Spring 11 Phys 121 - HW 05

Spring 11 Phys 121 - HW 05 - dzifanu(sd26397 – hw 5 –...

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Unformatted text preview: dzifanu (sd26397) – hw 5 – opyrchal – (121102) 1 This print-out should have 15 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 52 m length of coaxial cable has a solid cylindrical wire inner conductor with a di- ameter of 1 . 234 mm and carries a charge of 5 . 75 μ C. The surrounding conductor is a cylindrical shell and has an inner diameter of 6 . 076 mm and a charge of- 5 . 75 μ C. Assume the region between the conductors is air. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . What is the capacitance of this cable? Correct answer: 1 . 81476 nF. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 5 . 75 μ C , ℓ = 52 m , a = 1 . 234 mm , and b = 6 . 076 mm . The charge per unit length is λ ≡ Q ℓ . V =- integraldisplay b a vector E · dvectors =- 2 k e λ integraldisplay b a dr r =- 2 k e Q ℓ ln parenleftbigg b a parenrightbigg . The capacitance of a cylindrical capacitor is given by C ≡ Q V = ℓ 2 k e 1 ln parenleftbigg b a parenrightbigg = 52 m 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) × 1 ln parenleftbigg 6 . 076 mm 1 . 234 mm parenrightbigg · parenleftbigg 1 × 10 9 nF 1 F parenrightbigg = 1 . 81476 nF 002 (part 2 of 2) 10.0 points What is the potential difference between the two conductors? Correct answer: 3 . 16846 kV. Explanation: Therefore V = Q C = 5 . 75 μ C 1 . 81476 nF parenleftbigg 1 × 10 9 nF 1 F parenrightbigg × parenleftbigg 1 C 1 × 10 6 μ C parenrightbiggparenleftbigg 1 kV 1000 V parenrightbigg = 3 . 16846 kV 003 (part 1 of 2) 10.0 points The potential difference between a pair of oppositely charged parallel plates is 422 V. a) If the spacing between the plates is dou- bled without altering the charge on the plates, what is the new potential difference between the plates? Correct answer: 844 V. Explanation: Let : d 2 = 2 d 1 and Δ V 1 = 422 V . The charge remains the same, so Q = C 1 Δ V 1 = C 2 Δ V 2 ε A d 1 Δ V 1 = ε A d 2 Δ V 2 Δ V 1 d 1 = Δ V 2 d 2 Δ V 1 d 1 = Δ V 2 2 d 1 Δ V 2 = 2Δ V 1 = 2(422 V) = 844 V . dzifanu (sd26397) – hw 5 – opyrchal – (121102) 2 004 (part 2 of 2) 10.0 points b) If the plate spacing is doubled while the potential difference between the plates is kept constant, what is the ratio of the final charge on one of the plates to the original charge?...
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Spring 11 Phys 121 - HW 05 - dzifanu(sd26397 – hw 5 –...

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