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11 - Kapoor(mk9499 homework11 Turner(60230 This print-out...

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Kapoor (mk9499) – homework11 – Turner – (60230) 1 This print-out should have 13 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 5 . 1 cm 2 , sepa- rated by a distance 4 . 1 mm . A 29 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. E = d V . 2. E = V d . correct 3. E = parenleftbigg d V parenrightbigg 2 . 4. E = 1 ( V d ) 2 . 5. E = 1 V d . 6. E = parenleftbigg V d parenrightbigg 2 . 7. E = ( V d ) 2 . 8. E = V d . 9. None of these 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. σ = ǫ 0 ( V d ) 2 . 2. None of these 3. σ = ǫ 0 parenleftbigg d V parenrightbigg 2 . 4. σ = ǫ 0 ( V d ) 2 . 5. σ = ǫ 0 V d . correct 6. σ = ǫ 0 V d 7. σ = ǫ 0 parenleftbigg V d parenrightbigg 2 . 8. σ = ǫ 0 V d . 9. σ = ǫ 0 d V . 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 0 E = ǫ 0 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
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Kapoor (mk9499) – homework11 – Turner – (60230) 2 Calculate the capacitance. Correct answer: 1 . 10137 pF. Explanation: Let : A = 0 . 00051 m 2 , d = 0 . 0041 m , V = 29 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 . 00051 m 2 0 . 0041 m = 1 . 10137 × 10 - 12 F = 1 . 10137 pF . 004 (part 4 of 4) 10.0 points Calculate plate charge; i.e. , the magnitude of the charge on each plate. Correct answer: 31 . 9399 pC. Explanation: The charge Q on one of the plates is simply Q = C V = (1 . 10137 × 10 - 12 F) (29 V) = 3 . 19399 × 10 - 11 C = 31 . 9399 pC .
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