Lemuel physics 2 8 - Wong Lemuel Homework 8 Due 11:00 pm...

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Wong, Lemuel – Homework 8 – Due: Mar 23 2007, 11:00 pm – Inst: Matt Mackie 1 This print-out should have 31 questions. Multiple-choice 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 one-quarter, as shown U f = 1 2 Q 2 2 C = 1 2 Q 2 4 C = 1 4 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 = 0 . 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 0 2 2 V 0 = 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
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Wong, Lemuel – Homework 8 – Due: Mar 23 2007, 11:00 pm – Inst: Matt Mackie 2 When a certain air-filled parallel-plate 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 . keywords: 005 (part 1 of 2) 10 points A capacitor is constructed from two square metal plates. A dielectric κ = 3 . 49 fills the upper half of the capacitor and a dielectric κ = 9 . 3 fills the lower half of the capacitor. Neglect edge effects. 18 cm 0 . 55 mm dielectric constant bottom 9 . 3 dielectric constant top 3 . 49 Calculate the capacitance C of the device.
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