This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: nguyen (jmn727) – oldmidterm 02 – Turner – (59070) 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. 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 righthand 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 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 4 π ǫ r 2. E = Q π ǫ r 2 3. E = Q π ǫ r 4. E = λ π ǫ r 5. E = Q 2 π ǫ r 6. E = λ 2 π ǫ r correct 7. E = Q 4 π ǫ r 8. E = Q 2 π ǫ r 2 9. E = λ 2 π ǫ r 2 10. E = λ π ǫ r 2 Explanation: Apply Gauss’ Law to a cylindrical surface of radius r and length ℓ , to obtain 2 π r ℓ E = λ ℓ ǫ nguyen (jmn727) – oldmidterm 02 – Turner – (59070) 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 = k e ℓ ln parenleftbigg b a parenrightbigg 2. C = ℓ k e ln parenleftBig a b parenrightBig 3. C = ℓ 2 k e ln parenleftbigg b a parenrightbigg correct 4. C = ℓ 2 k e 5. C = ℓ k e 6. C = ℓ a 2 k e b 7. C = 2 ℓ k e ln parenleftBig a b parenrightBig 8. C = ℓ k e ln parenleftbigg b a parenrightbigg 9. C = k e ℓ 2 ln parenleftbigg b a parenrightbigg 10. C = 2 ℓ k e ln parenleftbigg b a parenrightbigg 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 2 . 7 μ F 61 . 4 μ F 46 . 3 μ F 7 9 . 2 μ F 97 . 1 V a b c d Find the capacitance between points a and b of the entire capacitor network....
View
Full
Document
This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Work

Click to edit the document details