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HW4-Solutions

# HW4-Solutions - Version PREVIEW HW4 feng(58230 This...

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Version PREVIEW – HW4 – feng – (58230) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Capacitance 001 10.0 points A capacitor that is connected to a 49 V source contains 99 μ C of charge. What is its capacitance? Correct answer: 2 . 02041 μ F. Explanation: Let : V = 49 V and q = 99 μ C . The capacitance is C = q V = 99 μ C 49 V = 2 . 02041 μ F Potential of a Capacitor 002 10.0 points A 5 . 1 μ F capacitor is charged with 3 . 3 × 10 5 C. What potential difference exists across it? Correct answer: 6 . 47059 × 10 10 V. Explanation: Let : C = 5 . 1 μ F = 5 . 1 × 10 6 F and q = 3 . 3 × 10 5 C . The capacitance is C = q V , so V = q C = 3 . 3 × 10 5 C 5 . 1 × 10 6 F = 6 . 47059 × 10 10 V Capacitance of a Sphere 003 (part 1 of 2) 10.0 points An isolated conducting sphere can be consid- ered as one element of a capacitor (the other element being a concentric sphere of infinite radius). If k = 1 4 π ǫ 0 , the capacitance of the system is C , and the charge on the sphere is Q , what is the radius r of the sphere? 1. r = Q 2 C k 2. r = k C Q 3. r = k Q C 4. r = C k 5. r = k C correct 6. r = k Q Explanation: The definition of capacitance is C = Q V Q = C V . Let V ( ) = 0. Then the potential of the conducting sphere is V = k Q r = k C V r , so r = k C . 004 (part 2 of 2) 10.0 points The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . If the potential on the surface of the sphere is 6000 V and the capacitance is 8 . 94 × 10 11 F, what is the surface charge density? Correct answer: 66 . 1182 nC / m 2 . Explanation:

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Version PREVIEW – HW4 – feng – (58230) 2 Let : k = 8 . 98755 × 10 9 N · m 2 / C 2 , V = 6000 V , and C = 8 . 94 × 10 11 F . The charge is Q = C V . Assuming a uniform surface charge density we have σ = Q 4 π r 2 = C V 4 π ( k C ) 2 = 1 4 π k 2 V C = 1 4 π (8 . 98755 × 10 9 N · m 2 / C 2 ) 2 × 6000 V 8 . 94 × 10 11 F = 66 . 1182 nC / m 2 . Variable Capacitor 005 10.0 points A variable air capacitor used in tuning circuits is made of N semicircular plates each of radius R and positioned d from each other. A second identical set of plates that is free to rotate is enmeshed with the first set. R θ d Determine the capacitance as a function of the angle of rotation θ , where θ = 0 corre- sponds to the maximum capacitance. 1. C = ǫ 0 (2 N ) R 2 θ d 2. C = ǫ 0 N R 2 π d 3. C = ǫ 0 N R 2 ( π - θ ) d 4. C = ǫ 0 N d 2 θ R 5. C = ǫ 0 (2 N - 1) R 2 ( π - θ ) d correct 6. C = ǫ 0 (2 N ) d θ R 2 7. C = 2 ǫ 0 R 2 (2 π - θ ) d 8. C = ǫ 0 N R 2 θ d 9. C = ǫ 0 R 2 ( π - θ ) d 10. C = N R 2 ( π - θ ) d Explanation: Considering the situation of θ = 0, the two sets of semicircular plates in fact form 2 N - 1 capacitors connected in parallel, with each one having capacitance C = ǫ 0 A d 2 = ǫ 0 π R 2 2 d 2 = ǫ 0 π R 2 d . So the total capacitance would be C = (2 N - 1) ǫ 0 π R 2 d . Note: The common area of the two sets of plates varies linearly when one set is rotating, so the capacitance at angle θ is C = ǫ 0 (2 N - 1) R 2 ( π - θ ) d .
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HW4-Solutions - Version PREVIEW HW4 feng(58230 This...

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