problem22_22

University Physics with Modern Physics with Mastering Physics (11th Edition)

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22.22: For points outside the sphere, the field is identical to that of a point charge of the same total magnitude located at the center of the sphere. The total charge is given by charge density × volume: C 10 60 . 1 m) 150 . 0 )( 3 4 )( m C n 50 . 7 ( 10 3 3 - × = = π q a) The field just outside the sphere is C N 4 . 42 m) (0.150 C) 10 (1.06 ) /C m N 10 9 ( 4 2 10 2 2 9 2 0 = × × = = - r πε q E b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will
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Unformatted text preview: be 1/4 as strong: 10.6 C N c) Inside the sphere, only the charge inside the radius in question affects the field. In this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to the field: C N 2 . 21 m) (0.075 C) 10 06 . 1 ( ) 8 / 1 ( ) C / m N 10 9 ( 2 10 2 2 9 = × ⋅ × =-E...
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This document was uploaded on 02/06/2008.

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