5 - karna(pk4534 – HW04 – li –(59050 1 This print-out...

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Unformatted text preview: karna (pk4534) – 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 4 . 1 cm 2 , sepa- rated by a distance 3 . 1 mm . A 16 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 = 1 ( V d ) 2 . 2. E = 1 V d . 3. E = V d . correct 4. E = ( V d ) 2 . 5. E = V d. 6. E = parenleftbigg d V parenrightbigg 2 . 7. None of these 8. E = parenleftbigg V d parenrightbigg 2 . 9. E = d V . 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 ( V d ) 2 . 2. None of these 3. σ = ǫ V d . correct 4. σ = ǫ parenleftbigg V d parenrightbigg 2 . 5. σ = ǫ parenleftbigg d V parenrightbigg 2 . 6. σ = ǫ d V . 7. σ = ǫ V d . 8. σ = ǫ ( V d ) 2 . 9. σ = ǫ V d 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 karna (pk4534) – HW04 – li – (59050) 2 Calculate the capacitance. Correct answer: 1 . 17104 pF. Explanation: Let : A = 0 . 00041 m 2 , d = 0 . 0031 m , V = 16 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 × . 00041 m 2 . 0031 m = 1 . 17104 × 10 − 12 F = 1 . 17104 pF . 004 (part 4 of 4) 10.0 points Calculate plate charge; i.e. , the magnitude of the charge on each plate. Correct answer: 18 . 7366 pC. Explanation: The charge Q on one of the plates is simply Q = C V = (1 . 17104 × 10 − 12 F) (16 V) = 1 . 87366 × 10 − 11 C = 18 . 7366 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 = k e b 2. C = 1 k e ( a- b ) 3. C = a k e 4. C = k e a 5. C = 1 k e ( a + b ) 6. C = a + b k e 7. C = b k e 8. C = b 2 4 k e ( b- a ) 9. C = b- a 2 k e ln parenleftbigg b a parenrightbigg 10. C = 1 k e parenleftbigg 1 a- 1 b parenrightbigg correct Explanation: Δ V = V a- V b = k e Q parenleftbigg 1 a- 1 b parenrightbigg- since V b is grounded. The charge on the inside of the shell doesn’t affect the grounded potential.potential....
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This note was uploaded on 11/02/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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5 - karna(pk4534 – HW04 – li –(59050 1 This print-out...

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