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# solution00_pdf - dang(rtd356 – HW04 – li –(59050 This...

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dang (rtd356) – HW04 – li – (59050) 1 This print-out should have 26 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 9 cm 2 , separated by a distance 4 . 3 mm . A 26 V potential differ- ence 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 = V d . 2. E = d V . 3. E = ( V d ) 2 . 4. None of these 5. E = 1 V d . 6. E = 1 ( V d ) 2 . 7. E = V d . correct 8. E = parenleftbigg d V parenrightbigg 2 . 9. E = parenleftbigg V d parenrightbigg 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. σ = ǫ 0 d V . 2. σ = ǫ 0 parenleftbigg V d parenrightbigg 2 . 3. σ = ǫ 0 ( V d ) 2 . 4. σ = ǫ 0 ( V d ) 2 . 5. σ = ǫ 0 parenleftbigg d V parenrightbigg 2 . 6. σ = ǫ 0 V d 7. σ = ǫ 0 V d . correct 8. σ = ǫ 0 V d . 9. None of these 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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dang (rtd356) – HW04 – li – (59050) 2 Calculate the capacitance. Correct answer: 1 . 8532 pF. Explanation: Let : A = 0 . 0009 m 2 , d = 0 . 0043 m , V = 26 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 . 0009 m 2 0 . 0043 m = 1 . 8532 × 10 12 F = 1 . 8532 pF . 004 (part 4 of 4) 10.0 points Calculate plate charge; i.e. , the magnitude of the charge on each plate. Correct answer: 48 . 1833 pC. Explanation: The charge Q on one of the plates is simply Q = C V = (1 . 8532 × 10 12 F) (26 V) = 4 . 81833 × 10 11 C = 48 . 1833 pC . 005 10.0 points Given a spherical capacitor with radius of the inner conducting sphere a and the outer shell b . The outer shell is grounded. The charges are + Q and - Q . A point C is located at r = R 2 , where R = a + b . a A B C + Q - Q b What is the capacitance of this spherical capacitor? 1. C = b 2 4 k e ( b - a ) 2. C = k e b 3. C = b k e 4. C = a + b k e 5. C = 1 k e ( a - b ) 6. C = k e a 7. C = 1 k e parenleftbigg 1 a - 1 b parenrightbigg correct 8. C = a k e 9. C = b - a 2 k e ln parenleftbigg b a parenrightbigg 10. C = 1 k e ( a + b ) Explanation: Δ V = V a - V b = k e Q parenleftbigg 1 a - 1 b parenrightbigg - 0 since V b is grounded. The charge on the inside of the shell doesn’t affect the grounded potential. The capacitance of this spherical capacitor
dang (rtd356) – HW04 – li – (59050) 3 is C = Q Δ V = Q k e Q parenleftbigg 1 a - 1 b parenrightbigg = 1 k e parenleftbigg 1 a - 1 b parenrightbigg .

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solution00_pdf - dang(rtd356 – HW04 – li –(59050 This...

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