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# 5 - roofner(bar784 Homework 5 Weathers(22202 This print-out...

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roofner (bar784) – Homework 5 – Weathers – (22202) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points On a clear, sunny day, there is a vertical elec- trical field of about 105 N / C pointing down over flat ground or water. What is the magnitude of the surface charge density on the ground for these conditions? Correct answer: 9 . 2969 × 10 10 C / m 2 . Explanation: Let : E = - 105 N / C . By Gauss’ law, contintegraldisplay vector E · d vector A = q ǫ 0 . If we consider the electric field as being im- mediately above the ground or water, then we can think of the ground or water as an infi- nite sheet of charge. The question is, should the electric field beneath the surface be the same as the field above? Should it be zero? The real answer is quite complicated. In or- der to work this problem, we assume the field is zero beneath the surface. This makes sense: since water is a conductor, the field inside the water should be zero. Using Gauss’ Law, E A = σ A ǫ 0 , we have σ = E ǫ 0 = ( - 105 N / C) × ( 8 . 85419 × 10 12 C 2 / N · m 2 ) = - 9 . 2969 × 10 10 C / m 2 , with a magnitude of 9 . 2969 × 10 10 C / m 2 . 002 (part 1 of 2) 10.0 points A conducting spherical shell having an inner radius of 4 . 4 cm and outer radius of 5 . 1 cm carries a net charge of 7 . 2 μ C. A conducting sphere is placed at the center of this shell having a radius of 1 . 3 cm carring a net charge of 2 . 4 μ C. 1 . 3 cm 4 . 4 cm , Q 2 inside 5 . 1 cm Q ′′ 2 outside 2 . 4 μ C 7 . 2 μ C P Determine the surface charge density on the inner surface of the shell. Correct answer: - 9 . 86498 × 10 5 C / m 2 . Explanation: Let : r = 1 . 3 cm , not required a = 4 . 4 cm = 0 . 044 m , b = 5 . 1 cm = 0 . 051 m , and q = 2 . 4 μ C = 2 . 4 × 10 6 C . Basic Concept: E = 0 inside a Conductor. Gauss’ Law, contintegraldisplay vector E · d vector A = Q ǫ 0 . Solution: Since the electric field is zero inside any conductor in electrostatic equi- librium, the net charge is zero inside any spherical Gaussian surface of radius r , where a < r < b . Thus the charge on the inner sur- face of the sphere must be - q . If we call the charge density on the inner surface σ in , then - q = 4 π a 2 σ in σ in = - q 4 π a 2 = - 2 . 4 × 10 6 C 4 π (0 . 044 m) 2 = - 9 . 86498 × 10 5 C / m 2 . 003 (part 2 of 2) 10.0 points Determine the surface charge density on the outer surface of the shell. Correct answer: 0 . 000293712 C / m 2 .

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roofner (bar784) – Homework 5 – Weathers – (22202) 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 find σ out in a manner similar to that used in Part 1 Q + q = 4 π b 2 σ out σ out = Q + q 4 π b 2 = 7 . 2 × 10 6 C + 2 . 4 × 10 6 C 4
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5 - roofner(bar784 Homework 5 Weathers(22202 This print-out...

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