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Unformatted text preview: montes (dam2772) – Homework2 – shih – (56310) 1 This print-out should have 20 questions. Multiple-choice 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 air-filled parallel-plate 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 air-filled 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.

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Homework 2 - montes(dam2772 – Homework2 – shih –(56310 1 This print-out should have 20 questions Multiple-choice questions may continue on

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