# hw5 - Homework 5 Solutions Spring 2010 Required Problems...

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Unformatted text preview: Homework 5 Solutions Spring 2010 Required Problems Chapter 22 9. The only contributions to the flux are from the faces perpendicular to the electric field. Over each of these two surfaces, the magnitude of the field is constant, so the flux is just E A r r g on each of these two surfaces. ( 29 ( 29 2 2 encl E right left right left Q E E ε Φ = + =- = → E A E A r r r r g g l l 18. See Example 22-3 for a detailed discussion related to this problem. ( a )Inside a solid metal sphere the electric field is 0 . ( b ) Inside a solid metal sphere the electric field is 0 . ( c ) Outside a solid metal sphere the electric field is the same as if all the charge were concentrated at the center as a point charge. ( 29 ( 29 ( 29 6 9 2 2 2 2 5.50 10 C 1 8.988 10 N m C 5140N C 4 3.10m Q E r πε- × = = × ⋅ = The field would point towards the center of the sphere. ( d ) Same reasoning as in part ( c ). ( 29 ( 29 ( 29 6 9 2 2 2 2 5.50 10 C 8.988 10 N m C 772N C 8.00m 1 4 Q E r πε- × = = × ⋅ = The field would point towards the center of the sphere. ( e ) The answers would be no different for a thin metal shell. ( f ) The solid sphere of charge is dealt with in Example 22-4. We see from that Example that the field inside the sphere is given by 3 1 . 4 Q E r r πε = Outside the sphere the field is no different. So we have these results for the solid sphere. ( 29 ( 29 ( 29 ( 29 6 9 2 2 3 5.50 10 C 0.250m 8.988 10 N m C 0.250m 458N C 3.00m E r- × = = × ⋅ = ( 29 ( 29 ( 29 ( 29 6 9 2 2 3 5.50 10 C 2.90m 8.988 10 N m C 2.90m 5310N C 3.00m E r- × = = × ⋅ = ( 29 ( 29 ( 29 6 9 2 2 2 5.50 10 C 3.10m 8.988 10 N m C 5140N C 3.10m E r- × = = × ⋅ = ( 29 ( 29 ( 29 6 9 2 2 2 5.50 10 C 8.00m 8.988 10 N m C 772N C 3.10m E r- × = = × ⋅ = All point towards the center of the sphere. 46. Because the slab is very large, and we are considering only distances from the slab much less than its height or breadth, the symmetry of the slab results in the field being perpendicular to the slab, with a constant magnitude for a constant distance from the center. We assume that E ρ and so the electric field points away from the center of the slab.and so the electric field points away from the center of the slab....
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## This note was uploaded on 04/23/2010 for the course PHYSICS ?? taught by Professor Zettle during the Spring '08 term at Berkeley.

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hw5 - Homework 5 Solutions Spring 2010 Required Problems...

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