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- Arizona
- Arizona State University - Tempe
- Physics
- Physics 131
- Adams/menendez
- [SOLUTIONS] Mastering Physics HW31.pdf

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HW31 Due: 11:59pm on Thursday, November 19, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] To get the full benefit from "The Magnetic Field in a Charging Capacitor", you should go through all of the hints. Please note that, in this problem, the assumption of constant current was not used in any way; the problem could have been done just as well with an instantaneous current I(t). GBA The Magnetic Field in a Charging Capacitor When a capacitor is charged, the electric field , and hence the electric flux , between the plates changes. This change in flux induces a magnetic field, according to Ampère's law as extended by Maxwell: . You will calculate this magnetic field in the space between capacitor plates, where the electric flux changes but the conduction current is zero. Part A A parallel-plate capacitor of capacitance with circular plates is charged by a constant current . The radius of the plates is much larger than the distance between them, so fringing effects are negligible. Calculate , the magnitude of the magnetic field inside the capacitor as a function of distance from the axis joining the center points of the circular plates. Hint A.1 How to approach the problem The Ampère-Maxwell law allows you to find the magnetic field in cases of high symmetry. Such a case is given in this problem. You should convince yourself that the magnetic field vector is everywhere tangential to circles around an imagined axis joining the centers of the two circular plates. Hence we can conveniently integrate around such a circle. We call the radius of this circle , and realize that only the change of flux through the area within the circle will contribute to the magnitude of the magnetic field vector on its perimeter. But the change of electric flux will be the same everywhere between the plates. Therefore, all we need to do is to calculate the change of flux across all the space between the capacitor plates and then multiply by a scaling factor. Hint A.2 What is the electric flux between the plates? Express the electric flux between the plates of the capacitor in terms of the parameters of the problem and the magnitude of the charge on each plate. Express your answer in terms of and given quantities. ANSWER: = Answer not displayed Hint A.3 What is the rate of change of the electric flux? Hint not displayed Hint A.4 Scaling for the area within the Ampèrian loop Hint not displayed Hint A.5 Evaluate the integral Hint not displayed Express your answer in terms of and given quantities. ANSWER: = Correct Note: The picture in Problem 29.35 shows a radius of 1.0 m; however the actual radius that you will get in the question is randomized. Please replace the 1.0-m radius shown in the picture with the radius as given in your problem statement. Thanks. GBA Problem 29.35 An electric field points into the page and occupies a circular region of radius 1.7 , as shown in the figure . There are no electric charges in the region, but there is a magnetic field forming closed loops pointing clockwise, as shown. The magnetic-field strength 60 from the center of the region is 1.6 . Part A What is the rate of change of the electric field? Express your answer using two significant figures. ANSWER: = 4.8×10 11 Correct Part B Is the electric field increasing or decreasing? ANSWER: increasing decreasing Correct Note: The assumption made by "B Fields Due to a Charging Capacitor", that the charging current is constant, is a reasonable assumption ONLY for a very short time span (we'll learn what "very short" means in Lectures 32 and 33. GBA B Fields Due to a Charging Capacitor In the figure the capacitor plates have area 35.00 and are separated by a distance of 1.90 (the figure is not drawn to scale). The plates are in vacuum. The charging current has a constant value of 2.10 (this may be considered true only for a very short time interval). At t = 0 the charge on the plates is zero. Part A Calculate the charge on the plates when 0.500 . Give your answer in nC. ANSWER: = 1.05 Correct Part B Calculate the electric field between the plates when 0.500 . ANSWER: = 3.39×10 4 Correct Part C Calculate the potential difference between the plates when 0.500 . ANSWER: = 64.4 Correct Part D Calculate , the time rate of change of the voltage between the plates (during the time while may be considered constant). ANSWER: 1.29×10 8 Correct V/s Part E Calculate , the time rate of change of the electric field between the plates (during the time while may be considered constant). ANSWER: = 6.78×10 10 = Correct Part F Calculate the rate of change of electric flux through the circular region defined by the circle passing through the points and in the figure (during the time while may be considered constant). Assume that the radius is 50.0 of the radius . ANSWER: 5.93×10 7 Correct Part G Calculate the displacement current through the circular region defined by the circle passing through the points and in the figure (during the time while may be considered constant). Assume that the radius is 50.0 of the radius . ANSWER: 5.25×10 ?4 Correct Part H Calculate the magnetic field strength at point during during the time while may be considered constant. Give your answer in nT. ANSWER: 6.29 Correct nT Part I Calculate the magnetic field strength at the point which is a distance from the centerline of the capacitor, and which is directly above point , during during the time while may be considered constant. Give your answer in nT. ANSWER: 12.6 Correct nT Part J Calculate the magnetic field strength at the point which is a distance 1.5 times from the centerline of the capacitor, and which is directly above point , during during the time while may be considered constant. Give your answer in nT. ANSWER: 8.39 Correct nT Charged Capacitor and Resistor Learning Goal: To study the behavior of a circuit containing a resistor and a charged capacitor when the capcaitor begins to discharge. A capacitor with capacitance is initially charged with charge . At time , a switch is thrown to close the circuit connecting the capacitor in series with a resistor of resistance . Part A What happens to the charge on the capacitor immediately after the switch is thrown? ANSWER: The electrons on the negative plate of the capacitor are held inside the capacitor by the positive charge on the other plate. Only the surface charge is held in the capacitor; the charge inside the metal plates flows through the resistor. The electrons on the negative plate immediately pass through the resistor and neutralize the charge on the positive plate. The electrons on the negative plate eventually pass through the resistor and neutralize the charge on the positive plate. Correct Part B What is the current that flows through the resistor immediately after the switch is thrown? Hint B.1How to approach the problem Hint not displayed Hint B.2Find the voltage at Hint not displayed Express your answer in terms of any or all of the quantities , , and . ANSWER: = Correct Note that since current is charge per time, the preceeding formula shows that the units of must be time. The combination of variables is called the time constant. It will occur frequently in problems involving a resistor and a capacitor. Exercise 26.38 A 5.10 capacitor that is initially uncharged is connected in series with a 6.80 resistor and an emf source with 125 negligible internal resistance. Part A Just after the circuit is completed, what is the voltage drop across the capacitor? ANSWER: = 0 Correct Part B Just after the circuit is completed, what is the voltage drop across the resistor? ANSWER: = 125 Correct Part C Just after the circuit is completed, what is the charge on the capacitor? ANSWER: = 0 Correct Part D Just after the circuit is completed, what is the current through the resistor? ANSWER: = 1.84×10 ?2 Correct Part E A long time after the circuit is completed (after many time constants) what is the voltage drop across the capacitor? ANSWER: = 125 Correct Part F A long time after the circuit is completed (after many time constants) what is the voltage drop across the resistor? ANSWER: = 0 Correct Part G A long time after the circuit is completed (after many time constants) what is the charge on the capacitor? ANSWER: = 6.38×10 ?4 Correct Part H A long time after the circuit is completed (after many time constants) what is the current through the resistor? ANSWER: ANSWER: = 0 Correct Problem 30.66 In the circuit shown in the figure , switch S is closed at time with no charge initially on the capacitor. Part A What is the reading of just after S is closed? ANSWER: 25 Correct V Part B What is the reading of just after S is closed? ANSWER: 50 Correct V Part C What is the reading of just after S is closed? ANSWER: 0 Correct V Part D What is the reading of just after S is closed? ANSWER: 50 Correct V Part E What is the reading of just after S is closed? ANSWER: 0.5 Correct A Part F What is the reading of just after S is closed? ANSWER: 0 Correct A Part G What is the reading of just after S is closed? ANSWER: 0.5 Correct A Part H What does read long after S is closed? ANSWER: 0 Correct V Part I What does read long after S is closed? ANSWER: 0 Correct V Part J What does read long after S is closed? ANSWER: 75 Correct V Part K What does read long after S is closed? ANSWER: 0 Correct V Part L What does read long after S is closed? ANSWER: 0 Correct A Part M What does read long after S is closed? ANSWER: 0 Correct A Part N What does read long after S is closed? ANSWER: 0 Correct A Part O What is the maximum charge on the capacitor? ANSWER: 5.63×10 ?6 Correct C Part P When does the maximum charge on the capacitor occur? ANSWER: just after the switch is closed long after the switch is closed Correct Score Summary: Your score on this assignment is 97.2%. You received 43.76 out of a possible total of 45 points. clockwork MasteringPhysics: Assignment Print View

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