Physics 2 - gilvin(jg47854 11 Capacitance meyers(21235 This...

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gilvin (jg47854) – 11. Capacitance – meyers – (21235) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This homework is due Friday, March 11, at midnight Tucson time. 001(part1of2)10.0points A 3 . 47 μ F capacitor is connected to a 16 V battery. What is the charge on each plate of the capacitor? Correct answer: 55 . 52 μ C. Explanation: Let : C = 3 . 47 μ F = 3 . 47 × 10 6 F and Δ V 1 = 16 V . q = C Δ V 1 = (3 . 47 × 10 6 F) (16 V) · 10 6 μ C 1 C = 55 . 52 μ C . 002(part2of2)10.0points If this same capacitor is connected to a 3 V battery, what charge is stored? Correct answer: 10 . 41 μ C. Explanation: Let : Δ V 2 = 3 V . The charge is q = C Δ V 2 = (3 . 47 × 10 6 F) (3 V) · 10 6 μ C 1 C = 10 . 41 μ C . 003 10.0points A parallel plate capacitor is connected to a battery. + Q Q d 2 d If we double the plate separation, 1. the charge on each plate is halved. cor- rect 2. the electric field is doubled. 3. the capacitance is doubled. 4. the potential difference is halved. 5. None of these. Explanation: The capacitance of a parallel plate capaci- tor is C = ǫ 0 A d . Hence doubling d halves the capacitance, and Q = C V is also halved parenleftbigg C = ǫ 0 A 2 d = 1 2 ǫ 0 A d = 1 2 C parenrightbigg . 004 10.0points A parallel-plate capacitor has a capacitance C 0 . A second parallel-plate capacitor has plates with twice the area and twice the sepa- ration. The capacitance of the second capacitor is most nearly 1. 1 2 C 0 . 2. 2 C 0 . 3. C 0 . correct
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gilvin (jg47854) – 11. Capacitance – meyers – (21235) 2 4. 1 4 C 0 . 5. 4 C 0 . Explanation: C = ǫ 0 A d = ǫ 0 2 A 2 d = ǫ 0 A d = C 0 . 005(part1of2)10.0points Consider Earth and a cloud layer 780 m above Earth to be the plates of a parallel-plate ca- pacitor. The permittivity of free space is 8 . 85419 × 10 12 C 2 / N · m 2 . If the cloud layer has an area of 0 . 7 km 2 = 7 × 10 5 m 2 , what is the capacitance? Correct answer: 7 . 94607 × 10 9 F. Explanation: Let : A = 0 . 7 km 2 = 7 × 10 5 m 2 , h = 780 m and ǫ 0 = 8 . 85419 × 10 12 C 2 / N · m 2 . The capacitance of the cloud layer is C = ǫ 0 A h = ( 8 . 85419 × 10 12 C 2 / N · m 2 ) × parenleftbigg 7 × 10 5 m 2 780 m parenrightbigg = 7 . 94607 × 10 9 F . 006(part2of2)10.0points If an electric field strength greater than 3 × 10 6 N / C causes the air to break down and conduct charge (lightning), what is the maximum charge the cloud can hold? Correct answer: 18 . 5938 C. Explanation: Let : E max = 3 × 10 6 N / C . The maximum charge is Q max = C V max = C E max h = (7 . 94607 × 10 9 F) (3 × 10 6 N / C) × (780 m) = 18 . 5938 C . 007 10.0points 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 0 ( V d ) 2 2. σ = ǫ 0 V d 3. σ = ǫ 0 V d correct 4. σ = ǫ 0 V d 5. σ = ǫ 0 parenleftbigg d V parenrightbigg 2 6. σ = ǫ 0 parenleftbigg V d parenrightbigg 2 7. σ = ǫ 0 d V 8. σ = ǫ 0 ( 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
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