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Unformatted text preview: Wong, Lemuel – Homework 8 – Due: Mar 23 2007, 11:00 pm – Inst: Matt Mackie 1 This printout should have 31 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. NOTE: HW is due 11pm central time, which is midnight our time. 001 (part 1 of 1) 10 points A fully charged capacitor stores 11 . 2 J of en ergy. How much energy remains when its charge has decreased to half its original value? Correct answer: 2 . 8 J. Explanation: Let : U = 11 . 2 J . Q = C V , so the energy is U = 1 2 C V 2 = 1 2 Q 2 C . If Q is halved, U falls to onequarter, as shown U f = 1 2 Q 2 2 C = 1 2 Q 2 4 C = 1 4 µ 1 2 Q 2 C ¶ = 1 4 U = 1 4 (11 . 2 J) = 2 . 8 J . keywords: 002 (part 1 of 2) 10 points Two capacitors of 26 . 4 μ F and 5 . 16 μ F are connected in parallel and charged with a 50 V power supply. Find the total energy stored in the capaci tors. Correct answer: 0 . 03945 J. Explanation: Given : C 1 = 26 . 4 μ F , C 2 = 5 . 16 μ F , and V = 50 V . The capacitors are in parallel, so C eq = C 1 + C 2 and V 1 = V 2 = V . Thus the energy stored is U = 1 2 ( C 1 + C 2 ) V 2 = 1 2 (26 . 4 μ F + 5 . 16 μ F)(50 V) 2 1 F 10 6 μ F = . 03945 J . 003 (part 2 of 2) 10 points What potential difference would be required across the same two capacitors connected in series in order for the combination to store the same amount of energy? Correct answer: 135 . 201 V. Explanation: With the capacitors in series, 1 C series = 1 C 1 + 1 C 2 = C 2 + C 1 C 1 C 2 . Since U = C series V 2 2 V = r 2 U C series = s 2 U ( C 1 + C 2 ) C 1 C 2 = s 2(0 . 03945 J)(26 . 4 μ F+5 . 16 μ F) (26 . 4 μ F)(5 . 16 μ F) · 10 6 μ F 1 F = 135 . 201 V . keywords: 004 (part 1 of 1) 10 points Wong, Lemuel – Homework 8 – Due: Mar 23 2007, 11:00 pm – Inst: Matt Mackie 2 When a certain airfilled parallelplate ca pacitor is connected across a battery, it ac quires a charge (on each plate) of magnitude 136 μ C. While the battery connection is maintained, a dielectric slab is inserted into the space between the capacitor plates and completely fills this region. This results in the accumulation of an additional charge of 318 μ C on each plate. κ κ What is the dielectric constant of the slab? Correct answer: 3 . 33824 . Explanation: Let : Q a = 136 μ C , Q d = 136 μ C + 318 μ C , and = 454 μ C . Since the capacitor is connected to the bat tery the whole time, we know that the poten tial drop across the capacitor is held constant. The charge changes only because the capaci tance does. When the capacitor is filled with air, we have Q a = C a V . When the dielec tric slab is inserted, the charge is given by Q d = C d V . The dielectric constant is then κ = C d C a = Q d Q a = 454 μ C 136 μ C = 3 . 33824 ....
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This note was uploaded on 02/17/2011 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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