College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Please provide the correct answer for the wrong questions. To be honest, I have just followed the solutions as attached. Please provide the correct answer.

15. (a) The distance between qi and qz is
(x, –x,)* +(y; -y.)° =/(-0.020 m–0.035 m) +(0.015 m–0.005 m) =0.056 m.
The magnitude of the force exerted by q, on q, is
E.zt 199.1 (8.99×10ʻN-m³/C°) (3.0×10°C) (4.0×10“C)
(0.056 m)
= 35 N.
(b) The vector F, is directed toward qi and makes an angle 0 with the +x axis, where
972
CHAPTER 2I
1.5 cm-0.5 cm
0= tan 2-y
=tan
-10.3°-10°.
-2.0 cm-3.5 cm,
(c) Let the third charge be located at (x3, y3), a distance r from q2. We note that q1, q2, and
q3 must be collinear; otherwise, an equilibrium position for any one of them would be
impossible to find. Furthermore, we cannot place q3 on the same side of q2 where we also
find q1, since in that region both forces (exerted on q2 by q3 and q1) would be in the same
direction (since q2 is attracted to both of them). Thus, in terms of the angle found in part
(a), we have x; = x2 -r cose and y; = y2 –r sin@ (which means y3 > y2 since 0 is negative).
The magnitude of force exerted on q2 by q3 is F,=k|4:4,/r, which must equal that of
the force exerted on it by q; (found in part (a)). Therefore,
= k
= 0.0645 m = 6.45 cm.
Consequently, x3 =x2 -r cos0=-2.0 cm – (6.45 cm) cos(-10°) =-8.4 cm,
(d) and y; = y2 -r sine= 1.5 cm – (6.45 cm) sin(-10°) = 2.7 cm.
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Transcribed Image Text:15. (a) The distance between qi and qz is (x, –x,)* +(y; -y.)° =/(-0.020 m–0.035 m) +(0.015 m–0.005 m) =0.056 m. The magnitude of the force exerted by q, on q, is E.zt 199.1 (8.99×10ʻN-m³/C°) (3.0×10°C) (4.0×10“C) (0.056 m) = 35 N. (b) The vector F, is directed toward qi and makes an angle 0 with the +x axis, where 972 CHAPTER 2I 1.5 cm-0.5 cm 0= tan 2-y =tan -10.3°-10°. -2.0 cm-3.5 cm, (c) Let the third charge be located at (x3, y3), a distance r from q2. We note that q1, q2, and q3 must be collinear; otherwise, an equilibrium position for any one of them would be impossible to find. Furthermore, we cannot place q3 on the same side of q2 where we also find q1, since in that region both forces (exerted on q2 by q3 and q1) would be in the same direction (since q2 is attracted to both of them). Thus, in terms of the angle found in part (a), we have x; = x2 -r cose and y; = y2 –r sin@ (which means y3 > y2 since 0 is negative). The magnitude of force exerted on q2 by q3 is F,=k|4:4,/r, which must equal that of the force exerted on it by q; (found in part (a)). Therefore, = k = 0.0645 m = 6.45 cm. Consequently, x3 =x2 -r cos0=-2.0 cm – (6.45 cm) cos(-10°) =-8.4 cm, (d) and y; = y2 -r sine= 1.5 cm – (6.45 cm) sin(-10°) = 2.7 cm.
Your answer is partially correct.
The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 3.11 µC, x1 = 5.01 cm, y1 = 0.163 cm and
92 = -3.84 µC, x2 = -2.24 cm, y2 = 1.95 cm. Find the (a) magnitude and (b) direction (with respect to +x-axis in the range (-180°;180°) of
the electrostatic force on particle 2 due to particle 1. At what (c) x and (d) y coordinates should a third particle of charge q3 = 4.86 µC
be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?
(a) Number
19.3
Units
N
(b) Number
-13.8
Units
* (degrees)
(c) Number
-0.095
Units
(d) Number
0.037
Units
expand button
Transcribed Image Text:Your answer is partially correct. The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 3.11 µC, x1 = 5.01 cm, y1 = 0.163 cm and 92 = -3.84 µC, x2 = -2.24 cm, y2 = 1.95 cm. Find the (a) magnitude and (b) direction (with respect to +x-axis in the range (-180°;180°) of the electrostatic force on particle 2 due to particle 1. At what (c) x and (d) y coordinates should a third particle of charge q3 = 4.86 µC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero? (a) Number 19.3 Units N (b) Number -13.8 Units * (degrees) (c) Number -0.095 Units (d) Number 0.037 Units
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