Hw11_pdf - mcconnell (kam2342) homework11 Turner (60230) 1...

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Unformatted text preview: mcconnell (kam2342) homework11 Turner (60230) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points An air-filled capacitor consists of two parallel plates, each with an area of 4 . 5 cm 2 , sepa- rated by a distance 3 . 6 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 = ( V d ) 2 . 2. E = d V . 3. E = parenleftbigg V d parenrightbigg 2 . 4. E = V d . 5. E = parenleftbigg d V parenrightbigg 2 . 6. E = 1 ( V d ) 2 . 7. E = 1 V d . 8. E = V d . correct 9. None of these Explanation: Since E is constant between the plates, V = integraldisplay vector E d vector l = E d E = V d . 002 (part 2 of 4) 10.0 points The magnitude of the surface charge density on each plate is 1. = V d . correct 2. None of these 3. = ( V d ) 2 . 4. = parenleftbigg V d parenrightbigg 2 . 5. = V d . 6. = ( V d ) 2 . 7. = parenleftbigg d V parenrightbigg 2 . 8. = d V . 9. = V d Explanation: Use Gausss Law. We find that a pillbox of cross section S which sticks through the sur- face on one of the plates encloses charge S. The flux through the pillbox is only through the top, so the total flux is E S. Gauss Law gives = E = V d Alternatively, we could just recall this result for an infinite conducting plate (meaning we neglect edge effects) and apply it. 003 (part 3 of 4) 10.0 points mcconnell (kam2342) homework11 Turner (60230) 2 Calculate the capacitance. Correct answer: 1 . 10677 pF. Explanation: Let : A = 0 . 00045 m 2 , d = 0 . 0036 m , V = 16 V , and = 8 . 85419 10- 12 C 2 / N m 2 . The capacitance is given by C = A d = 8 . 85419 10- 12 C 2 / N m 2 . 00045 m 2 . 0036 m = 1 . 10677 10- 12 F = 1 . 10677 pF ....
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Hw11_pdf - mcconnell (kam2342) homework11 Turner (60230) 1...

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