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Physhw 5-solutions

# Physhw 5-solutions - billing(cab4763 hw 5 opyrchal(121104...

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billing (cab4763) – hw 5 – opyrchal – (121104) 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(part1of2)10.0points A 27 m length of coaxial cable has a solid cylindrical wire inner conductor with a di- ameter of 3 . 489 mm and carries a charge of 10 . 03 μ C. The surrounding conductor is a cylindrical shell and has an inner diameter of 7 . 755 mm and a charge of - 10 . 03 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 . 8806 nF. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 10 . 03 μ C , = 27 m , a = 3 . 489 mm , and b = 7 . 755 mm . The charge per unit length is λ Q . V = - integraldisplay b a vector E · dvectors = - 2 k e λ integraldisplay b a dr r = - 2 k e Q ln parenleftbigg b a parenrightbigg . The capacitance of a cylindrical capacitor is given by C Q V = 2 k e 1 ln parenleftbigg b a parenrightbigg = 27 m 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) × 1 ln parenleftbigg 7 . 755 mm 3 . 489 mm parenrightbigg · parenleftbigg 1 × 10 9 nF 1 F parenrightbigg = 1 . 8806 nF 002(part2of2)10.0points What is the potential difference between the two conductors? Correct answer: 5 . 33341 kV. Explanation: Therefore V = Q C = 10 . 03 μ C 1 . 8806 nF parenleftbigg 1 × 10 9 nF 1 F parenrightbigg × parenleftbigg 1 C 1 × 10 6 μ C parenrightbigg parenleftbigg 1 kV 1000 V parenrightbigg = 5 . 33341 kV 003(part1of2)10.0points The potential difference between a pair of oppositely charged parallel plates is 364 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: 728 V. Explanation: Let : d 2 = 2 d 1 and Δ V 1 = 364 V . The charge remains the same, so Q = C 1 Δ V 1 = C 2 Δ V 2 ε 0 A d 1 Δ V 1 = ε 0 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(364 V) = 728 V .

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billing (cab4763) – hw 5 – opyrchal – (121104) 2 004(part2of2)10.0points 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? Correct answer: 0 . 5. Explanation: Solution: The potential difference is the same, so Δ V = Q 1 C 1 = Q 2 C 2 Q 1 C 2 = Q 2 C 1 Q 1 ε 0 A d 2 = Q 2 ε 0 A d 1 Q 1 d 2 = Q 2 d 1 Q 1 2 d 1 = Q 2 d 1 Q 2 Q 1 = 1 2 = 0 . 5 .
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Physhw 5-solutions - billing(cab4763 hw 5 opyrchal(121104...

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