practice_14-solutions.pdf

# practice_14-solutions.pdf - agha(msa2448 practice 14...

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agha (msa2448) – practice 14 – turner – (90130) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points An air-filled cylindrical capacitor has a capac- itance of 7 pF and is 2 . 8 cm in length. If the radius of the outside conductor is 2 . 1 cm, what is the radius of the in- ner conductor? The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 1 . 68103 cm. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , = 2 . 8 cm = 0 . 028 m , b = 2 . 1 cm , and C = 7 pF = 7 × 10 12 F . For an inner radius of a and an outer radius of b , The capacitance of a cylindrical capacitor is given by C = 2 k e ln parenleftbigg b a parenrightbigg ln parenleftbigg b a parenrightbigg = 2 k e C b a = e ℓ/ (2 k e C ) a = b e ℓ/ (2 k e C ) Since 2 k e C = 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) × (7 × 10 12 F) = 0 . 125826 m , a = (2 . 1 cm) e (0 . 028 m) / (0 . 125826 m) = 1 . 68103 cm . 002 10.0points If the plate separation of an isolated parallel plate capacitor is doubled, 1. None of these correct 2. the electric field is doubled. 3. the charge density on each plate is dou- bled. 4. the charge on the each plate is halved. 5. the potential difference is halved. Explanation: Q and A are constant, so E = Q A ǫ 0 = constant . E = Q A ǫ 0 = (2 Q ) (2 A ) ǫ 0 and V = Q C = 2 Q C = 2 Q C = 2 V . 003(part1of2)10.0points Consider the following system of equivalent capacitors. E B 3 μ F 3 μ F 3 μ F 3 μ F 3 μ F 3 μ F Find the equivalent capacitance of the circuit.

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