Capacitance-solutions

Capacitance-solutions - gilvin (jg47854) - 11. Capacitance...

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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 (part 1 of 2) 10.0 points A 3.47 µF capacitor is connected to a 16 V battery. What is the charge on each plate of the capacitor? Correct answer: 55.52 µC. Explanation: Let : C = 3.47 µF = 3.47 × 10 −6 F and ΔV 1 = 16 V . q = C ΔV 1 = (3.47 × 10 −6 F) (16 V) · 10 6 µC 1 C = 55.52 µC . +Q −Q b b b b 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 002 (part 2 of 2) 10.0 points If this same capacitor is connected to a 3 V battery, what charge is stored? C = ϵ 0 A d . Hence doubling d halves the capacitance, and Q = C V is also halved ( Correct answer: 10.41 µC. Explanation: C = ϵ 0 A 2 d = 2 ϵ 0 A d = 2 C) . Let : ΔV 2 = 3 V . The charge is q = C ΔV 2 = (3.47 × 10 −6 F) (3 V) · 10 6 µC 1 C = 10.41 µC . 003 10.0 points A parallel plate capacitor is connected to a battery. 004 10.0 points 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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2 d = ? 0 A 4. 1 4 C 0 . 5.4 C 0 . Explanation: C = ϵ 0 A gilvin (jg47854) - 11. Capacitance - meyers - (21235) 2 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 . = ϵ 0 2 A d d = C 0 . 005 (part 1 of 2) 10.0 points 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: 007 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 = ϵ 0 (V d) 2 2. σ = ϵ 0 V d ϵ 0 V 3. σ = Let : A = 0.7 km 2 = 7 × 10 5 m 2 , d correct h = 780 m and ϵ 0 = 8.85419 × 10 −12 C 2 /N · m 2 . The capacitance of the cloud layer is C = ϵ 0 A ( h 4. σ = ϵ 0 V d ( 5. σ = ϵ 0 ( ) 2 d V ) 2 V = 8.85419 × 10 −12 C 2 /N · m 2 ) 6. σ = ϵ 0 ( ) d × 7 × 10 5 m 2 780 m 7. σ = ϵ 0 d = 7.94607 × 10 −9 F . 006 (part 2 of 2) 10.0 points
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Capacitance-solutions - gilvin (jg47854) - 11. Capacitance...

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