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Unformatted text preview: m ) − ( 0.22 m )2
8.99 × 109 N ⋅ m 2 /C 2 +8.4 × 10 −6 C 2 ( ) 952 13. ELECTRIC FORCES AND ELECTRIC FIELDS y
SSM REASONING AND SOLUTION The net
electrostatic force on charge 3 at x = +3.0 m is the
1
+3.0 m
+18 µC
vector sum of the forces on charge 3 due to the other
two charges, 1 and 2. According to Coulomb's law
(Equation 18.1), the magnitude of the force on charge θ = 45°
+45 µC
3 due to charge 1 is
–12 µC
k q1 q3
3
02
F13 =
+3.0 m
2
r13 x Figure 1 where the distance between charges 1 and 3 is r13.
2
According to the Pythagorean theorem, r13 = x 2 + y 2 . Therefore, F13 (8.99 ×109 N ⋅ m2 / C2 )(18 ×10−6 C)( 45 ×10−6 C) = 0.405 N
=
( 3.0 m )2 + ( 3.0 m )2 Charges 1 and 3 are equidistant from the origin, so that θ = 45° (see Figure 1). Since
charges 1 and 3 are both positive, the force on charge 3 due to charge 1 is repulsive and
along the line that connects them, as shown in Figure 2. The components of F13 are:
F13 x = F13 cos 45° = 0.286 N and F13 y = – F13 sin 45° = –0.286 N The second force on charge 3 is the attractive force
(opposite signs) due to its interaction with charge 2
located at the origin. The magnitude of the force on
charge 3 due to charge 2 is, according to Coulomb's law ,
F23 = k q2 q3
2
r23 = k q2 q3
x y
1 45° 2 2 0 (8.99 ×109 N ⋅ m2 / C2 )(12 ×10−6 C )( 45 ×10−6 C )
=
( 3.0 m )2 F23 3
45° x F13
F Figure 2 13 = 0.539 N Since charges 2 and 3 have opposite signs, they attract each other, and charge 3 experiences
a force to the left as shown in Figure 2. Taking up and to the right as the positive directions,
we have
F3 x = F13 x + F23 x = +0.286 N − 0.539 N = −0.253 N F3 y = F13 y = −0.286 N Chapter 18 Problems Using the Pythagorean theorem, we find the magnitude of F3 to
be
F3 = F32x + F32y = (−0.253 N) + (−0.286 N) = 0.38 N
2 2 953 0.253 N φ
0.286 N
F3 The direction of F3 relative to the –x axis is specified by the
angle φ (see Figure 3), where
Figure 3
N
φ = tan = 49° below the − x axis 0.253 N ______________________________________________________________________________
−1 0.286 14. REASONING The electrical force that each charge exerts on charge 2 is shown in the
following drawings. F21 is the force exerted on 2 by 1, and F23 is the force exerted on 2
by 3. Each force has the same magnitude, because the charges have the same magnitude
and the distances are equal.
−q F21 +q 1 F23 2 +q +q F23 +q 3 1 2 (a) +q +q 3 F21 F21 −q 1 2
F23 (b)
3 +q
(c)
The net electric force F that acts on charge 2 is shown in the following diagrams. F21 F23
F (a) F21 F23 F=0N (b) F23
F F21 (c) It can be seen from the diagrams that the largest electric force occurs in (a), followed by (c),
and then by (b). SOLUTION The magnitude F21 of the force exerted on 2 by 1 is the same as the
magnitude F23 of the force exerted on 2 by 3, since the magnitudes of the charges are the
same and the distances are the same. Coulomb’s law gives the magnitudes as 954 ELECTRIC FORCES AND ELECTRIC FIELDS F21 = F23 = kqq
r2 (8.99 × 109 N ⋅ m2 /C2 )(8.6 × 10−6 C)(8.6 × 10−6 C) = 4.6 × 104 N
=
2
(3.8 × 10−3 m )
In part (a) of the drawing showing the net electric force acting on charge 2, both F21 and F23
point to the left, so the net force has a magnitude of ( ) F = 2 F12 = 2 4.6 × 104 N = 9.2 × 104 N
In part (b) of the drawing showing the net electric force acting on charge 2, F21 and F23
point in opposite directions, so the net force has a magnitude of 0 N .
In part (c) showing the net electric force acting on charge 2, the magnitude of the net force
can be obtained from the Pythagorean theorem: ( 4.6 × 104 N ) + ( 4.6 × 104 N )
2 2
2
F = F21 + F23 = 2 = 6.5 × 10 4 N ______________________________________________________________________________
15. SSM REASONING AND SOLUTION
a. Since the gravitational force between the spheres is one of attraction and the electrostatic
force must balance it, the electric force must be one of repulsion. Therefore, the charges
must have the same algebraic signs, both positive or both negative . b. There are two forces that act on each sphere; they are the gravitational attraction FG of
one sphere for the other, and the repulsive electric force FE of one sphere on the other.
From the problem statement, we know that these two forces balance each other, so that
FG = FE. The magnitude of FG is given by Newton's law of gravitation (Equation 4.3:
FG = Gm1m2 / r 2 ), while the magnitude of FE is given by Coulomb's law (Equation 18.1: FE = k q1 q2 / r 2 ). Therefore, we have Gm1m2
r 2 = k q1 q2
r 2 or Gm 2 = k q 2 Chapter 18 Problems 955 since the spheres have the same mass m and carry charges of the same magnitude q .
Solving for q , we find G
6.67 ×10 –11 N ⋅ m 2 /kg 2
= (2.0 ×10 –6 kg)
= 1.7 ×10 –16 C
k
8.99 ×109 N ⋅ m 2 /C2
______________________________________________________________________________
q =m 16. REASONING AND SOLUTION The electrostatic forces decreases with the square of the
distance separating the charges. If this distance is...
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 Spring '13
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