hw07b - aljabr (faa335) Hw07 Ross (89251) 1 This print-out...

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Unformatted text preview: 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. = V d 2. = parenleftbigg d V parenrightbigg 2 ....
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hw07b - aljabr (faa335) Hw07 Ross (89251) 1 This print-out...

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