER: = 0 Correct Part B What work is done by the electric force when the charge moves a distance of 0.700 upward? ANSWER: = 8.06×10 ?4 Correct Part C What work is done by the electric force when the charge moves a distance of 2.60 at an angle of 59 downward from the easterly horizontal? ANSWER: = ?2.57×10 ?3 Correct Problem 23.53 A particle with a charge of 9.90 is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved a distance of 9.00 , the additional force has done an amount of work equal to 7.90×10 ?5 and the particle has kinetic energy equal to 3.20×10 ?5 . Part A What work was done by the electric force? ANSWER: = ?4.70×10 ?5 Correct Part B What is the potential of the starting point with respect to the endpoint? ANSWER: -4750 Correct Part C What is the magnitude of the electric field? ANSWER: = 5.27×10 4 Correct Stopping the Proton An infinitely long line of charge has a linear charge density of 4.50×10 ?12 . A proton is at distance 19.0 from the line and is moving directly toward the line with speed 2300 . Part A How close does the proton get to the line of charge? Hint A.1 How to approach the problem Since the only force acting on the proton is the conservative electric force, the best way to approach the problem is to use energy conservation. The initial speed of the proton when it is at distance 19.0 from the line of charge is given, and its final speed, when it comes to a stop, will be zero. Thus, its initial and final kinetic energy can easily be calculated. The change in potential energy of the proton can be found from the potential difference between the starting and ending points of the proton's path. Note that you can calculate the potential difference from the electric field. In fact, the linear charge density is given, so the electric field around the line of charge can be calculated using the symmetry of the system. Hint A.2 Calculate the final kinetic energy Calculate , the final kinetic energy of the proton when it comes to a stop. Hint A.2.1The final speed of the proton Hint not displayed Express your answer numerically in joules. ANSWER: = 0 Correct The proton has zero kinetic energy only at the instant it stops. Then it will reverse its direction of motion and start to move away from the line of charge due to the repulsive electric force. Hint A.3 Calculate the initial kinetic energy Calculate , the initial kinetic energy of the proton when it is at distance 19.0 from the line of charge. Hint A.3.1The initial speed of the proton Hint not displayed Express your answer numerically in joules. ANSWER: = 4.42×10 ?21 Correct Hint A.4 Electric field around a line of charge An infinite line of charge will generate a field with only a radial component. The electric field around a line of charge is given by the equation , where 4.50×10 ?12 is the linear charge density of the line and is the radial distance from the line at which the electric field is evaluated. Hint A.5 Potential difference over the path of the proton If the electric field is known, the potential difference between two points and is given by the equation , where is an infinitesimal displacement along a path between and . For this problem, since the electric field caused by an infinitely long line of charge has only a radial component, the equation above can be rewritten as , where is the potential difference between the initial and the final locations of the proton, which are at 19.0 and from the line of charge. You already know values for 4.50×10 ?12 , 8.85×10 ?12 , and , but the other variables, , , and , are all still unknown. Hint A.6 Putting it all together Since the total energy of the proton must be constant, the sum of its initial kinetic and potential energies must be equal to the sum of its final kinetic and potential energies; that is, , which can be rewritten as . Furthermore, the potential energy of the proton will be related to the potential at any point by , where 1.60×10 ?19 is the charge of the proton. Thus, you can relate the change in potential energy of the proton to the potential difference that you calculate by integrating the electric field between the initial and final locations of the proton. Substitute your result into the equation of conservation of energy and solve for , the final distance of the proton from the line of charge. Express your answer in meters. ANSWER: 0.135 Correct Another way you could have solved this problem is by defining the point where to be at either the initial or final location of the proton, and calculating the corresponding potential at the other point. However, since only the potential difference is needed (or potential energy difference) to solve the problem, this would be one more unnecessary step to worry about. You may wish to wait until after Lecture 11 to answer the following two questions. For "Relationship between Electric Force and Electric Potential Conceptual Question" it is useful to have a ruler. In that problem, the 40-V equipotential line may be considered to go to the lower right-hand corner of the figure. GBA Electric Fields and Equipotential Surfaces The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1- increments. Part A What is the work done by the electric force to move a 1- charge from A to B? Hint A.1 Find the potential difference between A and B Hint not displayed Hint A.2 Potential difference and work Hint not displayed Express your answer in joules. ANSWER: = 0 Correct Part B What is the work done by the electric force to move a 1- charge from A to D? Hint B.1Find the potential difference between A and D Hint not displayed Hint B.2Potential difference and work Hint not displayed Express your answer in joules. ANSWER: = 1 Correct Part C The magnitude of the electric field at point C is Hint C.1 Electric field and equipotential surfaces Hint not displayed ANSWER: greater than the magnitude of the electric field at point B. less than the magnitude of the electric field at point B. equal to the magnitude of the electric field at point B. unknown because the value of the electric potential at point C is unknown. Correct Relationship between Electric Force and Electric Potential Conceptual Question Three points (A, B, and C) are located on equipotential lines as shown. Part A A proton is released from Point A. Indicate the direction of the electric force vector acting on the proton. Hint A.1 Determining the direction of the electric force Hint not displayed ANSWER: The electric force vector at Point A points to the left. Correct Part B An electron is released from Point B. Indicate the direction of the electric force vector acting on the electron. Hint B.1Determining the direction of the electric force Hint not displayed ANSWER: The electric force vector at Point B points to the right. Correct Part C An electron is released from Point B and a second electron is released from Point C. What can you say about the electric forces experienced by these electrons the instant they are released? Hint C.1 Relating electric field strength to change in electric potential Hint not displayed ANSWER: The electron released at Point B experiences a greater force. The electron released at Point C experiences a greater force. Electrons released from Points B and C would experience equal forces. The relationship between the two forces cannot be determined. Correct In conclusion, recall that the electric field is always perpendicular to the equipotential line and points in the direction of decreasing potential. The electric force acting on a charge will then depend on the sign of the charge. On positive charges, the electric force is parallel to the field; on negative charges, the electric force is in the opposite direction as the field. Score Summary: Your score on this assignment is 99.1%. You received 49.54 out of a possible total of 50 points. clockwork MasteringPhysics: Assignment Print View

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