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MasteringPhysics: Assignment Print View 1 of 5 http://session.masteringphysics.com/myct/assignmentPrint?assignmentI... Chapter 28 Homework Due: 8:00am on Wednesday, March 3, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] Calculating Electric Flux through a Disk Suppose a disk with area is placed in a uniform electric field of magnitude . The disk is oriented so that the vector normal to its surface, , makes an angle with the electric field, as shown in the figure. Part A What is the electric flux through the surface of the disk that is facing right (the normal vector to this surface is shown in the figure)? Assume that the presence of the disk does not interfere with the electric field. Hint A.1 Definition of electric flux Hint not displayed Hint A.2 Simplifying the integrand Hint not displayed Hint A.3 Evaluate the scalar product Hint not displayed Express your answer in terms of ANSWER: , , and = Correct Problem 28.9 Part A What is the electric flux through the surface shown in the figure? ANSWER: 1.00 Correct Problem 28.13 A rectangle lies in the xz-plane. What is the electric flux through the rectangle if Part A ANSWER: 0 Correct Part B ANSWER: 6.00×10−2 Correct Flux out of a Cube A point charge of magnitude is at the center of a cube with sides of length . Part A 3/2/2010 12:23 PM MasteringPhysics: Assignment Print View 2 of 5 What is the electric flux Hint A.1 http://session.masteringphysics.com/myct/assignmentPrint?assignmentI... through each of the six faces of the cube? How to approach the problem Hint not displayed Hint A.2 Calculate the total electric flux Hint not displayed Hint A.3 Flux through a face Hint not displayed Use for the permittivity of free space. ANSWER: = Correct The shape of the surface enclosing a charge, in this case a cube, does not affect the total electric flux through the surface. The flux depends only on the total enclosed charge. Part B What would be the flux Hint B.1 through a face of the cube if its sides were of length ? How to approach the problem Gauss's law states that the flux through a closed surface depends only on the charge enclosed by that surface. Does changing the size of the cube affect the total charge enclosed by the cube? Use for the permittivity of free space. ANSWER: = Correct Just as the shape of the surface does not affect the total electric flux coming out of that surface, its size does not make any difference in the total electric flux either. The only relevant quantity is the total enclosed charge. Problem 28.4 The electric field is constant over each face of the cube shown in the figure . Part A Does the box contain positive charge, negative charge, or no charge? ANSWER: positive charge negative charge no charge Correct Problem 28.22 The net electric flux through a closed surface is 1600 . Part A How much charge is enclosed within the surface? ANSWER: -14.2 nC Correct Charged Insulating Spheres Two small insulating spheres with radius 5.50×10−2 positively charged, with net charge 4.50 are separated by a large center-to-center distance of 0.535 . One sphere is negatively charged, with net charge -1.00 , and the other sphere is . The charge is uniformly distributed within the volume of each sphere. Part A What is the magnitude Hint A.1 of the electric field midway between the spheres? How to approach the problem Draw a diagram of the system to keep track of the directions of the fields. Calculate the electric field at the point midway between the charged spheres separately for each sphere, using Gauss's law, and use vector addition to determine the net electric field. Hint A.2 Using Gauss's law 3/2/2010 12:23 PM MasteringPhysics: Assignment Print View 3 of 5 http://session.masteringphysics.com/myct/assignmentPrint?assignmentI... You need to construct two separate Gaussian surfaces, each of which will surround one of the charged spheres. These surfaces also need to contain the point midway between the spheres so that the electric field at that point can be found. Therefore, for convenience, you should use a spherical Gaussian surface with one of the charged spheres in the center, since that way the electric field will be uniform and perpendicular to the surface at all points, so that the flux will be given by Hint A.3 Calculate . Also, be careful about the radii of the spherical surfaces. Calculate the field due to the negatively charged sphere , the magnitude of the electric field at the midway point due to the sphere with charge -1.00 Hint A.3.1 only. Using the flux to calculate the field Since the flux through the surface will be (due to Gauss's law), you can solve for the electric field using the equation for the surface area of the (Gaussian) sphere, , and the value of the enclosed charge. Take the permittivity of free space to be ANSWER: Hint A.4 = 8.85×10−12 . 5 = 1.26×10 Correct Determine the direction of the electric field from the first sphere Hint not displayed Hint A.5 Calculate the field due to the positively charged sphere Hint not displayed Hint A.6 Determine the direction of the electric field from the positively charged sphere Hint not displayed Hint A.7 Vector addition Hint not displayed = 8.85×10−12 Take the permittivity of free space to be ANSWER: . 5 = 6.91×10 Correct Part B What is the direction of the electric field midway between the spheres? ANSWER: toward the positively charged sphere toward the negatively charged sphere upward perpendicular to the line connecting the centers of the spheres downward perpendicular to the line connecting the centers of the spheres Correct Since the electric field will point toward a negative charge and away from a positive charge, the electric field from each sphere separately will point toward the negatively charged sphere, and so the total field will also point in that direction. The Electric Field of a Ball of Uniform Charge Density A solid ball of radius has a uniform charge density . Part A What is the magnitude of the electric field Hint A.1 at a distance from the center of the ball? Gauss's law Hint not displayed Hint A.2 Find Hint not displayed Express your answer in terms of , , , and . ANSWER: = Correct Notice that this result is identical to that reached by applying Coulomb's law to a point charge centered at the origin with . The field outside of a uniformly charged sphere does not depend on the size of the sphere, only on its charge. A uniformly charged sphere generates an electric field as if all the charge were concentrated at its center. Part B What is the magnitude of the electric field Hint B.1 at a distance from the center of the ball? How does this situation compare to that of the field outside the ball? Hint not displayed Express your answer in terms of , , , and . ANSWER: = Correct Part C 3/2/2010 12:23 PM MasteringPhysics: Assignment Print View 4 of 5 Let http://session.masteringphysics.com/myct/assignmentPrint?assignmentI... represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true? Hint C.1 Plot the electric field The figure shows a plot of the electric field as a function of . Check all that apply. ANSWER: . . . The maximum electric field occurs when . The maximum electric field occurs when . The maximum electric field occurs as Answer Requested A Conducting Shell around a Conducting Rod An infinitely long conducting cylindrical rod with a positive charge infinitely long) with a charge per unit length of and radius per unit length is surrounded by a conducting cylindrical shell (which is also , 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 Hint A.1 from the axis of the cylindrical rod? 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.1 Apply Gauss's law The magnitude of the net force on charges within a conductor is always zero. This implies that the magnitude of the electric field within the conductor is zero. Think about a cylindrical Gaussian surface of length whose radius lies at the middle of the outer cylindrical shell. Since the electric field inside a conductor is zero and the Gaussian surface lies within the conductor, the electric flux across the Gaussian surface must be zero. What, then, must , the total charge inside this Gaussian surface, be? ANSWER: Hint B.2 What is =0 Correct Find the charge contribution from the surface , the total charge on the inner surface of the cylindrical shell that is contained within the Gaussian surface? Express your answer in terms of ANSWER: and . = Answer not displayed 3/2/2010 12:23 PM MasteringPhysics: Assignment Print View 5 of 5 http://session.masteringphysics.com/myct/assignmentPrint?assignmentI... 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, Hint D.1 , outside the shell? 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 Score Summary: Your score on this assignment is 103.6%. You received 62.69 out of a possible total of 70 points, plus 9.8 points of extra credit. 3/2/2010 12:23 PM