oldmidterm 02 - nguyen(jmn727 oldmidterm 02 Turner(59070...

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nguyen (jmn727) – oldmidterm 02 – Turner – (59070) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A capacitor network is shown in the following figure. 17 . 4 V 3 . 43 μ F 4 . 3 μ F 12 . 3 μ F a b What is the voltage across the 4 . 3 μ F upper right-hand capacitor? Correct answer: 7 . 72083 V. Explanation: Let : C 1 = 3 . 43 μ F , C 2 = 4 . 3 μ F , C 3 = 12 . 3 μ F , and V = 17 . 4 V . Since C 1 and C 2 are in series they carry the same charge C 1 V 1 = C 2 V 2 , and their voltages add up to V , voltage of the battery V 1 + V 2 = V C 2 V 2 C 1 + V 2 = V C 2 V 2 + C 1 V 2 = V C 1 V 2 = V C 1 C 1 + C 2 = (17 . 4 V)(3 . 43 μ F) 3 . 43 μ F + 4 . 3 μ F = 7 . 72083 V . 002 (part 1 of 2) 10.0 points A coaxial cable with length has an inner conductor that has a radius a and carries a charge of Q . The surrounding conductor has an inner radius b and a charge of - Q . Assume the region between the conductors is air. The linear charge density λ Q . radius = a + Q radius = b - Q What is the electric field halfway between the conductors? 1. E = λ 4 π ǫ 0 r 2. E = Q π ǫ 0 r 2 3. E = Q π ǫ 0 r 4. E = λ π ǫ 0 r 5. E = Q 2 π ǫ 0 r 6. E = λ 2 π ǫ 0 r correct 7. E = Q 4 π ǫ 0 r 8. E = Q 2 π ǫ 0 r 2 9. E = λ 2 π ǫ 0 r 2 10. E = λ π ǫ 0 r 2 Explanation: Apply Gauss’ Law to a cylindrical surface of radius r and length , to obtain 2 π r ℓ E = λ ℓ ǫ 0
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nguyen (jmn727) – oldmidterm 02 – Turner – (59070) 2 E = λ 2 π r ǫ 0 . 003 (part 2 of 2) 10.0 points What is the capacitance C of this coaxial cable? 1. C = k e ln parenleftbigg b a parenrightbigg 2. C = k e ln parenleftBig a b parenrightBig 3. C = 2 k e ln parenleftbigg b a parenrightbigg correct 4. C = 2 k e 5. C = k e 6. C = ℓ a 2 k e b 7. C = 2 k e ln parenleftBig a b parenrightBig 8. C = k e ln parenleftbigg b a parenrightbigg 9. C = k e 2 ln parenleftbigg b a parenrightbigg 10. C = 2 k e ln parenleftbigg b a parenrightbigg Explanation: First recall that k e = 1 4 π ǫ 0 so E = 2 k e λ r which we can integrate along a radial path from a to b to get the voltage difference, V = - integraldisplay b a E dr = 2 k e λ integraldisplay a b dr r = 2 k e λ ln r vextendsingle vextendsingle vextendsingle a b = 2 k e λ ln parenleftbigg b a parenrightbigg then C = Q V = λ ℓ V = 2 k e ln parenleftbigg b a parenrightbigg . keywords: 004 (part 1 of 4) 10.0 points Four capacitors are connected as shown in the figure. 12 . 7 μ F 61 . 4 μ F 46 . 3 μ F 79 . 2 μ F 97 . 1 V a b c d Find the capacitance between points a and b of the entire capacitor network. Correct answer: 118 . 296 μ F. Explanation: Let : C 1 = 12 . 7 μ F , C 2 = 46 . 3 μ F , C 3 = 61 . 4 μ F , C 4 = 79 . 2 μ F , and E = 97 . 1 V . C 1 C 3 C 2 C 4 E a b c d A good rule of thumb is to eliminate junc- tions connected by zero capacitance.
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nguyen (jmn727) – oldmidterm 02 – Turner – (59070) 3 C 2 C 3 C 1 C 4 a b The definition of capacitance is C Q V .
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