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Unformatted text preview: Chapter 21 Practice Pr solns: # 22,25,27,29, 30 ,31, 35, 36, 41, 47, 48, 59 22.(a) When the surface is perpendicular to the field, its normal is either parallel or antiparallel to . E r Then Equation 21.1 gives 2 2 cos(0 or 180 ) (850 N/C)(2 m ) 1.70 kN m /C. E A EA Φ = ⋅ = ° ° = ± = ± ⋅ r r (b) cos(45 or 135 ) E A EA Φ = ⋅ = ° ° = r r 2 2 (1.70 kN m /C)(0.866) 1.20 kN m /C. ± ⋅ = ± ⋅ (c) cos90 0. EA Φ = ° = 25. I NTERPRET This problem is about the electric flux through the surface of a sphere. D EVELOP The general expression for the electric flux Φ is given by Equation 21.1: cos , E A EA θ Φ = ⋅ = r r where θ is the angle between the normal vector A r and the electric field . E r The magnitude of the normal vector A r is 2 4 , A r π = the surface area of the sphere. E VALUATE For a sphere, with E r parallel or antiparallel to , A r Equation 21.1 gives 2 2 2 4 4 (0.1 m/2) (47 kN/C) 1.48 kN m /C r E π π Φ = ± = ± = ± ⋅ 27. I NTERPRET This problem is about applying Gauss’s law to find the electric flux through a closed surface. D EVELOP Gauss’s law given in Equation 21.3 states that the flux through any closed surface is proportional to the charge enclosed: enclosed q E dA ε Φ = ⋅ = r r r E VALUATE For the surfaces shown, the results are as follows: (a) enclosed ( 2 ) , q q q q = +  =  / . q ε Φ =  (b) enclosed ( 2 ) ( ) 3 ( 3 ) 2 q q q q q q q = +  +  + +  =  and 2 / . q ε Φ =  (c) enclosed q = and 0. Φ = (d) enclosed 3 ( 3 ) q q q = +  = and 0. Φ = A SSESS The flux through the closed surface depends only on the charge enclosed, and is independent of the shape of the surface. 29. I NTERPRET This problem is about applying Gauss’s law to find the electric flux through the surface of a cube which encloses a charge....
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This note was uploaded on 06/03/2010 for the course PHYS 299 taught by Professor Hoston,amahd,bakanowski during the Summer '08 term at University of Louisville.
 Summer '08
 Hoston,Amahd,Bakanowski
 Physics

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