OHW 11 - terry (ect328) oldhomework 11 Turner (59130) 1...

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Unformatted text preview: terry (ect328) oldhomework 11 Turner (59130) 1 This print-out should have 12 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 9 . 5 cm 2 , sepa- rated by a distance 4 . 1 mm . A 20 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 = 1 V d . 2. E = parenleftbigg d V parenrightbigg 2 . 3. E = parenleftbigg V d parenrightbigg 2 . 4. E = ( V d ) 2 . 5. None of these 6. E = d V . 7. E = 1 ( V d ) 2 . 8. E = V d . 9. E = V d . correct 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. None of these 2. = ( V d ) 2 . 3. = parenleftbigg V d parenrightbigg 2 . 4. = d V . 5. = parenleftbigg d V parenrightbigg 2 . 6. = V d . 7. = V d . correct 8. = V d 9. = ( V d ) 2 . 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 Calculate the capacitance. terry (ect328) oldhomework 11 Turner (59130) 2 Correct answer: 2 . 05158 pF. Explanation: Let : A = 0 . 00095 m 2 , d = 0 . 0041 m , V = 20 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 . 00095 m 2 . 0041 m = 2 . 05158 10 12 F = 2 . 05158 pF ....
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OHW 11 - terry (ect328) oldhomework 11 Turner (59130) 1...

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