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Course Number: PHYS 260, Fall 2008

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MasteringPhysics: Assignment Print View Page 1 of 15 Assignment Display Mode: View Printable Answers Physics 260 Fall 2008 HW08 Due at 5:00pm on Thursday, November 6, 2008 View Grading Details Problem 28.40 Description: The figure shows a solid metal sphere at the center of a hollow metal sphere. (a) What is the total charge on the exterior of the inner sphere? (b) What is the total charge on the inside...

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Assignment MasteringPhysics: Print View Page 1 of 15 Assignment Display Mode: View Printable Answers Physics 260 Fall 2008 HW08 Due at 5:00pm on Thursday, November 6, 2008 View Grading Details Problem 28.40 Description: The figure shows a solid metal sphere at the center of a hollow metal sphere. (a) What is the total charge on the exterior of the inner sphere? (b) What is the total charge on the inside surface of the hollow sphere? (c) What is the total charge on ... The figure shows a solid metal sphere at the center of a hollow metal sphere. Part A What is the total charge on the exterior of the inner sphere? ANSWER: = Part B What is the total charge on the inside surface of the hollow sphere? ANSWER: = Part C What is the total charge on the exterior surface of the hollow sphere? http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 2 of 15 ANSWER: = Problem 28.41 Description: The earth has a vertical electric field at the surface, pointing down, that averages E. This field is maintained by various atmospheric processes, including lightning. (a) What is the excess charge on the surface of the earth? The earth has a vertical electric field at the surface, pointing down, that averages 112 by various atmospheric processes, including lightning. Part A What is the excess charge on the surface of the earth? Register to View Answer. This field is maintained Problem 28.50 Description: The figure shows two very large slabs of metal that are parallel and distance l apart. Each slab has a total surface area (top + bottom) A. The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the ... The figure shows two very large slabs of metal that are parallel and distance apart. Each slab has a total surface area (top + bottom) . The thickness of each slab is so small in comparison to its lateral dimensions that the surface area around the sides is negligible. Metal has total charge Assume and metal is positive. has total charge . Part A Determine the electric field strength ANSWER: j k l m n j k l m n in region . http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 3 of 15 j k l m n i j k l m n Part B Determine the electric field strength ANSWER: i j k l m n j k l m n j k l m n j k l m n in region . Part C Determine the electric field strength ANSWER: j k l m n i j k l m n j k l m n j k l m n in region . Part D Determine the electric field strength ANSWER: i j k l m n j k l m n j k l m n j k l m n in region . Part E http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 4 of 15 Determine the electric field strength ANSWER: j k l m n j k l m n j k l m n i j k l m n in region . Part F Determine the surface charge density ANSWER: j k l m n j k l m n j k l m n j k l m n j k l m n j k l m n i j k l m n on the surface . Part G Determine the surface charge density ANSWER: j k l m n i j k l m n j k l m n j k l m n j k l m n j k l m n on the surface . http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 5 of 15 j k l m n Part H Determine the surface charge density ANSWER: j k l m n j k l m n i j k l m n j k l m n j k l m n j k l m n j k l m n on the surface . Part I Determine the surface charge density ANSWER: j k l m n j k l m n j k l m n j k l m n j k l m n j k l m n i j k l m n on the surface . Problem 28.55 Description: All examples of Gauss's law have used highly symmetrical surfaces where the flux integral is either zero or E net Phi_e=Q_in/ epsilon_0 is independent of the surface. This is worth checking. The figure shows a cube of edge ... http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 6 of 15 All examples of Gauss's law have used highly symmetrical surfaces where the flux integral is either zero or is independent of the surface. This is worth checking. The figure shows a cube of edge length charge density . The flux through one face of the cube is not simply because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. . Yet we've centered on a Part A Consider the face parallel to the yz-plane. Define area as a strip of width and height with the vector pointing in th through this little area. Your is located at position . Use the known electric field of a wire to calculate the electric flux in terms of , which is a variable, and various constants. It should not explicitly contain any angles. Express your answer in terms of the variables ANSWER: , ,, , and appropriate constants. = Part B Now integrate to find the total flux through this face. , , and appropriate constants. Express your answer in terms of the variables ANSWER: = Part C Finally, show that the net flux through the cube is . http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 7 of 15 ANSWER: Answer Key: Because there are four faces through which the flux flows, the net flux t Phi(e) = 4*lambda*L/4/epsilon_0 = lambda*L/epsilon_0 = (Q_in/L)*L/epsilon_0 = Q_in/epsilon_0. Problem 28.56 Description: An infinite cylinder of radius R has a linear charge density lambda. The volume charge density (C/m)^3 withi charge within a small volume dV is dq =... An infinite cylinder of radius Part A The charge within a small volume Hint: Let is . The integral of , radius , and thickness over a cylinder of length is the total charge has a linear charge density . The volume charge density within the cylinder be a cylindrical shell of length . What is the volume of such a shell? ANSWER: Answer Key: Consider the cylindrical shell of length L, radius r, and thickness dr. T dq = pho*dV = rho(0)*r/R * 2*pi*r*dr*L = 2*pi*rho(0)*L/R * r^2*dr Integrating this expression to obtain the total charge in the cylinder: Q = integral(dq) = 2*pi*rho(0)*L/R*integral{from 0 to R}(r^2*dr) = 2*pi*r => rho(0) = 3*lambda/(2*pi*R^2). Part B Use Gauss's law to find an expression for the electric field ANSWER: j k l m n j k l m n i j k l m n j k l m n inside the cylinder, . Part C Does your expression have the expected value at the surface, ANSWER: Answer Key: The expression for the electric field strength simplifies at r=R t ? Explain. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 8 of 15 E = lambda/(2*pi*epsilon0)*(R^2/R^3) = lambda/(2*pi*epsilon0*R). This is the same result as for a long, charged wire. Prob Description: A sphere of radius R has total charge Q. The volume charge density (C/m)^3 within the sphere is phantom(xx sphere. (a... A sphere of radius has total charge . The volume charge density within the sphere is This charge density decreases linearly from Part A Show that . at the center to zero at the edge of the sphere. Hint: You'll need to do a volume integral. ANSWER: Answer Key: Consider the thin shell of width dr at a distance r from the center. The Q = integral(dq) = integral{from 0 to R}(4*pi*r^2*rho*dr) = integral{from = 4*pi*rho(0)*integral{from 0 to R}((r^2-r^3/R)*dr) = 4*pi*rho(0)*(r^3/3=> rho(0)=3*Q/(pi*R^3) Part B Show that the electric field inside the sphere points radially outward with magnitude ANSWER: Answer Key: Gauss's law for the spherical Gaussian surface of radius r at r < integral(E*dA)=Q(in)/epsilon0 => E*(4*pi*r^2) = Q(in)/epsilon0 The charge inside the Gaussian surface is Q(in) = integral(dq) = integral{from 0 to R}(4*pi*r^2*rho*dr) = i = 4*pi*rho(0)*(r^3/3-r^4/(4*R)){from 0 to r} = 4*pi*rho(0)*(r^3/3 Gauss's law becomes E*(4*pi*r^2) = 4*pi*rho(0)/epsilon0*(r^3/3-r^4/(4*R)) => E = rho(0)/epsilon0*(r/3-r^2/(4*R)) = Q*r/(4*pi*epsilon0*R^3)* Part C Show that your result of part ANSWER: has the expected value at . Answer Key: At r=R, the above expression for the electric field reduces to http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 9 of 15 E = Q*R/(4*pi*epsilon0*R^3)*(4-3*R/R) = Q/(4*pi*epsilon0*R^2) This is an expected result, because all charge Q is inside R and An Electron in a Diode Description: Calculate the final speed of an electron (initially at rest) that travels from the cathode to the anode in a cylindrical diode, given the potential difference between the electrodes. Before the advent of solid-state electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists essentially of two electrodes within a highly evacuated enclosure. One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode, with the anode at the higher potential. Part A , mounted coaxially within a Suppose a diode consists of a cylindrical cathode with a radius of 6.200102 cylindrical anode with a radius of 0.5580 . The potential difference between the anode and cathode is 315 . An electron leaves the surface of the cathode with zero initial speed ( strikes the anode. Hint A.1 How to approach the problem Try to draw a simple diagram of the diode, with the path of the electron going from the central cathode to the outer anode. Since the diode is a cylinder, the symmetry implies that only radial motion needs to be considered for the electron, since any motion of the electron around the center will not change its potential energy. Note that only a potential difference is given between the plates, so you will need to be careful about your definitions. Use the equation for the conservation of energy to find the final speed of the electron. Part A.2 Calculate the change in potential energy of the electron Calculate the change in the potential energy Hint anode. A.2.a Potential energy and potential Recall that the potential energy of a charge in an electric field is related to the potential of that field, of the electron as it moves from the inner cathode to the outer ). Find its speed when it http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 10 of 15 evaluated at the position of the charge, by the equation either the anode or cathode. However, the potential difference relation above to find the change in the potential energy Express your answer numerically in joules. ANSWER: = . In this problem, the potential is not known at is given, so use the of the electron. Part A.3 Calculate the initial kinetic energy of the electron Calculate the initial kinetic energy of the electron at the cathode. Hint A.3.a Initial velocity of the electron Recall that the electron is emitted from the surface of the cathode with zero initial speed. Express your answer in joules. ANSWER: = Hint A.4 Putting it all together The equation of conservation of energy, . Since find the final speed of the electron. Express your answer numerically in meters per second. ANSWER: = , can be rewritten as , you now have enough information to Note that the size of the diode makes no difference, as long as the potential difference between the two electrodes is a known constant. Also, note that the potential at the surface of the anode and cathode are not known separately, but the potential difference is enough for these calculations. In general, the potential at a particular point is not physically important. Only potential differences are important, just as only the change in potential energy is important to mechanics problems. Bouncing Electrons Description: Two electrons are propelled toward each other with different velocities. Find the distance of closest approach using conservation of energy and momentum. Two electrons, each with mass and charge , are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed toward electron B in the positive x direction, and electron B has initial speed toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 11 of 15 Part A Which of the following statements about the motion of the electrons in the given reference frame will be true at the instant the two electrons reach their minimum separation? ANSWER: j k l m n Electron A is moving faster than electron B. j k l m n Electron B is moving faster than electron A. j k l m n Both electrons are moving at the same (nonzero) speed in opposite directions. i j k l m n Both electrons are moving at the same (nonzero) speed in the same direction. j k l m n Both electrons are momentarily stationary. If at a given moment the electrons are still moving toward each other, then they will be closer in the next instant. If at a given moment the electrons are moving away from each other, then they were closer in the previous instant. The electrons will be traveling in the same direction at the same speed at the moment they reach their minimum separation. Only in a reference frame in which the total momentum is zero (the center of momentum frame) would the electrons be stationary at their minimum separation. Part B What is the minimum separation Hint B.1 that the electrons reach? How to approach the problem Since no external or nonconservative forces act on the system of the two electrons, both momentum and total energy (kinetic plus potential) are conserved. Find one expression for the energy when the electrons are far apart, and another when they reach their minimum separation . This will give you an equation in which the only unknown is the speed of the electrons at the moment of their minimum separation. Apply conservation of momentum, using the same initial and final states, to obtain a second equation involving the speed of the electrons. Solve the simultaneous energy and momentum equations to obtain . Part B.2 Find the initial energy What is the total energy of the two electrons when they are initially released? Assume that the electrons and . are so far apart that their potential energy is zero. Express your answer in terms of ANSWER: = Part B.3 Find the final energy What is the total energy of the electrons when they reach their minimum separation . ? Assume that the (identical) speed of the two electrons is Part B.3.a Find the final kinetic energy What is the final kinetic energy (both electrons)? Express your answer in terms of ANSWER: = , and . http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 12 of 15 Part B.3.b Find the final potential energy What is the final potential energy of this 2-electron system? Express your answer in terms of , , and ANSWER: . = Express your answer in terms of ANSWER: , ,, , and (where ). = Part B.4 Find the initial momentum What is the total momentum of the two electrons when they are initially released? , , and . Express your answer as a vector in terms of ANSWER: = Part B.5 Find the final momentum What is the total momentum of the two electrons when they reach their minimum separation . and . ? Assume that the (identical) velocity of the two electrons is Express your answer as a vector in terms of ANSWER: = If you have found both the initial and final linear momenta, you should be able to obtain a simple relation between and . Hint B.6 find . Some math help ; that is, . Substitute for in the energy conservation equation to From the momentum equations, Express your answer in term of , ANSWER: , , and (where ). = http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 13 of 15 An experienced physicist might approach this problem by considering the system of electrons in a reference frame in which the initial momentum is zero. In this frame the initial speed of each electron is . Try solving the problem this way. Make sure that you obtain the same result for approach is easier. , and decide for yourself which Electric Potential Energy of Three Point Charges Description: Calculate the electric potential energy of three identical charges at the corners of an equilateral triangle. Part A Three equal point charges, each with charge 1.85 sides are of length 0.250 , are placed at the vertices of an equilateral triangle whose of the system? (Take as zero the potential . What is the electric potential energy energy of the three charges when they are infinitely far apart.) Hint A.1 How to approach the problem Use the equation for the electric potential energy between two point charges to calculate the energy for each interaction between two of the three point charges. The sum of these energies will be the total electric potential energy. Be careful to avoid double counting. Part A.2 Find the electric potential energy of one pair Assume that one charge is interacting with a second charge, ignoring any effects from the third charge. What is the electric potential energy for this single interaction? Hint A.2.a Electric potential energy of a pair of charges Recall that the electric potential energy the formula . Express your answer in joules to three significant figures. ANSWER: = between two charges and separated by a distance is given by Since the charges are at the vertices of an equilateral triangle, each pair of charges will be 0.250 apart, no matter which two charges are selected. This fact, coupled with the fact that the charges are identical, means that all pair interactions are identical. Part A.3 How many interactions are there? http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 14 of 15 How many pair interactions are there for the three charges? Hint A.3.a Double counting It is important to keep in mind that a pair of charges can interact only once, so if the first charge is interacting with the second charge for one pair, the interaction of the second charge with the first charge cannot also be used, since the pair has already been counted. ANSWER: Use = 8.851012 = for the permittivity of free space. ANSWER: The potential energy is usually written . This means that all pairs of charges (1-2, 1-3, and 2-3) will interact, but no charge can interact with itself ( ), nor can any pair be counted twice as a result of the condition for all possible pairs. For example, , will be counted, while , will not. Description: In the figure, a proton is fired with a speed of 200,000m/s from the midpoint of the capacitor toward the posit In the figure, a proton is fired with a speed of 200,000 from the midpoint of the capacitor toward the positive plate. Part A Show that this is insufficient speed to reach the positive plate. http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008 MasteringPhysics: Assignment Print View Page 15 of 15 ANSWER: Answer Key: The electric potential at the midpoint of the capacitor is 250 V. This is Delta_U = e*Delta_V = e*(250 V) = (1.6*10^(-9) C)*(250 V) = 4.00*10^(-17) if it moves all the way to the positive plate. This increase in potential K = m*v^2/2 = (1.67*10^(-27) kg)*(2.00*10^5 m/s)^2/2 = 3.34*10^(-17) J. This available kinetic energy is not enough to provide for the increase i Part B What is the proton's speed as it collides with the negative plate? ANSWER: = Write up to hand in problems 28.55, 28.58 and 29.45. Note, they are due at the beginning of lecture! You do not have to type in your explanations! Summary 0 of 10 items complete (0% avg. score) 0 of 100 points http://session.masteringphysics.com/myct/assignmentPrint?assignmentID=1165792 11/4/2008
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