P22_042 - addition becomes (0 . 518 6 37 ) + (0 . 518 6 37...

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42. (a) Charge Q 1 =+80 × 10 9 Cisonthe y axis at y =0 . 003 m, and charge Q 2 =+80 × 10 9 Cison the y axis at y = 0 . 003 m. The force on particle 3 (which has a charge of q =+18 × 10 9 C) is due to the vector sum of the repulsive forces from Q 1 and Q 2 . In symbols, ~ F 31 + ~ F 32 = ~ F 3ne t , where | ~ F 31 | = k q 3 | q 1 | r 2 31 and | ~ F 32 | = k q 3 q 2 r 2 32 . Using the Pythagorean theorem, we have r 31 = r 32 =0 . 005 m. In magnitude-angle notation
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Unformatted text preview: addition becomes (0 . 518 6 37 ) + (0 . 518 6 37 ) = (0 . 829 6 ) . Therefore, the net force is 0 . 829 N in the + x direction. (b) Switching the sign of Q 2 amounts to reversing the direction of its force on q . Consequently, we have (0 . 518 6 37 ) + (0 . 518 6 143 ) = (0 . 621 6 90 ) . Therefore, the net force is 0 . 621 N in the y direction....
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