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**Unformatted text preview: **fierro (jmf2547) – HW13 – berk – (60290) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A parallel-plate capacitor of dimensions 1 . 51 cm × 3 . 39 cm is separated by a 1 . 67 mm thickness of paper. Find the capacitance of this device. The dielectric constant κ for paper is 3.7. Correct answer: 10 . 0418 pF. Explanation: Let : κ = 3 . 7 , d = 1 . 67 mm = 0 . 00167 m , and A = 1 . 51 cm × 3 . 39 cm = 0 . 00051189 m 2 . We apply the equation for the capacitance of a parallel-plate capacitor and find C = κǫ A d = (3 . 7) (8 . 85419 × 10 − 12 C 2 / N · m 2 ) × parenleftbigg . 00051189 m 2 . 00167 m parenrightbigg 1 × 10 12 pF 1 F = 10 . 0418 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 . 268316 μ c. Explanation: Let : E max = 1 . 6 × 10 7 V / m . Since the thickness of the paper is 0 . 00167 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 . 0418 pF)(26720 V) · 1 × 10 − 12 F 1 pF · 1 × 10 6 μ C 1 C = . 268316 μ c . 003 (part 1 of 2) 10.0 points Four capacitors are connected as shown in the figure. 1 7 μ F 57 μ F 32 μ F 7 4 μ F 99 V a b c d Find the capacitance between points a and b . Correct answer: 111 . 494 μ F. Explanation: Let : C 1 = 17 μ F , C 2 = 32 μ F , C 3 = 57 μ F , C 4 = 74 μ F , and E = 99 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 fierro (jmf2547) – HW13 – berk – (60290) 2 The definition of capacitance is C ≡ Q V ....

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