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21-2 - p—oso{Tl— 0.20 m—-| 91 = Q = QB = —3.0 so...

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Unformatted text preview: p—oso {Tl—+0.20 m—-| 91 = Q: = QB = —3.0 so +3.0 ac —4.0 he (a) y —. - L x F32;J F31 93 ('3) FIGURE 21-17 Example 21—2. Qcaunon Each charge exerts its own ﬁlms. No charge blocks the effect of the others Three charges in a line. Three charged particles are arranged in a line, as shown in Fig. 21—1? Calculate the net electrostatic force on particle 3 (the —4. 0 ,uC on the right) due to the other two charges. APPROACH The net force on particle 3 IS the vector sum of the force F31 exerted on 3 by particle 1 and the force F32 exerted on 3 by particle 2: F= F31 + F32. SOLUTION The magnitudes of these two forces are obtained using Coulomb’s law, Eq. 21—1: F31_ stQl r3] _ (9.0 x 109N-m2/C2)(4.0 x 10-6 c)(ao x 111-6 C) _ 1 N (0.510111)2 ' ’ where r31 = 0.50 m is the distance from Q3 to Q1. Similarly, PB: 2 kegs): r32 = (9.0 x 109N-m2/Cz][4.0 x 10‘6 C)(3.0 x 10‘6 C) = 27N (0.20 m)2 ' ‘ Since we were calculating the magnitudes of the forces, we omitted the signs of the charges. But we must be aware of them to get the direction of each force. Let the line Joining the particles be the x axis, and we take it positive to the right Then because F31 15 repulsive and F3; 15 attractive, the directions of the forces are as shown 111 Fig. 21—17b: F31 points in the positive .1: direction and .8,2 points in the negative x direction. The net force on particle 3 is then F = —F32 + F31 = —2.’?N + 1.2N = —1.5N. The magnitude of the net force is 1.5 N, and it points to the left. NOTE Charge Q1 acts on charge Q3 just as if Q2 were not there (this is the principle of superposition). That is, the charge in the middle, {22 , in no way blocks the effect of charge Q1 acting on Q3. Naturally, Q3 exerts its own force on Q3. ...
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