Unformatted text preview: on 18.2,
F = qE. Since the charge q is negative, the electric field E must point downward, as the
product qE in the expression F = qE must be positive, since the force F points upward. The
magnitudes of the two forces must be equal, so that mg = q E . This expression can be
solved for E.
SOLUTION The magnitude of the electric field E is E= mg (0.012 kg)(9.80 m/s 2 )
=
= 6.5 × 103 N/C
–6
q
18 ×10 C As discussed in the reasoning, this electric field points downward .
______________________________________________________________________________ 998 67. ELECTRIC FORCES AND ELECTRIC FIELDS +z
SSM REASONING The drawing at the right shows
the setup. Here, the electric field E points along the +y
–F
–q
axis and applies a force of +F to the +q charge and a
force of –F to the –q charge, where q = 8.0 µC denotes
+y
the magnitude of each charge. Each force has the same
magnitude of F = E q , according to Equation 18.2.
E
+q
The torque is measured as discussed in Section 9.1.
According to Equation 9.1, the torque produced by each
+F
force has a magnitude given by the magnitude of the
+x
force times the lever arm, which is the perpendicular
distance between the point of application of the force
and the axis of rotation. In the drawing the z axis is the axis of rotation and is midway
between the ends of the rod. Thus, the lever arm for each force is half the length L of the rod
or L/2, and the magnitude of the torque produced by each force is (E q )(L/2).
SOLUTION The +F and the –F force each cause the rod to rotate in the same sense about
the z axis. Therefore, the torques from these forces reinforce one another. Using the
expression (E q )(L/2) for the magnitude of each torque, we find that the magnitude of the
net torque is
Magnitude of = E q L + E q L = E q L net torque
2
2 ( )( ) = 5.0 × 103 N/C 8.0 × 10 –6 C ( 4.0 m ) = 0.16 N ⋅ m
______________________________________________________________________________
68. REASONING The magnitude of the electrostatic force that acts on particle 1 is given by
Coulomb’s law as F = k q1 q2 / r 2 . This equation can be used to find the magnitude q2
of the charge. SOLUTION Solving Coulomb’s law for the magnitude q2 of the charge gives ( 3.4 N )( 0.26 m )
F r2
=
= 7.3 × 10−6 C
9
2
2
−6
k q1
8.99 × 10 N ⋅ m /C 3.5 × 10 C
2 q2 = ( )( ) (18.1) Since q1 is positive and experiences an attractive force, the charge q2 must be negative .
______________________________________________________________________________
69. SSM WWW REASONING Each particle will experience an electrostatic force due to
the presence of the other charge. According to Coulomb's law (Equation 18.1), the Chapter 18 Problems 999 magnitude of the force felt by each particle can be calculated from F = k q1 q2 / r 2 , where
q1 and q2 are the respective charges on particles 1 and 2 and r is the distance between them. According to Newton's second law, the magnitude of the force experienced by each
particle is given by F = ma , where a is the acceleration of the particle and we have assumed
that the electrostatic force is the only force acting.
SOLUTION
a. Since the two particles have identical positive charges, q1 = q2 = q , and we have, using
the data for particle 1,
kq
r2 2 = m1a1 Solving for q , we find that q= m1a1r 2
k = (6.00 × 10 –6 kg) (4.60 × 103 m/s 2 ) (2.60 × 10 –2 m) 2
8.99 × 109 N ⋅ m 2 /C 2 = 4.56 × 10 –8 C b. Since each particle experiences a force of the same magnitude (From Newton's third law),
we can write F1 = F2, or m1a1 = m2a2. Solving this expression for the mass m2 of particle 2,
we have
(6.00 ×10 –6 kg)(4.60 × 103 m/s 2 )
= 3.25 ×10 –6 kg
a2
8.50 × 103 m/s 2
______________________________________________________________________________
m2 = m1a1 = 70. REASONING AND SOLUTION The electric field is defined by Equation 18.2: E = F/q0.
Thus, the magnitude of the force exerted on a charge q in an electric field of magnitude E is
given by
F=qE
(1)
The magnitude of the electric field can be determined from its x and y components by using
the Pythagorean theorem:
2
2
E = Ex + E y = ( 6.00 ×103 N/C) + (8.00 ×103 N/C )
2 2 = 1.00 × 104 N/C a. From Equation (1) above, the magnitude of the force on the charge is
F = (7.50 × 10–6 C)(1.00 × 104 N/C) = 7.5 × 10 –2 N 1000 ELECTRIC FORCES AND ELECTRIC FIELDS b. From the defining equation for the electric field, it follows that the direction of the force
on a charge is the same as the direction of the field, provided that the charge is positive.
Thus, the angle that the force makes with the x axis is given by
3 Ey −1 8.00 × 10 N/C = tan 6.00 × 103 N/C = 53.1° Ex ______________________________________________________________________________ θ = tan −1 71. REASONING The two charges lying on the x axis produce no net electric field at the
coordinate origin. This is because they have identical charges, are located the same distance
from the origin, and produce electric fields that point in opposite directions. The electric
field produced by q3 at the origin points away from the charge, or along the −y direction.
The electric field produced by q4 at the origin points toward the charge, or along the +y
direction. The net electric field is, then, E = –E3 + E4, where E3 and E4 can be determined
by using Equation 18.3.
SOLUTION The net electric field at th...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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