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Unformatted text preview: Felder, Jacob Homework 6 Due: Oct 17 2006, 9:00 pm Inst: Vitaly 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 4) 10 points An airfilled capacitor consists of two parallel plates, each with an area of 6 . 6 cm 2 , sepa rated by a distance 3 mm . A 23 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 = d V 2 . 3. E = 1 ( V d ) 2 . 4. E = V d . correct 5. E = V d. 6. E = d V . 7. E = V d 2 . 8. E = ( V d ) 2 . 9. None of these Explanation: Since E is constant between the plates, V = Z ~ E d ~ l = E d E = V d . 002 (part 2 of 4) 10 points The magnitude of the surface charge density on each plate is 1. = d V 2 . 2. = V d 3. = V d . correct 4. None of these 5. = ( V d ) 2 . 6. = ( V d ) 2 . 7. = d V . 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 points Felder, Jacob Homework 6 Due: Oct 17 2006, 9:00 pm Inst: Vitaly 2 Calculate the capacitance. Correct answer: 1 . 94792 pF. Explanation: Let : A = 0 . 00066 m 2 , d = 0 . 003 m , V = 23 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 . 00066 m 2 . 003 m = 1 . 94792 10 12 F = 1 . 94792 pF . 004 (part 4 of 4) 10 points Calculate plate charge; i.e. , the magnitude of the charge on each plate. Correct answer: 44 . 8022 pC. Explanation: The charge Q on one of the plates is simply Q = C V = (1 . 94792 10 12 F)(23 V) = 4 . 48022 10 11 C = 44 . 8022 pC . keywords: 005 (part 1 of 2) 10 points An isolated conducting sphere can be consid ered as one element of a capacitor (the other element being a concentric sphere of infinite radius). If k = 1 4 , the capacitance of the system is C , and the charge on the sphere is Q , what is the radius of the sphere? 1. r = C k 2. r = k C Q 3. r = Q 2 C k 4. r = k QC 5. r = k Q 6. r = k C correct Explanation: If the radius of the sphere is r , then the potential difference for the charged sphere is V = k Q r and for a capacitor, C = Q V = r k ....
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This note was uploaded on 04/03/2009 for the course PHYS 121 taught by Professor Opyrchal during the Fall '08 term at NJIT.
 Fall '08
 Opyrchal
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