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 2117 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 109Nm2/C2)(4.0 x 106 c)(ao x 1116 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 109Nm2/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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 Spring '11
 Getahneh

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