solution_pdf old midterm 2

solution_pdf old midterm 2 - malsam(wgm329 oldmidterm 02...

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malsam (wgm329) – oldmidterm 02 – turner – (56705) 1 This print-out should have 19 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. 16 V 4 . 99 μ F 6 . 7 μ F 13 . 4 μ F a b What is the voltage across the 6 . 7 μ F upper right-hand capacitor? Correct answer: 6 . 82977 V. Explanation: Let : C 1 = 4 . 99 μ F , C 2 = 6 . 7 μ F , C 3 = 13 . 4 μ F , and V = 16 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 = (16 V)(4 . 99 μ F) 4 . 99 μ F + 6 . 7 μ F = 6 . 82977 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 = λ π ǫ 0 r 4. E = Q 2 π ǫ 0 r 2 5. E = λ 2 π ǫ 0 r 2 6. E = Q 2 π ǫ 0 r 7. E = λ π ǫ 0 r 2 8. E = Q 4 π ǫ 0 r 9. E = Q π ǫ 0 r 10. E = λ 2 π ǫ 0 r correct 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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malsam (wgm329) – oldmidterm 02 – turner – (56705) 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 parenleftBig a b parenrightBig 2. C = 2 k e ln parenleftBig a b parenrightBig 3. C = k e ln parenleftbigg b a parenrightbigg 4. C = 2 k e 5. C = k e 6. C = k e 2 ln parenleftbigg b a parenrightbigg 7. C = 2 k e ln parenleftbigg b a parenrightbigg correct 8. C = 2 k e ln parenleftbigg b a parenrightbigg 9. C = k e ln parenleftbigg b a parenrightbigg 10. C = ℓ a 2 k e b 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. 21 . 1 μ F 63 . 4 μ F 44 . 2 μ F 82 . 2 μ F 95 V a b c d Find the capacitance between points a and b of the entire capacitor network. Correct answer: 129 . 343 μ F. Explanation: Let : C 1 = 21 . 1 μ F , C 2 = 44 . 2 μ F , C 3 = 63 . 4 μ F , C 4 = 82 . 2 μ F , and E = 95 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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malsam (wgm329) – oldmidterm 02 – turner – (56705) 3 C 2 C 3 C 1 C 4 a b The definition of capacitance is C Q V .
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