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Unformatted text preview: Kapoor (mk9499) – homework13 – Turner – (60230) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A parallelplate capacitor of dimensions 1 . 73 cm × 3 . 68 cm is separated by a 1 . 93 mm thickness of paper. Find the capacitance of this device. The dielectric constant κ for paper is 3.7. Correct answer: 10 . 8065 pF. Explanation: Let : κ = 3 . 7 , d = 1 . 93 mm = 0 . 00193 m , and A = 1 . 73 cm × 3 . 68 cm = 0 . 00063664 m 2 . We apply the equation for the capacitance of a parallelplate capacitor and find C = κǫ A d = (3 . 7) (8 . 85419 × 10 − 12 C 2 / N · m 2 ) × parenleftbigg . 00063664 m 2 . 00193 m parenrightbigg 1 × 10 12 pF 1 F = 10 . 8065 pF . 002 (part 2 of 2) 10.0 points What is the maximum charge that can be placed on the capacitor? The electric strength of paper is 1 . 6 × 10 7 V / m. Correct answer: 0 . 333706 μ c. Explanation: Let : E max = 1 . 6 × 10 7 V / m . Since the thickness of the paper is 0 . 00193 m, the maximum voltage that can be applied before breakdown is V max = E max d. Hence, the maximum charge is Q max = C V max = C E max d = (10 . 8065 pF)(30880 V) · 1 × 10 − 12 F 1 pF · 1 × 10 6 μ C 1 C = . 333706 μ c . 003 (part 1 of 2) 10.0 points Four capacitors are connected as shown in the figure. 1 1 μ F 54 μ F 38 μ F 8 6 μ F 90 V a b c d Find the capacitance between points a and b . Correct answer: 119 . 304 μ F. Explanation: Let : C 1 = 11 μ F , C 2 = 38 μ F , C 3 = 54 μ F , C 4 = 86 μ F , and E = 90 V . C 1 C 3 C 2 C 4 E B 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 Kapoor (mk9499) – homework13 – Turner – (60230) 2 The definition of capacitance is C ≡ Q V . The series connection of C 2 and C 3 gives the equivalent capacitance C 23 = 1 1 C 2 + 1 C 3 = C 2 C 3 C 2 + C 3 = (38 μ F) (54 μ F) 38 μ F + 54 μ F = 22 . 3043 μ F . The total capacitance C ab between a and b can be obtained by calculating the capacitance in the parallel combination of the capacitors C 1 , C 4 , and C 23 ; i.e. , C ab = C 1 + C 4 + C 23 = 11 μ F + 86 μ F + 22 . 3043 μ F = 119 . 304 μ F . 004 (part 2 of 2) 10.0 points What is the charge on the 38 μ F upper centered capacitor? Correct answer: 2007 . 39 μ C. Explanation: The voltages across C 2 and C 3 , respectively, (the voltage between a and b ) are V ab = V 23 = 90 V , and we have Q 23 = Q 3 = Q 2 = V ab C 23 = (90 V) (22 . 3043 μ F) = 2007 . 39 μ C ....
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

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