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Unformatted text preview: benavides (jjb2356) – oldmidterm 02 – Turner – (59130) 1 This printout should have 16 questions. Multiplechoice 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 . 1 μ F 9 . 7 μ F 12 . 4 μ F a b What is the voltage across the 9 . 7 μ F upper righthand capacitor? Correct answer: 4 . 75362 V. Explanation: Let : C 1 = 4 . 1 μ F , C 2 = 9 . 7 μ F , C 3 = 12 . 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 . 1 μ F) 4 . 1 μ F + 9 . 7 μ F = 4 . 75362 V . 002 (part 1 of 2) 10.0 points A coaxial cable with length ℓ 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 ℓ . ℓ radius = a + Q radius = b Q b What is the electric field halfway between the conductors? 1. E = λ π ǫ r 2 2. E = Q π ǫ r 3. E = Q π ǫ r 2 4. E = λ π ǫ r 5. E = Q 2 π ǫ r 2 6. E = λ 4 π ǫ r 7. E = Q 2 π ǫ r 8. E = λ 2 π ǫ r correct 9. E = λ 2 π ǫ r 2 10. E = Q 4 π ǫ r Explanation: Apply Gauss’ Law to a cylindrical surface of radius r and length ℓ , to obtain 2 π r ℓ E = λ ℓ ǫ benavides (jjb2356) – oldmidterm 02 – Turner – (59130) 2 E = λ 2 π r ǫ . 003 (part 2 of 2) 10.0 points What is the capacitance C of this coaxial cable? 1. C = 2 ℓ k e ln parenleftbigg b a parenrightbigg 2. C = ℓ 2 k e ln parenleftbigg b a parenrightbigg correct 3. C = k e ℓ ln parenleftbigg b a parenrightbigg 4. C = ℓ k e ln parenleftBig a b parenrightBig 5. C = ℓ k e 6. C = ℓ 2 k e 7. C = 2 ℓ k e ln parenleftBig a b parenrightBig 8. C = k e ℓ 2 ln parenleftbigg b a parenrightbigg 9. C = ℓ 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 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 = ℓ 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. 1 7 μ F 67 μ F 40 . 1 μ F 8 3 . 5 μ F 92 . 1 V a b c d Find the capacitance between points a and b of the entire capacitor network....
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 Spring '08
 Turner
 Capacitance, Work, Correct Answer, E= E= E=, Qab Vab

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