47131_HW4

# 47131_HW4 - ragsdale(zdr82 – HW4 – ditmire –(58335 1 This print-out should have 23 questions Multiple-choice questions may continue on the

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Unformatted text preview: ragsdale (zdr82) – HW4 – ditmire – (58335) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points An air-filled capacitor consists of two parallel plates, each with an area of 1 . 3 cm 2 , sepa- rated by a distance 1 . 8 mm . A 25 V potential difference is applied to these plates. The permittivity of a vacuum is 8 . 85419 × 10 − 12 C 2 / N · m 2 . 1 pF is equal to 10 − 12 F . The magnitude of the electric field between the plates is 1. None of these 2. E = parenleftbigg V d parenrightbigg 2 . 3. E = V d . correct 4. E = ( V d ) 2 . 5. E = V d . 6. E = d V . 7. E = 1 V d . 8. E = parenleftbigg d V parenrightbigg 2 . 9. E = 1 ( V d ) 2 . Explanation: Since E is constant between the plates, V = integraldisplay vector E · d vector l = E d E = V d . 002 (part 2 of 4) 10.0 points The magnitude of the surface charge density on each plate is 1. None of these 2. σ = ǫ V d . correct 3. σ = ǫ V d 4. σ = ǫ V d . 5. σ = ǫ ( V d ) 2 . 6. σ = ǫ parenleftbigg V d parenrightbigg 2 . 7. σ = ǫ d V . 8. σ = ǫ parenleftbigg d V parenrightbigg 2 . 9. σ = ǫ ( V d ) 2 . Explanation: Use Gauss’s Law. We find that a pillbox of cross section S which sticks through the sur- face on one of the plates encloses charge σ S. The flux through the pillbox is only through the top, so the total flux is E S. Gauss’ Law gives σ = ǫ E = ǫ V d Alternatively, we could just recall this result for an infinite conducting plate (meaning we neglect edge effects) and apply it. 003 (part 3 of 4) 10.0 points ragsdale (zdr82) – HW4 – ditmire – (58335) 2 Calculate the capacitance. Correct answer: 0 . 639469 pF. Explanation: Let : A = 0 . 00013 m 2 , d = 0 . 0018 m , V = 25 V , and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 . The capacitance is given by C = ǫ A d = 8 . 85419 × 10 − 12 C 2 / N · m 2 × . 00013 m 2 . 0018 m = 6 . 39469 × 10 − 13 F = . 639469 pF . 004 (part 4 of 4) 10.0 points Calculate plate charge; i.e. , the magnitude of the charge on each plate. Correct answer: 15 . 9867 pC. Explanation: The charge Q on one of the plates is simply Q = C V = (6 . 39469 × 10 − 13 F) (25 V) = 1 . 59867 × 10 − 11 C = 15 . 9867 pC . 005 10.0 points By what factor does the capacitance of a metal sphere increase if its volume is tripled? 1. 1.41 times 2. 1.5 times 3. 1.73 times 4. 1.33 times 5. 2 times 6. 2.5 times 7. 27 times 8. 3 times 9. 1.44 times correct 10. 9 times Explanation: Let : V ′ = 3 V . C = 4 π ǫ R , and V = 4 3 π R 3 , so C ∝ V 1 / 3 . Thus C ′ ∝ (3 V ) 1 / 3 = 3 √ 3 V 1 / 3 is larger than C by 3 √ 3 = 1 . 44 times....
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## This note was uploaded on 07/17/2011 for the course ECON 101 taught by Professor Garton during the Spring '10 term at Edison College.

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47131_HW4 - ragsdale(zdr82 – HW4 – ditmire –(58335 1 This print-out should have 23 questions Multiple-choice questions may continue on the

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