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121homework 5

# 121homework 5 - Felder Jacob Homework 6 Due 9:00 pm Inst...

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Felder, Jacob – Homework 6 – Due: Oct 17 2006, 9:00 pm – Inst: Vitaly 1 This print-out should have 23 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 4) 10 points An air-filled 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. σ = ² 0 d V 2 . 2. σ = ² 0 V d 3. σ = ² 0 V d . correct 4. None of these 5. σ = ² 0 ( V d ) 2 . 6. σ = ² 0 ( V d ) 2 . 7. σ = ² 0 d V . 8. σ = ² 0 V d . 9. σ = ² 0 V d 2 . Explanation: Use Gauss’s Law. We find that a pillbox of cross section S which sticks through the sur- face on one of the plates encloses charge σ S. The flux through the pillbox is only through the top, so the total flux is E S. Gauss’ Law gives σ = ² 0 E = ² 0 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

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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 ² 0 = 8 . 85419 × 10 - 12 C 2 / N · m 2 . The capacitance is given by C = ² 0 A d = 8 . 85419 × 10 - 12 C 2 / N · m 2 × 0 . 00066 m 2 0 . 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 π ² 0 , 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 Q C 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 . Thus r = k C . 006 (part 2 of 2) 10 points The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . If the potential on the surface of the sphere is 2603 V and the capacitance is 6 . 2 × 10 - 11 F, what is the surface charge density?
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121homework 5 - Felder Jacob Homework 6 Due 9:00 pm Inst...

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