# HW 11 - saldana (avs387) homework 11 turner (56705) 1 This...

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Unformatted text preview: saldana (avs387) homework 11 turner (56705) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A parallel-plate capacitor of dimensions 1 . 91 cm 4 . 31 cm is separated by a 1 . 78 mm thickness of paper. Find the capacitance of this device. The dielectric constant for paper is 3.7. Correct answer: 15 . 151 pF. Explanation: Let : = 3 . 7 , d = 1 . 78 mm = 0 . 00178 m , and A = 1 . 91 cm 4 . 31 cm = 0 . 00082321 m 2 . We apply the equation for the capacitance of a parallel-plate capacitor and find C = A d = (3 . 7) (8 . 85419 10 12 C 2 / N m 2 ) parenleftbigg . 00082321 m 2 . 00178 m parenrightbigg 1 10 12 pF 1 F = 15 . 151 pF . 002 (part 2 of 2) 10.0 points What is the maximum charge that can be placed on the capacitor? The electric strength of paper is 1 . 6 10 7 V / m. Correct answer: 0 . 4315 c. Explanation: Let : E max = 1 . 6 10 7 V / m . Since the thickness of the paper is 0 . 00178 m, the maximum voltage that can be applied before breakdown is V max = E max d. Hence, the maximum charge is Q max = C V max = C E max d = (15 . 151 pF)(28480 V) 1 10 12 F 1 pF 1 10 6 C 1 C = . 4315 c . 003 (part 1 of 4) 10.0 points Determine the total energy stored in a con- ducting sphere with charge Q . Hint: Use the capacitance formula for a spherical capacitor which consists of two spherical shells. Take the inner sphere to have a radius a and the outer shell to have an infinite radius. 1. U = Q 2 8 a 2 2. U = Q 2 a 3. U = Q 2 16 a 4. U = Q 2 4 a 5. U = Q 2 8 a correct 6. U = Q 8 a 7. U = Q 2 8 a 8. U = Q 2 a 4 Explanation: The capacitance formula for a spherical ca- pacitor of inner radius a and outer radius b is C = ab k e ( b- a ) . If we let b , we find we can neglect a in the denominator compared to b , so C a k e = 4 a. The total energy stored is U = Q 2 2 C = Q 2 8 a . saldana (avs387) homework 11 turner (56705) 2 004 (part 2 of 4) 10.0 points Find the energy stored in a capacitor of charge Q filled with dielectric, use C = C . 1. U = Q 2 2 C 2. U = Q 2 2 C correct 3. U = Q 2 2 ( - 1) C 4. U = Q 2 C 5. U = Q 2 3 ( - 1) C 6. U = Q 4 C 7. U = Q 2 C 8. U = Q 2 3 C 9. U = Q 2 3 C Explanation: The energy stored in a capacitor of capaci- tance C is U = Q 2 2 C = Q 2 2 C . 005 (part 3 of 4) 10.0 points Work = U f- U i , where i is the initial state where there is a slab in the gap and f is the final state where there is no slab in the gap. Find the work done in pulling a dielectric slab of dielectric constant from the gap of a parallel plate capacitor of plate charge Q and capacitance C ....
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## This note was uploaded on 10/26/2010 for the course PHYS 56705 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.

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HW 11 - saldana (avs387) homework 11 turner (56705) 1 This...

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