{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Hw12_pdf - mcconnell(kam2342 – homework12 – Turner...

This preview shows pages 1–3. Sign up to view the full content.

mcconnell (kam2342) – homework12 – Turner – (60230) 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. 85 V 27 . 2 μ F 27 . 2 μ F 27 . 2 μ F 27 . 2 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 . 48. Find the dielectric constant κ of the mate- rial inserted into the top right-hand capaci- tor. Correct answer: 2 . 84615. Explanation: Let : C 1 = C = 27 . 2 μ F , C 2 = C = 27 . 2 μ F , C 3 = C = 27 . 2 μ F , C 4 = C = 27 . 2 μ F , E B = 85 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 . 48 . Solving for κ , we have κ = α 2 - α = 1 . 48 2 - 1 . 48 = 2 . 84615 . 002 10.0 points Consider the two cases shown below. In Case

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
mcconnell (kam2342) – 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 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. None of these 2. C 12 C 12 = κ 2 . 3. C 12 C 12 = 1 + κ . 4. C 12 C 12 = 2 κ . 5. C 12 C 12 = 1 + κ 2 κ . 6. C 12 C 12 = 2 1 + κ . 7. C 12 C 12 = 2 κ 1 + κ . correct 8. C 12 C 12 = κ . 9. C 12 C 12 = 1 + κ 2 . 10. C 12 C 12 = 1 κ .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}