12 - Kapoor (mk9499) homework12 Turner (60230) This...

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Kapoor (mk9499) – homework12 – Turner – (60230) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A capacitor network with air-flled capacitors as shown below. 29 . 7 V 70 . 2 μ ± 70 . 2 μ ± 70 . 2 μ ± 70 . 2 μ ± b a c d When the top right-hand capacitor is flled with a material oF dielectric constant κ , the charge on this capacitor is increases by a Fac- tor oF 1 . 43. ±ind the dielectric constant κ oF the mate- rial inserted into the top right-hand capaci- tor. Correct answer: 2 . 50877. Explanation: Let : C 1 = C = 70 . 2 μ ± , C 2 = C = 70 . 2 μ ± , C 3 = C = 70 . 2 μ ± , C 4 = C = 70 . 2 μ ± , E B = 29 . 7 V , and Q = 1 . 5 Q . E B C 1 C 3 C 2 C 4 b a c d The capacitors C 3 and C 4 have nothing to do with this problems. In addition, the capac- itances are all equal and their specifc values are immaterial. ±urthermore, the electric po- tential oF the battery is not required. C 1 = C 2 = C 3 = C 4 , where Q and Q are the initial and fnal charges on C 2 and Q Q α =ratio oF fnal to initial charge on C 2 . We know the charges on C 1 and C 2 are the same. Initially, V ab = V 1 + V 2 = Q C 1 + Q C 2 = Q C + Q C = 2 Q C . (1) ThereFore Q = 1 2 V ab C . AFter the dielectric material is inserted in C 2 , the capacitance becomes C 2 = κ C . There- Fore, V ab = V 1 + V 2 = Q C 1 + Q C 2 = Q C + Q κ C = κ + 1 κ Q C , and using Eq. (1) and solving For Q , we have 2 Q C = κ + 1 κ Q C Q = κ κ + 1 V ab C = κ κ + 1 2 Q Q Q α = 2 κ κ + 1 = 1 . 43 . Solving For κ , we have κ = α 2 - α = 1 . 43 2 - 1 . 43 = 2 . 50877 . 002 10.0 points Consider the two cases shown below. In Case
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Kapoor (mk9499) – homework12 – Turner – (60230) 2 One two identical capacitors are connected to a battery with emf V . In Case Two, a di- electric slab with dielectric constant κ Flls the gap of capacitor C 2 . Let C 12 be the resultant capacitance for Case One and C 12 the resul- tant capacitance for Case Two. Case One V C 1 C 2 Case Two V C 1 C 2 κ The ratio C 12 C 12 of the resultant capacitances is 1. C 12 C 12 = κ 2 . 2. C 12 C 12 = 1 + κ . 3.
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This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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12 - Kapoor (mk9499) homework12 Turner (60230) This...

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