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# solution6_pdf - maini(nm7637 hw6 Shneidman(12108 This...

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maini (nm7637) – hw6 – Shneidman – (12108) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points Two conductors having net charges oF 5 . 5 μ C and - 5 . 5 μ C have a potential di±erence oF 15 . 3 V. Determine the capacitance oF the system. Correct answer: 3 . 59477 × 10 7 ². Explanation: Let : Q = 5 . 5 μ C = 5 . 5 × 10 6 C and V = 15 . 3 V . The capacitance is C = Q V = 5 . 5 × 10 6 C 15 . 3 V = 3 . 59477 × 10 7 ² . 002 (part 2 oF 2) 10.0 points Determine the potential di±erence between the two conductors iF the charges on each are increased to 122 μ C magnitude. Correct answer: 339 . 382 V. Explanation: Let : Q 2 = 122 μ C = 0 . 000122 C . The new potential di±erence between the two conductors is V 2 = Q 2 C = 0 . 000122 C 3 . 59477 × 10 7 ² = 339 . 382 V . 003 (part 1 oF 4) 10.0 points An air-flled capacitor consists oF two parallel plates, each with an area oF 2 . 1 cm 2 , sepa- rated by a distance 4 . 8 mm . A 25 V potential di±erence is applied to these plates. The permittivity oF a vacuum is 8 . 85419 × 10 12 C 2 / N · m 2 . 1 p² is equal to 10 12 ² . The magnitude oF the electric feld between the plates is 1. E = d V . 2. E = 1 ( V d ) 2 . 3. E = ( V d ) 2 . 4. E = p V d P 2 . 5. E = V d . correct 6. E = V d . 7. E = p d V P 2 . 8. None oF these 9. E = 1 V d . Explanation: Since E is constant between the plates, V = i v E · d v l = E d E = V d . 004 (part 2 oF 4) 10.0 points The magnitude oF the surFace charge density on each plate is 1. σ = ǫ 0 ( V d ) 2 . 2. σ = ǫ 0 p d V P 2 . 3. σ = ǫ 0 V d . correct

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maini (nm7637) – hw6 – Shneidman – (12108) 2 4. σ = ǫ 0 ( V d ) 2 . 5. σ = ǫ 0 V d 6. σ = ǫ 0 V d . 7. σ = ǫ 0 d V . 8. σ = ǫ 0 p V d P 2 . 9. None of these Explanation: Use Gauss’s Law. We Fnd that a pillbox of cross section S which sticks through the sur- face on one of the plates encloses charge σ S. The ±ux through the pillbox is only through the top, so the total ±ux is E S. Gauss’ Law gives σ = ǫ 0 E = ǫ 0 V d Alternatively, we could just recall this result for an inFnite conducting plate (meaning we neglect edge e²ects) and apply it. 005 (part 3 of 4) 10.0 points Calculate the capacitance. Correct answer: 0 . 387371 p³. Explanation: Let : A = 0 . 00021 m 2 , d = 0 . 0048 m , V = 25 V , and ǫ 0 = 8 . 85419 × 10 12 C 2 / N · m 2 . The capacitance is given by C = ǫ 0 A d = 8 . 85419 × 10 12 C 2 / N · m 2 × 0 . 00021 m 2 0 . 0048 m = 3 . 87371 × 10 13 ³ = 0 . 387371 p³ . 006 (part 4 of 4) 10.0 points Calculate plate charge; i.e. , the magnitude of the charge on each plate.
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solution6_pdf - maini(nm7637 hw6 Shneidman(12108 This...

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