P22_012

# P22_012 - x 3 = x 2 r cos and y 3 = y 2 r sin (which means...

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12. (a) The distance between q 1 and q 2 is r 12 = p ( x 2 x 1 ) 2 +( y 2 y 1 ) 2 = p ( 0 . 020 0 . 035) 2 +(0 . 015 0 . 005) 2 =0 . 0559 m . The magnitude of the force exerted by q 1 on q 2 is F 21 = k | q 1 q 2 | r 2 12 = ( 8 . 99 × 10 9 )( 3 . 0 × 10 6 )( 4 . 0 × 10 6 ) 0 . 0559 2 =34 . 5N . The vector ~ F 21 is directed towards q 1 and makes an angle θ with the + x axis, where θ =tan 1 µ y 2 y 1 x 2 x 1 =tan 1 µ 1 . 5 0 . 5 2 . 0 3 . 5 = 10 . 3 . (b) Let the third charge be located at ( x 3 ,y 3 ), a distance r from q 2 .W eno t etha t q 1 , q 2 and q 3 must be colinear; otherwise, an equilibrium position for any one of them would be impossible to Fnd. ±urthermore, we cannot place q 3 on the same side of q 2 wherewea lsoFnd q 1 , since in that region both forces (exerted on q 2 by q 3 and q 1
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Unformatted text preview: x 3 = x 2 r cos and y 3 = y 2 r sin (which means y 3 &gt; 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.

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