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solution_pdf 11 - malsam(wgm329 homework 11 turner(56705...

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malsam (wgm329) – 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 . 75 cm × 5 . 97 cm is separated by a 0 . 54 mm thickness of paper. Find the capacitance of this device. The dielectric constant κ for paper is 3.7. Correct answer: 63 . 3825 pF. Explanation: Let : κ = 3 . 7 , d = 0 . 54 mm = 0 . 00054 m , and A = 1 . 75 cm × 5 . 97 cm = 0 . 00104475 m 2 . We apply the equation for the capacitance of a parallel-plate capacitor and find C = κ ǫ 0 A d = (3 . 7) (8 . 85419 × 10 12 C 2 / N · m 2 ) × parenleftbigg 0 . 00104475 m 2 0 . 00054 m parenrightbigg 1 × 10 12 pF 1 F = 63 . 3825 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 . 547624 μ c. Explanation: Let : E max = 1 . 6 × 10 7 V / m . Since the thickness of the paper is 0 . 00054 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 = (63 . 3825 pF)(8640 V) · 1 × 10 12 F 1 pF · 1 × 10 6 μ C 1 C = 0 . 547624 μ 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 16 π ǫ 0 a 2. U = Q 2 4 π ǫ 0 a 3. U = Q 2 8 π ǫ 0 a correct 4. U = Q 8 π ǫ 0 a 5. U = Q 2 a 4 π ǫ 0 6. U = Q 2 π 8 ǫ 0 a 7. U = Q 2 8 π ǫ 0 a 2 8. U = Q 2 a Explanation: The capacitance formula for a spherical ca- pacitor of inner radius a and outer radius b is C = a b 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 π ǫ 0 a . The total energy stored is U = Q 2 2 C = Q 2 8 π ǫ 0 a .
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malsam (wgm329) – 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 κ = κ C . 1. U = Q 2 3 κ C 2. U = Q 2 3 C 3. U = Q 2 2 κ C correct 4. U = Q 4 κ C 5. U = Q 2 κ C 6. U = Q 2 3 ( κ - 1) C 7. U = Q 2 2 ( κ - 1) C 8. U = Q 2 2 C 9. U = Q 2 κ C Explanation: The energy stored in a capacitor of capaci- tance C κ is U = Q 2 2 C κ = Q 2 2 κ 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 κ from the gap of a parallel plate capacitor of plate charge Q and capacitance C . 1. W if = Q 2 2 C parenleftbigg κ + 1 κ parenrightbigg 2. W if = Q 2 2 C 3. W if = Q 2 2 C (1 - κ ) 4. W if = Q 2 2 κ C (1 - κ ) 5. W if = Q 2 2 C parenleftbigg κ - 1 κ parenrightbigg 6. W if = Q 2 2 C parenleftbigg 1 - 1 κ parenrightbigg correct 7. W if = Q 2 2 C parenleftbigg 1 κ - 1 parenrightbigg 8. W if = Q 2 5 κ C 9. W if = Q 2 2 κ C Explanation: The energy stored in the capacitor with the dielectric is U i = Q 2 2 κ C .
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