solutions 5

# solutions 5 - Patel Kinal Homework 5 Due Feb 5 2008 7:00 pm...

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Patel, Kinal – Homework 5 – Due: Feb 5 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points On a clear, sunny day, there is a vertical elec- trical feld o± about 146 N / C pointing down over ²at ground or water. What is the magnitude o± the sur±ace charge density on the ground ±or these conditions? Correct answer: 1 . 29271 × 10 - 9 C / m 2 . Explanation: Let : E = - 146 N / C . By Gauss’ law, I ~ E · d ~ A = q ² 0 . I± we consider the electric feld as being im- mediately above the ground or water, then we can think o± the ground or water as an inf- nite sheet o± charge. The question is, should the electric feld beneath the sur±ace be the same as the feld above? Should it be zero? The real answer is quite complicated. In or- der to work this problem, we assume the feld is zero beneath the sur±ace. This makes sense: since water is a conductor, the feld inside the water should be zero. Using Gauss’ Law, E A = σ A ² 0 , we have σ = E ² 0 = ( - 146 N / C) × ( 8 . 85419 × 10 - 12 C 2 / N · m 2 ) = - 1 . 29271 × 10 - 9 C / m 2 , with a magnitude o± 1 . 29271 × 10 - 9 C / m 2 . keywords: 002 (part 1 o± 2) 10 points A conducting spherical shell having an inner radius o± 4 . 1 cm and outer radius o± 5 . 4 cm carries a net charge o± 7 . 2 μ C. A conducting sphere is placed at the center o± this shell having a radius o± 0 . 9 cm carring a net charge o± 1 . 7 μ C. 0 . 9 cm 4 . 1 cm , Q 0 2 inside 5 . 4 cm Q 00 2 outside 1 . 7 μ C 7 . 2 μ C P Determine the sur±ace charge density on the inner sur±ace o± the shell. Correct answer: - 8 . 04769 × 10 - 5 C / m 2 . Explanation: Let : r = 0 . 9 cm , not required a = 4 . 1 cm = 0 . 041 m , b = 5 . 4 cm = 0 . 054 m , and q = 1 . 7 μ C = 1 . 7 × 10 - 6 C . Basic Concept: E = 0 inside a Conductor. Gauss’ Law, I ~ E · d ~ A = Q ² 0 . Solution: Since the electric feld is zero inside any conductor in electrostatic equi- librium, the net charge is zero inside any spherical Gaussian sur±ace o± radius r , where a < r < b . Thus the charge on the inner sur- ±ace o± the sphere must be - q . I± we call the charge density on the inner sur±ace σ in , then - q = 4 π a 2 σ in σ in = - q 4 π a 2 = - 1 . 7 × 10 - 6 C 4 π (0 . 041 m) 2 = - 8 . 04769 × 10 - 5 C / m 2 .

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Patel, Kinal – Homework 5 – Due: Feb 5 2008, 7:00 pm – Inst: Weathers 2 003 (part 2 of 2) 10 points Determine the surface charge density on the outer surface of the shell. Correct answer: 0 . 000242881 C / m 2 . Explanation: Let : Q = 7 . 2 μ C = 7 . 2 × 10 - 6 C . The total charge contained in any spherical Gaussian surface of radius r > b must be Q + q . Since the charge at the origin and the charge on the inner surface of the sphere are equal and opposite, the charge on the outer surface must be Q + q . We ±nd σ out in a manner similar to that used in Part 1 Q +
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solutions 5 - Patel Kinal Homework 5 Due Feb 5 2008 7:00 pm...

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