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Unformatted text preview: Platt, David Homework 8 Due: Sep 26 2005, 4:00 am Inst: Ken Shih 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. For 003 you need give your answer in unit of pF (1pF=e12*F). 001 (part 1 of 2) 10 points Consider a airfilled parallelplate capacitor with plates of length 8 . 8 cm, width 6 . 78 cm, spaced a distance 1 . 86 cm apart. Now imagine that a dielectric slab with di electric constant 2 . 7 is inserted a length 4 . 8 cm into the capacitor. The slab has the same width as the plates. The capacitor is com pletely filled with dielectric material down to a length of 4 . 8 cm. A battery is connected to the plates so that they are at a constant potential 0 . 83 V while the dielectric is inserted. 8 . 8 cm 4 . 8 cm 1 . 86 cm dielectric none  air bottom 1 . dielectric constant top 2 . 7 What is the ratio of the new potential en ergy to the potential energy before the inser tion of the dielectric? Edge effects can be neglected. Correct answer: 1 . 92727 . Explanation: Let : w = 6 . 78 cm , d = 1 . 86 cm , L = 8 . 8 cm , = 4 . 8 cm , = 2 . 7 , and V = 0 . 83 V . The given capacitor can be considered as two capacitors in parallel, as shown below. C 1 C 2 The potential energy in a capacitor is U = 1 2 C V 2 , so the ratio is U U = C V 2 C V 2 . Since the electric potential difference (volt age) across the capacitor is constant, U U = C C . (1) The capacitance of a parallel plate capaci tor is C = area separation = w L d . (2) After the insertion of the dielectric, the ca pacitor can be considered in two parts: the part filled with dielectric, and the part with no dielectric. Since the dielectric is inserted with a length , C = C 1 + C 2 , where (3) C 1 = w d C 2 = w ( L ) d . The total capacitance of the new system is C = C 1 + C 2 = ( L + ) w d . (4) Therefore, we have U U = C C = n L + [  1] o w d L w d Platt, David Homework 8 Due: Sep 26 2005, 4:00 am Inst: Ken Shih 2 = 1 + L h  1 i (5) = 1 + (4 . 8 cm) (8 . 8 cm) h (2 . 7) 1 i = 1 . 92727 . When the dielectric is fully inserted = L , so U U = 1 + L L h  1 i = . (6) 002 (part 2 of 2) 10 points From this, we can deduce that if a person inserts the slab part way into the conductor, and then releases it, the capacitor will exert a force on the dielectric in what direction?...
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This note was uploaded on 04/10/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics, Work

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