Phys121-6sol

# Phys121-6sol - Anderson, Matthew Homework 6 Due: Mar 4...

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Unformatted text preview: Anderson, Matthew Homework 6 Due: Mar 4 2008, noon Inst: Vitaly 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Two conductors having net charges of 4 . 8 C and- 4 . 8 C have a potential difference of 4 . 4 V. Determine the capacitance of the system. Correct answer: 1 . 09091 10- 6 F. Explanation: Let : Q = 4 . 8 C = 4 . 8 10- 6 C and V = 4 . 4 V . The capacitance is C = Q V = 4 . 8 10- 6 C 4 . 4 V = 1 . 09091 10- 6 F . 002 (part 2 of 2) 10 points Determine the potential difference between the two conductors if the charges on each are increased to 40 C magnitude. Correct answer: 36 . 6667 V. Explanation: Let : Q 2 = 40 C = 4 10- 5 C . The new potential difference between the two conductors is V 2 = Q 2 C = 4 10- 5 C 1 . 09091 10- 6 F = 36 . 6667 V . keywords: 003 (part 1 of 4) 10 points An air-filled capacitor consists of two parallel plates, each with an area of 1 . 6 cm 2 , sepa- rated by a distance 1 . 1 mm . A 15 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 2 . 3. E = V d . correct 4. None of these 5. E = 1 V d . 6. E = V d. 7. E = V d 2 . 8. E = 1 ( V d ) 2 . 9. E = d V . Explanation: Since E is constant between the plates, V = Z ~ E d ~ l = E d E = V d . 004 (part 2 of 4) 10 points The magnitude of the surface charge density on each plate is 1. = ( V d ) 2 . 2. = V d . correct Anderson, Matthew Homework 6 Due: Mar 4 2008, noon Inst: Vitaly 2 3. = V d 2 . 4. = ( V d ) 2 . 5. = d V . 6. = d V 2 . 7. = V d 8. = V d . 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 points Calculate the capacitance. Correct answer: 1 . 28788 pF....
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## This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.

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Phys121-6sol - Anderson, Matthew Homework 6 Due: Mar 4...

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