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Unformatted text preview: mcconnell (kam2342) – 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 . 69 cm × 5 . 91 cm is separated by a 1 . 4 mm thickness of paper. Find the capacitance of this device. The dielectric constant κ for paper is 3.7. Correct answer: 23 . 372 pF. Explanation: Let : κ = 3 . 7 , d = 1 . 4 mm = 0 . 0014 m , and A = 1 . 69 cm × 5 . 91 cm = 0 . 00099879 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 . 00099879 m 2 . 0014 m parenrightbigg 1 × 10 12 pF 1 F = 23 . 372 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 . 523534 μ c. Explanation: Let : E max = 1 . 6 × 10 7 V / m . Since the thickness of the paper is 0 . 0014 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 = (23 . 372 pF)(22400 V) · 1 × 10 − 12 F 1 pF · 1 × 10 6 μ C 1 C = . 523534 μ c . 003 (part 1 of 2) 10.0 points Four capacitors are connected as shown in the figure. 2 6 μ F 67 μ F 44 μ F 8 5 μ F 94 V a b c d Find the capacitance between points a and b . Correct answer: 137 . 559 μ F. Explanation: Let : C 1 = 26 μ F , C 2 = 44 μ F , C 3 = 67 μ F , C 4 = 85 μ F , and E = 94 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 mcconnell (kam2342) – 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 = (44 μ F) (67 μ F) 44 μ F + 67 μ F = 26 . 5586 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 = 26 μ F + 85 μ F + 26 . 5586 μ F = 137 . 559 μ F . 004 (part 2 of 2) 10.0 points What is the charge on the 44 μ F upper centered capacitor? Correct answer: 2496 . 5 μ C. Explanation: The voltages across C 2 and C 3 , respectively, (the voltage between a and b ) are V ab = V 23 = 94 V , and we have Q 23 = Q 3 = Q 2 = V ab C 23 = (94 V) (26 . 5586 μ F) = 2496 . 5 μ C ....
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This note was uploaded on 11/22/2010 for the course PHYS 303 taught by Professor Turner during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Turner

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