CHAPTER 18 ELECTRIC FORCES AND
ELECTRIC FIELDS
PROBLEMS
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7.
SSM WWW REASONING Initially, the two spheres are neutral. Since negative
charge is removed from the sphere which loses electrons, it then carries a net positive charge.
Furthermore, the neutral sphere to which the electrons are added is then negatively charged.
Once the charge is transferred, there exists an electrostatic force on each of the two spheres,
the magnitude of which is given by Coulomb's law (Equation 18.1), F kq1q2 / r 2 .
SOLUTION
a. Since each electron carries a charge of –1.60 u 10 –19 C, the amount of negative charge
removed from the first sphere is
§ 1.60 u 1019 C ·
3.0 u 1013 electrons ¨
¸
© 1 electron ¹
4.8 u 106 C
Thus, the first sphere carries a charge +4.8 u 10–6 C, while the second sphere carries a
6
charge –4.8 u10 C . The magnitude of the electrostatic force that acts on each sphere is,
therefore,
kq 1 q 2 (8.99 u 10 9 N ˜ m 2 /C 2 )(4.8 u 10 – 6 C) 2
F
0.83 N
r2
(0.50 m) 2
b. Since the spheres carry charges of opposite sign, the force is attractive .
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15. SSM WWW REASONING Each particle will experience an electric force due to the
presence of the other charge. According to Coulomb's law (Equation 18.1), the magnitude
of the force felt by each particle can be calculated from F kq1q2 / 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.
SOLUTION
a. Since the two particles have identical positive charges, q1
the data for particle 1,
kq 2
m1a1
2
r
q2
q , and we have, using
664
ELECTRIC FORCES AND ELECTRIC FIELDS
Solving for q, we find that
m 1 a1 r 2
q
(6.00 u 10 –6 kg) (4.60 u 10 3 m/s 2 ) (2.60 u 10 –2 m) 2
8.99 u 10 9 N ˜ m 2 /C 2
k
4.56 u 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 m2 a2 . Solving this expression for the mass m 2 of particle
2, we have
(6.00 u 10 –6 kg)(4.60 u 10 3 m/s 2 )
m2
3.25 u 10 –6 kg
3
2
a2
8.50 u 10 m/s
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m 1 a1
E2
REASONING
The
39. SSM WWW
figure shows the arrangement of the three
charges. Let E q represent the electric
field at the empty corner due to the –q
charge. Furthermore, let E1 and E2 be
the electric fields at the empty corner due
to charges +q1 and +q2, respectively.
E1
+q1
Eq
d 5
T
–q
d
+q2
2d
According to the Pythagorean theorem, the distance from the charge –q to the empty corner
along the diagonal is given by (2d ) 2 d 2
5d 2 d 5 . The magnitude of each
electric field is given by Equation 18.3, E
electric
fields
at
the
empty
Eq
kq / r
2
2
kq / d 5
2
kq / r . Thus, the magnitudes of each of the
corner
are
given
as
follows:
2
kq /(5d ) , since the length of the diagonal is d 5 ;
2
2
E1 kq1 /(4d ) ; and E2 kq2 / d . The angle T that the diagonal makes with the
horizontal is T tan 1 (d / 2d) 26.57q . Since the net electric field Enet at the empty corner
is zero, the horizontal component of the net field must be zero, and we have
E1 – E q cos 26.57q
0
or
kq 1 / 4 d 2 – kq cos 26.57q / 5d 2
Similarly, the vertical component of the net field must be zero, and we have
E 2 – E q sin 26.57q
0
or
kq 2 / d 2 – kq sin 26.57q / 5d 2
0
These last two expressions can be solved for the charges q1 and q2 .
SOLUTION
Solving the last two expressions for the charges q1 and q2 , we have
0
Chapter 18 Problems
q1
4
q
5
cos 26.57q
0.716 q
q2
1
q
5
sin 26.57q
0.0895 q
665
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59. SSM WWW REASONING Two forces act on the charged ball (charge q); they are
the downward force of gravity mg and the electric force F due to the presence of the charge
q in the electric field E. In order for the ball to float, these two forces must be equal in
magnitude and opposite in direction, so that the net force on the ball is zero (Newton's
second law). Therefore, F must point upward, which we will take as the positive direction.
According to Equation 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 qE .
This expression can be solved for E.
SOLUTION The magnitude of the electric field E is
E
mg
q
(0.012 kg)(9.80 m/s 2 )
18 u 10 –6 C
6.5 u 10 3 N/C
As discussed in the reasoning, this electric field points downward .
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