ch23_p51 - 51. (a) We integrate the volume charge density...

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51. (a) We integrate the volume charge density over the volume and require the result be equal to the total charge: 2 0 4 . R dx dy dz dr r Q ρρ = ³³³ ³ Substituting the expression ρ = s r / R , with s = 14.1 pC/m 3 , and performing the integration leads to 4 4 4 s R Q R §· π= ¨¸ ©¹ or 31 2 3 3 1 5 (14.1 10 C/m )(0.0560 m) 7.78 10 C. s QR πρ π −− ==× = × (b) At r = 0, the electric field is zero ( E = 0) since the enclosed charge is zero. At a certain point within the sphere, at some distance r from the center, the field (see Eq. 23-8 through Eq. 23-10) is given by Gauss’ law: enc 2 0 1 4 q E r ε = π where q enc is given by an integral similar to that worked in part (a): 4 2 enc 0 44. 4 r
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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ch23_p51 - 51. (a) We integrate the volume charge density...

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