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Unformatted text preview: maini (nm7637) hw6 Shneidman (12108) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points Two conductors having net charges of 5 . 5 C and- 5 . 5 C have a potential difference of 15 . 3 V. Determine the capacitance of the system. Correct answer: 3 . 59477 10 7 F. Explanation: Let : Q = 5 . 5 C = 5 . 5 10 6 C and V = 15 . 3 V . The capacitance is C = Q V = 5 . 5 10 6 C 15 . 3 V = 3 . 59477 10 7 F . 002 (part 2 of 2) 10.0 points Determine the potential difference between the two conductors if the charges on each are increased to 122 C magnitude. Correct answer: 339 . 382 V. Explanation: Let : Q 2 = 122 C = 0 . 000122 C . The new potential difference between the two conductors is V 2 = Q 2 C = . 000122 C 3 . 59477 10 7 F = 339 . 382 V . 003 (part 1 of 4) 10.0 points An air-filled capacitor consists of two parallel plates, each with an area of 2 . 1 cm 2 , sepa- rated by a distance 4 . 8 mm . A 25 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 = d V . 2. E = 1 ( V d ) 2 . 3. E = ( V d ) 2 . 4. E = parenleftbigg V d parenrightbigg 2 . 5. E = V d . correct 6. E = V d . 7. E = parenleftbigg d V parenrightbigg 2 . 8. None of these 9. E = 1 V d . Explanation: Since E is constant between the plates, V = integraldisplay vector E d vector l = E d E = V d . 004 (part 2 of 4) 10.0 points The magnitude of the surface charge density on each plate is 1. = ( V d ) 2 . 2. = parenleftbigg d V parenrightbigg 2 . 3. = V d . correct maini (nm7637) hw6 Shneidman (12108) 2 4. = ( V d ) 2 . 5. = V d 6. = V d . 7. = d V . 8. = parenleftbigg V d parenrightbigg 2 . 9. None of these 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. 005 (part 3 of 4) 10.0 points Calculate the capacitance....
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This note was uploaded on 01/31/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.

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solution6_pdf - maini (nm7637) hw6 Shneidman (12108) 1 This...

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