problem21_32

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

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21.32: a) ( 29 j j E ˆ ) C N 10 813 . 2 ( m 0400 . 0 C) 10 00 . 5 ( ) C m N 10 9 ( ˆ 4 4 2 9 2 2 9 2 1 0 1 1 × - = × - × = = - r πε q ( 29 C N 10 08 . 1 m) 0400 . 0 ( 0.0300m C) 10 00 . 3 ( ) C m N 10 9 ( 4 2 2 9 2 2 9 2 2 2 2 × = + × × = = - r q E The angle of , 2 E measured from the axis, - x is ( 29 ° = - - 9 . 126 tan 180 cm 3.00 cm 00 . 4 1 Thus j i j i E ˆ ) C N 10 (8.64 ˆ ) C N 10 6.485 ( ) 9 . 126 sin ˆ 9 . 126 cos ˆ ( ) C N 10 080 . 1 ( 3 3 4 2 × + × - = ° + ° × = b) The resultant field is
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