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Unformatted text preview: x 3 = x 2 r cos and y 3 = y 2 r sin (which means y 3 > y 2 since is negative). The magnitude of force exerted on q 2 by q 3 is F 23 = k  q 2 q 3  /r 2 , which must equal that of the force exerted on it by q 1 (found in part (a)). Therefore, k  q 2 q 3  r 2 = k  q 1 q 2  r 2 12 = r = r 12 r q 3 q 1 = 0 . 0645 cm . Consequently, x 3 = x 2 r cos = 2 . 0 cm (6 . 45 cm) cos( 10 . 3 ) = 8 . 4 cm and y 3 = y 2 r sin = 1 . 5 cm (6 . 45 cm) sin( 10 . 3 ) = 2 . 7 cm ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Force

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