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Unformatted text preview: song (shs546) – Homework 8 – weathers – (22202) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points When a certain airfilled parallelplate ca pacitor is connected across a battery, it ac quires a charge (on each plate) of magnitude 163 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 266 μ C on each plate. κ κ What is the dielectric constant of the slab? Correct answer: 2 . 6319. Explanation: Let : Q a = 163 μ C , Q d = 163 μ C + 266 μ C , and = 429 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 = 429 μ C 163 μ C = 2 . 6319 . 002 (part 1 of 4) 10.0 points Four capacitors are connected as shown in the figure. 1 6 . 2 μ F 5 7 . 1 μ F 8 8 . 2 μ F 98 . 2 V 46 . 8 μ F a b c d Find the capacitance between points a and b of the entire capacitor network. Correct answer: 118 . 153 μ F. Explanation: Let : C 1 = 16 . 2 μ F , C 2 = 46 . 8 μ F , C 3 = 57 . 1 μ F , C 4 = 88 . 2 μ F , and E B = 98 . 2 V . C 1 C 3 C 4 E B C 2 a b c d A good rule of thumb is to eliminate junc tions connected by zero capacitance. C 2 C 3 C 1 C 4 a b c d The definition of capacitance is C ≡ Q V . The parallel connection of C 1 and C 2 gives the equivalent capacitance C 12 = C 1 + C 2 = 16 . 2 μ F + 46 . 8 μ F = 63 μ F . song (shs546) – Homework 8 – weathers – (22202) 2 C 12 C 3 C 4 a b The series connection of C 12 and C 3 gives the equivalent capacitance C 123 = 1 1 C 12 + 1 C 3 = C 12 C 3 C 12 + C 3 = (63 μ F) (57 . 1 μ F) 63 μ F + 57 . 1 μ F = 29 . 9525 μ F . C 123 C 4 a b The parallel connection of C 123 and C 4 gives the equivalent capacitance C ab = C 123 + C 4 = 29 . 9525 μ F + 88 . 2 μ F = 118 . 153 μ F , or combining the above steps, the equivalent capacitance is C ab = ( C 1 + C 2 ) C 3 C 1 + C 2 + C 3 + C 4 = (16 . 2 μ F + 46 . 8 μ F) (57 . 1 μ F) 16 . 2 μ F + 46 . 8 μ F + 57 . 1 μ F + 88 . 2 μ F = 118 . 153 μ F ....
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
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