# 6 - kavo (ak22862) – HW06 – Janow – (12121) 1 This...

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Unformatted text preview: kavo (ak22862) – HW06 – Janow – (12121) 1 This print-out should have 17 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 14 . 9 μ C and- 14 . 9 μ C have a potential difference of 8 . 4 V. Determine the capacitance of the system. Correct answer: 1 . 77381 × 10 − 6 F. Explanation: Let : Q = 14 . 9 μ C = 1 . 49 × 10 − 5 C and V = 8 . 4 V . The capacitance is C = Q V = 1 . 49 × 10 − 5 C 8 . 4 V = 1 . 77381 × 10 − 6 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 54 μ C magnitude. Correct answer: 30 . 443 V. Explanation: Let : Q 2 = 54 μ C = 5 . 4 × 10 − 5 C . The new potential difference between the two conductors is V 2 = Q 2 C = 5 . 4 × 10 − 5 C 1 . 77381 × 10 − 6 F = 30 . 443 V . 003 (part 1 of 4) 10.0 points An air-filled capacitor consists of two parallel plates, each with an area of 9 . 2 cm 2 , sepa- rated by a distance 2 . 5 mm . A 16 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 = parenleftbigg d V parenrightbigg 2 . 2. E = 1 V d . 3. E = V d . 4. E = d V . 5. E = 1 ( V d ) 2 . 6. E = parenleftbigg V d parenrightbigg 2 . 7. E = V d . correct 8. E = ( V d ) 2 . 9. None of these 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. σ = ǫ V d . correct 2. σ = ǫ V d . 3. None of these kavo (ak22862) – HW06 – Janow – (12121) 2 4. σ = ǫ ( V d ) 2 . 5. σ = ǫ parenleftbigg V d parenrightbigg 2 . 6. σ = ǫ d V . 7. σ = ǫ parenleftbigg d V parenrightbigg 2 . 8. σ = ǫ V d 9. σ = ǫ ( 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 = ǫ E = ǫ V d Alternatively, we could just recall this result for an infinite conducting plate (meaning we neglect edge effects) and apply it....
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## This note was uploaded on 09/16/2009 for the course PHYS 121 taught by Professor Opyrchal during the Fall '08 term at NJIT.

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6 - kavo (ak22862) – HW06 – Janow – (12121) 1 This...

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