t2fall02sol - PHYSICS 2306 TEST #2 1 November 2002; 5:00 PM...

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PHYSICS 2306 TEST #2 1 November 2002; 5:00 PM in McBryde 129 Lecture Section 5:30 – 6:45 M,W 3”x5” Note Card and Calculator Allowed 90 Minutes Allowed SOLUTIONS _________________________ ____________________ ________________________ (Name - Printed) (Student I.D.) (Signed Pledge - See Below) I pledge that the answers submitted by me represent only my work. Further, I pledge to refrain from communicating any information about this test to anyone until after the results of this test are returned. Please fill in the three entries specified above. Please record your answers to problems #1 through #20 on the op-scan answer sheet. Please remember to darken the entries corresponding to your student I.D on the op-scan sheet. Do not put your name on the op- scan sheet , unless you do not mind your I.D. and name being associated with each other. Please return the question sheets in a stapled packet to the appropriate stack on the front table; and return the op-scan answer sheet to the other appropriate stack on the front table. No credit will be given if any question or answer sheets are missing from your exam. Constants and Conversion Factors: charge of electron(magnitude) or proton: e = 1 . 6 × 10 - 19 C mass of electron m e = 9 . 11 × 10 -31 kg mass of proton m p = 1 . 67 × 10 -27 kg Coulomb’s law constant: k e = 1/4 πε o = 8 . 99 × 10 9 (N m 2 /C 2 ) Permittivity of free space: ε o = 8 . 854 × 10 - 12 (C 2 /N m 2 ) Use as the speed of sound in air: v = 340 m/s Permeability of free space: µ o = 4 π × 10 -7 (T m/A)
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1. An ideal parallel-plate capacitor, with plate separation of 4 mm, has its plates connected across a 100 Volt battery. If the plates are circular with radius 4 cm, then the capacitance, when air is in the gap between the plates, is about: (1) 1 pF (2) 10 pF C = ε o A/d = 8.85 10 -12 ⋅π (0.04) 2 /[4 10 -3 ] = 11.1 10 -12 10 pF (3) 100 pF (4) 1 nF (5) 10 nF (6) 100 nF 2. For the situation described in problem #1, if the battery remains connected, and a dielectric material with dielectric constant 2.5 is inserted to fill the space between the plates, then the electric field in the dielectric will be: (1) 100 V/m (2) 250 V/m (3) 1000 V/m (4) 2500 V/m (5) 10000 V/m Since the battery remains connected the potential change is still 100 V; hence (6) 25000 V/m in the dielectric E = V/d = 100/4 10 -3 = 25 10 +3 = 25000 V/m 3. Consider the system of capacitors as shown in the figure below. The equivalent capacitance is: (1) 3 µ F 3 µ F 6 µ F (2) 4 µ F a 12 µ F b (3) 6 µ F (4) 8 µ F 4 µ F (5) 12 µ F 1/C 36 = 1/3 + 1/6 = 3/6; C 36 = 2 µ F . C 364 = C 36 + C 4 = 2 + 4 = 6 µ F (6) 18 µ F 1/C eq = 1/C 364 + 1/C 12 = 1/6 + 1/12 = 3/12; C eq = 4 µ F 4. For the situation described in problem #3, when a potential difference is applied between a and b, the charge on the 4 µ F capacitor is 32 µ C. Then the charge on the 3 µ F capacitor is: (1) 16 µ C V 4 = Q 4 /C 4 = 32/4 = 8 V; then V 36 = 8 V, so Q
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t2fall02sol - PHYSICS 2306 TEST #2 1 November 2002; 5:00 PM...

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