The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 2.37 µC, x1 = 5.87 cm, y1 = 0.494 cm and 92 = -6.49 µC, x2 = -2.14 cm, y2 = 1.12 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 o charge q3 = 6.63 µC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero? (a) Number i Units (b) Number i Units (c) Number i Units (d) Number i Units > >
The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 2.37 µC, x1 = 5.87 cm, y1 = 0.494 cm and 92 = -6.49 µC, x2 = -2.14 cm, y2 = 1.12 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 o charge q3 = 6.63 µC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero? (a) Number i Units (b) Number i Units (c) Number i Units (d) Number i Units > >
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Transcribed Image Text:The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 2.37 µC, x1 = 5.87 cm, y1 = 0.494 cm and
92 = -6.49 µC, x2 = -2.14 cm, y2 = 1.12 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 = 6.63 µC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?
(a) Number
i
Units
(b) Number
i
Units
(c) Number
i
Units
(d) Number
i
Units
>
>
![•15 Go The charges and coordinates of two charged particles held
fixed in an xy plane are q = +3.0 µC, x1 = 3.5 cm, y = 0.50 cm,
and q2 = -4.0 µC, x, = -2.0 cm, y, = 1.5 cm. Find the (a) magni-
tude and (b) direction of the electrostatic force on particle 2 due to
particle 1. At what (c) x and (d) y coordinates should a third parti-
cle of charge q3 = +4.0 µC be placed such that the net electrostatic
force on particle 2 due to particles 1 and 3 is zero?
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
15. (a) The distance between qi and q2 is
(b) Number
-13.8
Units
* (degrees)
rz =(x, -x,)* +(y, -x) =(-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
(c) Number
Units
-0.095
m
(8.99×10°N-m²/C) (3.0×10 C) (4.0×10ʻC)
(0.056 m)
F =k
=35 N.
(d) Number
0.037
Units
m
(b) The vector Ē, is directed toward gj and makes an angle Owith the +x axis, where
972
СНАРТER 21
1.5 cm-0.5 cm
0= tan -
= tan
=-10.3°-10°.
-2.0 cm-3.5 cm
(c) Let the third charge be located at (x3, ya), a distance r from q2. We note that qı, 92, 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 qi) would be in the same
direction (since qz is attracted to both of them). Thus, in terms of the angle found in part
(a), we have x =x -r coso and y "y2 -r sino(which means y3 > y2 since O is negative).
The magnitude of force exerted on q2 by q3 is F=k|q:9, Vr, which must equal that of
the force exerted on it by q, (found in part (a)). Therefore,
= 0.0645m = 6.45 cm
Consequently, x3 =x2 -r cos0=-2.0 cm - (6.45 cm) cos(-10°) =-8.4 cm,
(d) and y; = y2 -r sino= 1.5 cm – (6.45 cm) sin(-10°) = 2.7 cm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3809466-6183-4e33-a80c-3c84e9082183%2Fd6bd9e2b-0908-4304-b636-22b8a2630887%2Fho4zhxo_processed.png&w=3840&q=75)
Transcribed Image Text:•15 Go The charges and coordinates of two charged particles held
fixed in an xy plane are q = +3.0 µC, x1 = 3.5 cm, y = 0.50 cm,
and q2 = -4.0 µC, x, = -2.0 cm, y, = 1.5 cm. Find the (a) magni-
tude and (b) direction of the electrostatic force on particle 2 due to
particle 1. At what (c) x and (d) y coordinates should a third parti-
cle of charge q3 = +4.0 µC be placed such that the net electrostatic
force on particle 2 due to particles 1 and 3 is zero?
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
15. (a) The distance between qi and q2 is
(b) Number
-13.8
Units
* (degrees)
rz =(x, -x,)* +(y, -x) =(-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
(c) Number
Units
-0.095
m
(8.99×10°N-m²/C) (3.0×10 C) (4.0×10ʻC)
(0.056 m)
F =k
=35 N.
(d) Number
0.037
Units
m
(b) The vector Ē, is directed toward gj and makes an angle Owith the +x axis, where
972
СНАРТER 21
1.5 cm-0.5 cm
0= tan -
= tan
=-10.3°-10°.
-2.0 cm-3.5 cm
(c) Let the third charge be located at (x3, ya), a distance r from q2. We note that qı, 92, 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 qi) would be in the same
direction (since qz is attracted to both of them). Thus, in terms of the angle found in part
(a), we have x =x -r coso and y "y2 -r sino(which means y3 > y2 since O is negative).
The magnitude of force exerted on q2 by q3 is F=k|q:9, Vr, which must equal that of
the force exerted on it by q, (found in part (a)). Therefore,
= 0.0645m = 6.45 cm
Consequently, x3 =x2 -r cos0=-2.0 cm - (6.45 cm) cos(-10°) =-8.4 cm,
(d) and y; = y2 -r sino= 1.5 cm – (6.45 cm) sin(-10°) = 2.7 cm.
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