P22_041 - ~ F 3 net = 0. Writing r 3 1 = x and r 3 2 = x ....

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41. Charge q 1 = 80 × 10 6 C is at the origin, and charge q 2 =+40 × 10 6 Cisat x =0 . 20 m. The force on q 3 =+20 × 10 6 C is due to the attractive and repulsive forces from q 1 and q 2 , respectively. In symbols, ~ F 3ne t = ~ F 31 + ~ F 32 ,where | ~ F 31 | = k q 3 | q 1 | r 2 31 and | ~ F 32 | = k q 3 q 2 r 2 32 . (a) In this case r 31 =0 . 40 m and r 32 =0 . 20 m, with ~ F 31 directed towards x and ~ F 32 directed in the + x direction. Using the value of k in Eq. 22-5, we obtain ~ F 3ne t =89 . 9 90 N in the + x direction. (b) In this case r 31 =0 . 80 m and r 32 =0 . 60 m, with ~ F 31 directed towards x and ~ F 32 towards + x . Now we obtain ~ F 3ne t =2 . 5Ninthe
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Unformatted text preview: ~ F 3 net = 0. Writing r 3 1 = x and r 3 2 = x . 20 m, we equate | ~ F 3 1 | and | ~ F 3 2 | , and after canceling common factors, arrive at | q 1 | x 2 = q 2 ( x . 2) 2 . This can be further simpliFed to ( x . 2) 2 x 2 = q 2 | q 1 | = 1 2 . Taking the (positive) square root and solving, we obtain x = 0 . 68 m. If one takes the negative root and solves, one Fnds the location where the net force would be zero if q 1 and q 2 were of like sign (which is not the case here)....
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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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