HW4-solutions - – HW4 – Tsoi –(60250 1 This print-out should have 20 questions Multiple-choice questions may continue on the next column or

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Unformatted text preview: – HW4 – Tsoi – (60250) 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 An air-filled capacitor consists of two parallel plates, each with an area A , separated by a distance d . A potential difference V is applied to the two plates. The magnitude of the surface charge density on the inner surface of each plate is 1. σ = ǫ ( V d ) 2 2. σ = ǫ V d correct 3. σ = ǫ parenleftbigg V d parenrightbigg 2 4. σ = ǫ V d 5. σ = ǫ V d 6. σ = ǫ parenleftbigg d V parenrightbigg 2 7. σ = ǫ d V 8. σ = ǫ ( V d ) 2 Explanation: Use Gauss’s Law. We find that a pillbox of cross section S which sticks through the surface on one of the plates encloses charge σ S . The flux through the pillbox is only through the top, so the total flux is E S . Gauss’ Law gives σ = ǫ E = ǫ V d Alternatively, we could just recall this result for an infinite conducting plate (meaning we neglect edge effects) and apply it. 002 10.0 points By what factor does the capacitance of a metal sphere increase if its volume is tripled? 1. 9 times 2. 3 times 3. 1.33 times 4. 1.44 times correct 5. 1.5 times 6. 1.41 times 7. 2.5 times 8. 1.73 times 9. 27 times 10. 2 times Explanation: Let : V ′ = 3 V . C = 4 π ǫ R , and V = 4 3 π R 3 , so C ∝ V 1 / 3 . Thus C ′ ∝ (3 V ) 1 / 3 = 3 √ 3 V 1 / 3 is larger than C by 3 √ 3 = 1 . 44 times. 003 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. – HW4 – Tsoi – (60250) 2 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 (2 N ) R 2 θ d 2. C = ǫ (2 N- 1) R 2 ( π- θ ) d correct 3. C = ǫ N R 2 π d 4. C = ǫ R 2 ( π- θ ) d 5. C = ǫ N R 2 θ d 6. C = ǫ N R 2 ( π- θ ) d 7. C = N R 2 ( π- θ ) d 8. C = 2 ǫ R 2 (2 π- θ ) d 9. C = ǫ N d 2 θ R 10. C = ǫ (2 N ) d θ R 2 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 A d 2 = ǫ π R 2 2 d 2 = ǫ π R 2 d . So the total capacitance would be C = (2 N- 1) ǫ 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 (2 N- 1) R 2 ( π- θ ) d . 004 10.0 points A 59 m length of coaxial cable has a solid cylindrical wire inner conductor with a di- ameter of 4 . 084 mm and carries a charge of 7 . 73 μ C. The surrounding conductor is a cylindrical shell and has an inner diameter of 6 . 388 mm and a charge of- 7 . 73 μ C....
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This note was uploaded on 10/12/2008 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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HW4-solutions - – HW4 – Tsoi –(60250 1 This print-out should have 20 questions Multiple-choice questions may continue on the next column or

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