HW 5 - nguyen(akn452 – Homework#5 – yu –(44003 1 This...

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Unformatted text preview: nguyen (akn452) – Homework #5 – yu – (44003) 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 A capacitor network is shown in the following figure. 12 V 5 . 28 μ F 6 . 8 μ F 10 . 2 μ F a b What is the voltage across the 6 . 8 μ F upper right-hand capacitor? Correct answer: 5 . 24503 V. Explanation: Let : C 1 = 5 . 28 μ F , C 2 = 6 . 8 μ F , C 3 = 10 . 2 μ F , and V = 12 V . Since C 1 and C 2 are in series they carry the same charge C 1 V 1 = C 2 V 2 , and their voltages add up to V , voltage of the battery V 1 + V 2 = V C 2 V 2 C 1 + V 2 = V C 2 V 2 + C 1 V 2 = V C 1 V 2 = V C 1 C 1 + C 2 = (12 V)(5 . 28 μ F) 5 . 28 μ F + 6 . 8 μ F = 5 . 24503 V . 002 (part 1 of 2) 10.0 points A 54 m length of coaxial cable has a solid cylindrical wire inner conductor with a di- ameter of 2 . 59 mm and carries a charge of 10 . 47 μ C. The surrounding conductor is a cylindrical shell and has an inner diameter of 8 . 36 mm and a charge of- 10 . 47 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: 2 . 56371 nF. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 10 . 47 μ C , ℓ = 54 m , a = 2 . 59 mm , and b = 8 . 36 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 = 54 m 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) × 1 ln parenleftbigg 8 . 36 mm 2 . 59 mm parenrightbigg · parenleftbigg 1 × 10 9 nF 1 F parenrightbigg = 2 . 56371 nF 003 (part 2 of 2) 10.0 points nguyen (akn452) – Homework #5 – yu – (44003) 2 What is the potential difference between the two conductors? Correct answer: 4 . 08393 kV. Explanation: Therefore V = Q C = 10 . 47 μ C 2 . 56371 nF parenleftbigg 1 × 10 9 nF 1 F parenrightbigg × parenleftbigg 1 C 1 × 10 6 μ C parenrightbiggparenleftbigg 1 kV 1000 V parenrightbigg = 4 . 08393 kV 004 10.0 points The electron volt is a measure of 1. velocity. 2. impulse. 3. momentum. 4. charge. 5. energy. correct Explanation: Electron volt (eV) is a unit commonly used in atomic and nuclear physics. It is defined as the energy that an electron gains or loses by moving through a potential of 1 V....
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This note was uploaded on 03/27/2012 for the course PHYSICS 1444 taught by Professor Jarafee during the Spring '10 term at UT Arlington.

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HW 5 - nguyen(akn452 – Homework#5 – yu –(44003 1 This...

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