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- Arizona
- Arizona State University - Tempe
- Physics
- Physics 131
- Adams/menendez
- [SOLUTIONS] Mastering Physics HW13.pdf

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HW13 Due: 11:59pm on Thursday, October 1, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] Altering a Capacitor A parallel-plate capacitor with circular plates and a capacitance of 10.4 is connected to a battery which provides a voltage of 11.2 . The radius of the plates is very large compared to the separation distance between the plates. Part A How much charge is held separated by these capacitor plates? ANSWER: = 1.16×10 −4 Correct Part B How much charge would be on the plates if their separation were multiplied by 3 while the capacitor remained connected to the battery? ANSWER: = 3.88×10 −5 Correct Part C How much charge would be on the plates if the capacitor were connected to the battery after the radius of each plate was multiplied by 2 without changing their separation? ANSWER: = 4.66×10 −4 Correct 24.9 is the "Cylindrical Capacitor" problem. To get the full benefit of this problem, you should begin with Gauss' Law, find the electric field, integrate to find the absolute value of the potential difference, and use the DEF of capacitance to find C. This procedure is what you will be expected to follow on a test. For applying Gauss's Law, you may assume that L is very long. GBA Problem 24.9 A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is . The inner cylinder has a radius of , the outer one has a radius of , and the length of each cylinder is . Part A What is the capacitance? Use for the permittivity of free space. ANSWER: Correct Part B What applied potential difference is necessary to produce these charges on the cylinders? ANSWER: Correct 24.13 is the "Spherical Capacitor" problem. To get the full benefit of this problem, you should begin with Gauss' Law, find the electric field, integrate to find the absolute value of the potential difference, and use the DEF of capacitance to find C. This procedure is what you will be expected to follow on a test. Note that for parts B and C, r_1 and r_2 are understood to be slightly different from r_a and r_b. GBA Problem 24.13 A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has a radius of and the outer sphere has a radius of . A potential difference of is applied to the capacitor. Part A What is the capacitance of the capacitor? Use for the permittivity of free space. ANSWER: Correct Part B What is the magnitude of at , just outside the inner sphere? ANSWER: Correct Part C What is the magnitude of at , just inside the outer sphere? ANSWER: Correct You may use your results from 24.13 in doing "Potential Difference and Electric-Field Energy of a Spherical Capacitor." GBA Potential Difference and Electric-Field Energy of a Spherical Capacitor A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 10.0 centimeters, and the separation between the spheres is 1.50 centimeters. The magnitude of the charge on each sphere is 3.30 nanocoulombs. Part A What is the magnitude of the potential difference between the two spheres? Hint A.1 How to approach the problem Hint not displayed Hint A.2 Choosing the Gaussian surface Hint not displayed Hint A.3 Find the electric field Hint not displayed Hint A.4 How to use the electric field to calculate the potential difference Hint not displayed ANSWER: = 38.7 Correct Part B What is the electric-field energy stored in the capacitor? Hint B.1How to calculate the electric-field energy The electric-field energy is calculated by the equation . Use the simplest of these equations, given the variables already known and/or calculated. ANSWER: 6.38×10 −8 Correct Air Capacitor An air capacitor is made from two flat parallel metal plates a distance 1.65 apart. When the potential difference between the plates is 215.0 , the absolute value of the charge on each plate is 0.0175 . Ignore edge effects. Part A What is the capacitance of this metal geometry? ANSWER: ANSWER: 8.14×10 −11 Correct Part B What is the area of each plate? ANSWER: 1.52×10 −2 Correct Part C If the plates are circular, what is the radius of each plate? ANSWER: 6.95×10 −2 Correct m Part D What maximum voltage can be applied without dielectric breakdown? Dielectric breakdown for air occurs at an electric field strength of about 3.0 million N/C. ANSWER: 4950 Correct V Part E What total energy is stored by this capacitor when the amount of charge separated is 0.0175 ? ANSWER: 1.88×10 −6 Correct Part F If the voltage across this capacitor were doubled without changing the geometry in any way, how would the energy stored be affected? ANSWER: The energy stored would be unaffected. The energy stored would be halved. The energy stored would be doubled. The energy stored would be quadrupled. The energy stored would be tripled. Correct You may use your results from 24.9 in doing 24.35. GBA Exercise 24.35 A cylindrical air capacitor with a length of 15.2 stores an amount of energy equal to 3.80×10 −9 when the potential difference between the two conductors is 3.20 . Part A Calculate the magnitude of the charge on each conductor. ANSWER: = 2.38×10 −9 Correct Part B Calculate the ratio of the radii of the inner and outer conductors. ANSWER: = 3.12 Correct Score Summary: Your score on this assignment is 98.5%. You received 44.33 out of a possible total of 45 points. clockwork MasteringPhysics: Assignment Print View

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