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Unformatted text preview: dang (rtd356) HW04 li (59050) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points An airfilled 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. = d V . 2. = parenleftbigg V d parenrightbigg 2 . 3. = ( V d ) 2 . 4. = ( V d ) 2 . 5. = parenleftbigg d V parenrightbigg 2 . 6. = V d 7. = V d . correct 8. = V d . 9. None of these Explanation: Use Gausss Law. We find that a pillbox of cross section S which sticks through the sur face on one of the plates encloses charge S. The flux through the pillbox is only through the top, so the total flux is E S. Gauss Law gives = E = 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 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 = 8 . 85419 10 12 C 2 / N m 2 . The capacitance is given by C = A d = 8 . 85419 10 12 C 2 / N m 2 . 0009 m 2 . 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: 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 doesnt affect the grounded potential....
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This note was uploaded on 06/15/2010 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner

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