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**Unformatted text preview: **fierro (jmf2547) – HW12 – berk – (60290) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A capacitor network with air-filled capacitors as shown below. 37 . 7 V 75 . 5 μ F 75 . 5 μ F 75 . 5 μ F 75 . 5 μ F b a c d When the top right-hand capacitor is filled with a material of dielectric constant κ , the charge on this capacitor is increases by a fac- tor of 1 . 46. Find the dielectric constant κ of the mate- rial inserted into the top right-hand capaci- tor. Correct answer: 2 . 7037. Explanation: Let : C 1 = C = 75 . 5 μ F , C 2 = C = 75 . 5 μ F , C 3 = C = 75 . 5 μ F , C 4 = C = 75 . 5 μ F , E B = 37 . 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 specific values are immaterial. Furthermore, 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 final charges on C 2 and Q ′ Q ≡ α =ratio of final 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 . 46 . Solving for κ , we have κ = α 2- α = 1 . 46 2- 1 . 46 = 2 . 7037 . 002 10.0 points Consider the two cases shown below. In Case fierro (jmf2547) – HW12 – berk – (60290) 2 One two identical capacitors are connected to a battery with emf V . In Case Two, a di- electric slab with dielectric constant κ fills 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. C ′ 12 C 12 = 2 κ. 3. C ′ 12 C 12 = 1 + κ ....

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