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Unformatted text preview: side 1.20 m (Fig. 2162). The charges are µC, µC, and µC Calculate the magnitude and direction of the net force on each due to the other two. Solution:
Q1 x
α + 60° F31 F1 y F2 1 y F2 x F3 2 F1 2
β F13 y a = 1 .2 m F3 Q2 Q3 γ x F2 3 Q
3 23 3 32 µC Q
6 µC Q 2N N " " µC " means attraction " means attraction " means repulsion F N F N" 1. The total force acting on the particle with charge Q is F cos 60 cos sin sin N , , ,
, N angle from axis: 265.3° F
2 direction: tan , 2. The total force acting on the particle with charge Q2 is F2 F12 cos N 2 12 2 12 cos 60 sin sin N 2, , ,
2 ,
, N angle from axis: F23 ° direction: tan 2. The total force acting on the particle with charge Q2 is F3 F13 25 N 3 13 23 13 cos 120 23 cos sin 120 sin 0.130 N 3, 3, 23, 3 23
3 3 3, 25 130 0.2...
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 Spring '10
 TSOI
 Physics, Charge, Force

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