ch21-p064 - 64. Charge q1 = 80 106 C is at the origin, and...

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Taking the (positive) square root and solving, we obtain x = 0.683 m. If one takes the negative root and ‘solves’, one finds 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). (d) From the above, we see that y = 0. 64. Charge q 1 = –80 × 10 –6 C is at the origin, and charge q 2 = +40 × 10 –6 C is at 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, G G G FF F 33 1 3 2 net =+ , where 31 3 2 31 32 22 32 || ,| | . qq q q Fk F k rr == GG (a) In this case r 31 = 0.40 m and r 32 = 0.20 m, with F G directed towards – x and F G directed in the + x direction. Using the value of k in Eq. 21-5, we obtain 3 2 12 3net 3 2 2 66 92 2 6 ˆˆ ˆ ˆ i| | i i i 80 10 C 40 10 C ˆ (8.99 10 N m C )(20 10 C) i (0.40m) (0.20m) ˆ (89.9 N)i . q q F k k k q r r −− ⎛⎞ =− + = − + = + ⎜⎟ ⎝⎠ −× =×⋅ × + = G (b) In this case r 31 = 0.80 m and r 32 = 0.60 m, with 31 F G directed towards –
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