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Unformatted text preview: montes (dam2772) – Homework2 – shih – (56310) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two conductors insulated from each other are charged by transferring electrons from one conductor to the other. After 2 . 4882 × 10 14 have been transferred, the potential difference between the conductors is 12 . 3 V. The charge on an electron is 1 . 60218 × 10 − 19 C . What is the capacitance of the system? Correct answer: 3 . 24109 × 10 − 6 F. Explanation: Let : e = 1 . 60218 × 10 − 19 C , N = 2 . 4882 × 10 14 electrons , and V = 12 . 3 V . One conductor gains a charge of Q = N e while the other gains a charge of Q. The capacitance of the system is C = q V = N e V = (2 . 4882 × 10 14 ) ( 1 . 60218 × 10 − 19 C) 12 . 3 V = 3 . 24109 × 10 − 6 F . 002 10.0 points If we double the plate separation of a par allel plate capacitor which is connected to a battery, 1. the potential difference is halved. 2. the charge density on each plate is dou bled. 3. the charge on the each plate is halved. correct 4. None of these 5. the electric field is doubled. Explanation: The potential difference is V = constant and E = V d E ′ = V d ′ = V 2 d , so the electric field is halved. C = κ ǫ A d C ′ = κ ǫ A d ′ = κ ǫ A 2 d = C 2 and the capacitance is halved. Thus Q = C V Q ′ = C ′ V = C 2 V , and the charge on the each plate is also halved. 003 10.0 points An airfilled parallelplate capacitor is to have a capacitance of 0 . 4 F. If the distance between the plates is 1 . 4 mm, calculate the required surface area of each plate. The permittivity of free space is 8 . 85419 × 10 − 12 C 2 / N · m 2 . Correct answer: 63 . 2469 km 2 . Explanation: Let : C = 0 . 4 F , d = 1 . 4 mm = 0 . 0014 m , and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 . montes (dam2772) – Homework2 – shih – (56310) 2 The capacitance is C = ǫ A d A = C d ǫ = (0 . 4 F) (0 . 0014 m) 8 . 85419 × 10 − 12 C 2 / N · m 2 × parenleftbigg 1 km 2 10 6 m 2 parenrightbigg = 63 . 2469 km 2 . 004 (part 1 of 2) 10.0 points A(n) 24 μ F airfilled capacitor is charged to a potential difference of 1632 V. What is the energy stored in it? Correct answer: 31 . 9611 J. Explanation: Let : C = 24 μ F = 2 . 4 × 10 − 5 F and V = 1632 V . The energy stored in the capacitor is U = 1 2 C V 2 = 1 2 ( 2 . 4 × 10 − 5 F ) (1632 V) 2 = 31 . 9611 J . 005 (part 2 of 2) 10.0 points If the voltage source remains connected to the capacitor, but the air is replaced by a layer of plastic with dielectric constant of 4, what is the new value of energy stored in it? Correct answer: 127 . 844 J....
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This note was uploaded on 03/22/2012 for the course PHYS 302L taught by Professor Tsoi during the Spring '10 term at University of Texas at Austin.
 Spring '10
 TSOI
 Physics, Charge, Work

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