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- [SOLUTIONS] Mastering Physics HW8.pdf

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HW8 Due: 11:59pm on Thursday, September 17, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] NOTE: To get the full benefit from doing "The Electric Field and Surface Charge at a Conductor", you should go through all of the hints. GBA The Electric Field and Surface Charge at a Conductor Learning Goal: To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge. Part A Which of the following describes the electric field inside this conductor? ANSWER: It is in the same direction as the original external field. It is in the opposite direction from that of the original external field. It has a direction determined entirely by the charge on its surface. It is always zero. Correct The net electric field inside a conductor is always zero. If the net electric field were not zero, a current would flow inside the conductor. This would build up charge on the exterior of the conductor. This charge would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero. Part B The charge density inside the conductor is: ANSWER: 0 non-zero; but uniform non-zero; non-uniform infinite Correct You already know that there is a zero net electric field inside a conductor; therefore, if you surround any internal point with a Gaussian surface, there will be no flux at any point on this surface, and hence the surface will enclose zero net charge. This surface can be imagined around any point inside the conductor with the same result, so the charge density must be zero everywhere inside the conductor. This argument breaks down at the surface of the conductor, because in that case, part of the Gaussian surface must lie outside the conducting object, where there is an electric field. Part C Assume that at some point just outside the surface of the conductor, the electric field has magnitude and is directed toward the surface of the conductor. What is the charge density on the surface of the conductor at that point? Hint C.1 How to approach the problem Which of the following is the best way to solve this problem? ANSWER: Use Coulomb's law for the electric field from each charge element. Use Gauss's law with a short and flat cylindrical surface (picture a coin) with one end just below (inside) and the other just above (outside) the surface of the conductor. Use Gauss's law with a long and thin cylindrical surface (picture a straw capped at both ends) with one end far below (but inside the object) and the other far above the surface of the conductor. None of these: You need to know the surface charge distribution everywhere on the surface. None of these: You need to know the electric field at all points on the surface. Correct A straightforward way to solve this problem is to choose a Gaussian surface one end of which is just above and the other just below the surface of the conductor. Hint C.2 Calculate the flux through the top of the cylinder Using a flat cylinder with large top and bottom each of area just above and just below the surface of the conductor, find the flux generated through the top surface of the cylinder by the electric field of magnitude that points into the surface. Express your answer in terms of , , and any needed constants. ANSWER: ANSWER: = Correct Hint C.3 Calculate the flux through the bottom of the box Using a flat cylinder with large top and bottom of area just above and just below the surface of the conductor, find the flux generated through the bottom surface of the cylinder by the electric field inside the conductor (keep in mind that positive flux is outward through the cylinder's surface, which is downward into the conductor). Answer in terms of , , and any needed constants. ANSWER: = 0 Correct Hint C.4 What is the charge inside the Gaussian surface? Find the net charge inside this Gaussian surface. Express your answer in terms of the charge density and other given quantities. ANSWER: = Correct Notice that the surface charge density has dimensions of charge per area. Therefore, when is multiplied by an area, the result is charge, which is needed for Gauss's law. Hint C.5 Apply Gauss's law Now apply Gauss's law, neglecting any contribution to the flux due to the very short sides of the cylinder. Gauss's law states that . The area should cancel out of your result. Express your answer in terms of and . ANSWER: = Correct Holey Sphere, Batman! A spherical cavity of radius 7.00 is at the center of a metal sphere of radius 20.0 . A point charge 7.00 rests at the very center of the cavity, whereas the metallic spherical shell carries a net charge of -20.0 . Part A Determine the electric field strength at a point 5.00 from the center of the cavity. ANSWER: = 2.52×10 7 Correct Part B Determine the electric field strength at a point 10.0 from the center of the cavity. ANSWER: = 0 Correct Part C What is the surface charge density (with proper sign) on the inner surface of the metallic spherical shell? ANSWER: ?1.14×10 ?4 Correct Part D What is the electric field strength within the cavity at a point almost touching the inner metal surface? ANSWER: 1.28×10 7 Correct Part E What is the surface charge density (with proper sign) on the outer surface of the metallic spherical shell? ANSWER: ?2.59×10 ?5 Correct Part F What is the electric field strength at a point just barely outside the outer surface of the metal spherical shell? ANSWER: 2.92×10 6 Correct Part G Determine the electric field strength at a point 50.0 from the center. ANSWER: 4.68×10 5 ANSWER: = 4.68×10 5 Correct Problem 22.38 A very long conducting tube (hollow cylinder) has inner radius and outer radius . It carries charge per unit length , where is a positive constant with units of . A line of charge lies along the axis of the tube. The line of charge has charge per unit length . Part A Calculate the electric field in terms of and the distance from the axis of the tube for . Express your answer in terms of the variables , , and constant . ANSWER: = Correct Part B Calculate the electric field in terms of and the distance from the axis of the tube for . Express your answer in terms of the variables , , and constant . ANSWER: = 0 Correct Part C Calculate the electric field in terms of and the distance from the axis of the tube for . Express your answer in terms of the variables , , and constant . ANSWER: = Correct Part D What is the charge per unit length on the inner surface of the tube? ANSWER: = -1.00 Correct Part E What is the charge per unit length on the outer surface of the tube? ANSWER: = 2.00 Correct A Conducting Shell around a Conducting Rod An infinitely long conducting cylindrical rod with a positive charge per unit length is surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge per unit length of and radius , as shown in the figure. Part A What is , the radial component of the electric field between the rod and cylindrical shell as a function of the distance from the axis of the cylindrical rod? Hint A.1 The implications of symmetry Hint not displayed Hint A.2 Apply Gauss' law Hint not displayed Hint A.3 Find the charge inside the Gaussian surface Hint not displayed Hint A.4 Find the flux Hint not displayed Express your answer in terms of , , and , the permittivity of free space. ANSWER: = Correct Part B What is , the surface charge density (charge per unit area) on the inner surface of the conducting shell? Hint B.1Apply Gauss's law Hint not displayed Hint B.2Find the charge contribution from the surface Hint not displayed ANSWER: = Correct Part C What is , the surface charge density on the outside of the conducting shell? (Recall from the problem statement that the conducting shell has a total charge per unit length given by .) Hint C.1 What is the charge on the cylindrical shell? Hint not displayed ANSWER: = Correct Part D What is the radial component of the electric field, , outside the shell? Hint D.1 How to approach the problem Hint not displayed Hint D.2 Find the charge within the Gaussian surface Hint not displayed Hint D.3 Find the flux in terms of the electric field Hint not displayed ANSWER: = Correct Parallel Plates - Unequal Charges Two large, thin, metallic plates are placed parallel to each other, separated by 1.80 . The top plate carries a uniform positive charge density of 24.0 , while the bottom plate carries a uniform negative charge density of -34.0 . Part A What is the magnitude of the electric field halfway between the plates? ANSWER: 3.28×10 6 Correct N/C Part B What is the direction of the electric field halfway between the plates? ANSWER: towards the top plate towards the bottom plate the electric field is zero Correct Part C What is the magnitude of the electric field above the two plates? ANSWER: 5.65×10 5 Correct N/C Part D What is the direction of the electric field above the two plates? ANSWER: away from the top plate towards the top plate the electric field is zero Correct Part E What is the magnitude of the electric field below the two plates? ANSWER: 5.65×10 5 Correct N/C Part F What is the direction of the electric field below the two plates? ANSWER: towards the bottom plate away from the bottom plate the electric field is zero Correct Part G What is the surface charge density on the top surface of the top plate (with proper sign)? ANSWER: -5.00 Correct C/m Part H What is the surface charge density on the bottom surface of the top plate (with proper sign)? ANSWER: 29.0 Correct C/m Part I What is the surface charge density on the top surface of the bottom plate (with proper sign)? ANSWER: -29.0 Correct C/m Part J What is the surface charge density on the bottom surface of the bottom plate (with proper sign)? ANSWER: -5.00 Correct C/m Score Summary: Your score on this assignment is 99%. You received 49.51 out of a possible total of 50 points. clockwork MasteringPhysics: Assignment Print View

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