# hw07b - aljabr(faa335 Hw07 Ross(89251 This print-out should...

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aljabr (faa335) – Hw07 – Ross – (89251) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two conductors insulated from each other are charged by transferring electrons from one conductor to the other. After 1 . 3501 × 10 14 have been transferred, the potential difference between the conductors is 6 . 8 V. The charge on an electron is - 1 . 60218 × 10 - 19 C . What is the capacitance of the system? Correct answer: 3 . 18103 × 10 - 6 F. Explanation: Let : e = - 1 . 60218 × 10 - 19 C , N = 1 . 3501 × 10 14 electrons , and V = 6 . 8 V . One conductor gains a charge of Q = - N e while the other gains a charge of - Q. The capacitance of the system is C = q V = - N e V = - (1 . 3501 × 10 14 ) ( - 1 . 60218 × 10 - 19 C) 6 . 8 V = 3 . 18103 × 10 - 6 F . 002 (part 1 of 4) 10.0 points An air-filled capacitor consists of two parallel plates, each with an area of 7 . 3 cm 2 , sepa- rated by a distance 1 . 5 mm . A 16 V potential difference is applied to these plates. The permittivity of a vacuum is 8 . 85419 × 10 - 12 C 2 / N · m 2 . 1 pF is equal to 10 - 12 F . The magnitude of the electric field between the plates is 1. E = parenleftbigg V d parenrightbigg 2 . 2. E = parenleftbigg d V parenrightbigg 2 . 3. None of these 4. E = V d . correct 5. E = V d . 6. E = 1 V d . 7. E = 1 ( V d ) 2 . 8. E = d V . 9. E = ( V d ) 2 . Explanation: Since E is constant between the plates, V = integraldisplay vector E · d vector l = E d E = V d . 003 (part 2 of 4) 10.0 points The magnitude of the surface charge density on each plate is 1. σ = ǫ 0 V d 2. σ = ǫ 0 parenleftbigg d V parenrightbigg 2 .

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