Solutions for HW chapter 18
2- REASONING The law of conservation of electric charges states that the net electric
charge of an isolated system remains constant. Initially the plate-rod system has a net
charge of 3.0 μC + 2.0 μC = 1.0 μC. After the transfer this charge is shared equally
by both objects, so that each carries a charge of 0.50 μC. Therefore, 2.5 μC of
negative charge must be transferred from the plate to the rod. To determine how
many electrons this is, we will divide this charge magnitude by the magnitude of the
charge on a single electron.
SOLUTION The magnitude of the charge on an electron is e, so that the number N of
electrons transferred is
N
Magnitude of transferred charge
2.5 106 C

 1.6 1013

19
e
1.60 10
C
4- REASONING The law of conservation of electric charges states that the net electric
charge of an isolated system remains constant. Initially the plate-rod system has a net
charge of 3.0 μC + 2.0 μC = 1.0 μC. After the transfer this charge is shared equally
by both objects, so that each carries a charge of 0.50 μC. Therefore, 2.5 μC of
negative charge must be transferred from the plate to the rod. To determine how
many electrons this is, we will divide this charge magnitude by the magnitude of the
charge on a single electron.
SOLUTION The magnitude of the charge on an electron is e, so that the number N of
electrons transferred is
N
Magnitude of transferred charge
2.5 106 C

 1.6 1013
19
e
1.60 10
C
7- SSM REASONING AND SOLUTION The magnitude of the force of attraction
between the charges is given by Coulomb's law (Equation 18.1): F  k q1 q2 / r 2 ,
where q1 and q2 are the magnitudes of the charges and r is the separation of the
charges. Let FA and FB represent the magnitudes of the forces between the charges
when the separations are rA and rB = rA/9, respectively. Then
2
2
FB k q1 q2 / rB2  rA   rA 
2

    
  (9)  81
2
FA k q1 q2 / rA  rB   rA / 9 
Therefore, we can conclude that FB  81 FA  (81)(1.5 N)= 120 N .
10-
10. REASONING The drawing at the right
shows the set-up. The force on the +q
charge at the origin due to the other +q
charge is given by Coulomb’s law
(Equation 18.1), as is the force due to the
+2q charge. These two forces point to the
left, since each is repulsive. The sum of
the two is twice the force on the +q charge at the origin due to the other +q charge
alone.
SOLUTION Applying Coulomb’s law, we have
kq q
 0.50 m 2
Force due to +q
charge at x 0.50 m

k 2q q
 d 2
Force due to +2q
charge at x  d

2
kq q
 0.50 m 2
Twice the force due to
+ q charge at x 0.50 m
Rearranging this result and solving for d give
k 2q q
d 
2

kq q
 0.50 m 
2
or
d 2  2  0.50 m 
2
or
d  0.71 m
We reject the negative root, because a negative value for d would locate the +2q
charge to the left of the origin. Then, the two forces acting on the charge at the
origin would have different directions, contrary to the statement of the problem.
Therefore, the +2q charge is located at a position of x  0.71 m .
16- REASONING According to Newton’s second law, the centripetal acceleration
experienced by the orbiting electron is equal to the centripetal force divided by the
electron’s mass. Recall from Section 5.3 that the centripetal force Fc is the name
given to the net force required to keep an object on a circular path of radius r. For an
electron orbiting about two protons, the centripetal force is provided almost
exclusively by the electrostatic force of attraction between the electron and the
protons. This force points toward the center of the circle and its magnitude is given
by Coulomb’s law.
SOLUTION The magnitude ac of the centripetal acceleration is equal to the
magnitude Fc of the centripetal force divided by the electron’s mass: ac  Fc / m
(Equation 5.3). The centripetal force is provided almost entirely by the electrostatic
force, so Fc = F, where F is the magnitude of the electrostatic force of attraction
between the electron and the two protons, Thus, ac  F / m . The magnitude of the
electrostatic force is given by Coulomb’s law,
F  k q1 q2 / r 2 (Equation 18.1),
where q1  e and q2  2e are the magnitudes of the charges, r is the radius of
the orbit, and k  8.99 109 N  m2 / C2 . Substituting this expression for F into
ac  F / m , and using m = 9.11  1031 kg for the mass of the electron, we find that
ac 

F

m
k
e 2e
k e 2e
r2

m
m r2
8.99 109 N  m2 / C2  1.60
 1019 C 2 1.60  1019 C
 9.111031 kg  2.65 1011 m 
2
 7.19 1023 m/s 2
20- REASONING The kinetic energy of the orbiting electron is KE  12 mv 2 (Equation
6.2), where m and v are its mass and speed, respectively. We can obtain the speed by
noting that the electron experiences a centripetal force whose magnitude Fc is given
by Fc  mv 2 / r (Equation 5.3), where r is the radius of the orbit. The centripetal
force is provided almost exclusively by the electrostatic force of attraction F
between the electron and the protons, so Fc = F. The electrostatic force points
toward the center of the circle and its magnitude is given by Coulomb’s law as
F  k q1 q2 / r 2 (Equation 18.1), where q1 and q2 are the magnitudes of the
charges.
SOLUTION The kinetic energy of the electron is
KE  12 mv 2
(6.2)
Solving the centripetal-force expression, Fc  mv 2 / r (Equation 5.3), for the speed
v, and substituting the result into Equation 6.2 gives
 rF 
KE  12 mv 2  12 m  c   12 r Fc
 m 
(1)
The centripetal force is provided almost entirely by the electrostatic force, so Fc = F,
where F is the magnitude of the electrostatic force of attraction between the electron
and the three protons. This force is given by Coulomb’s law, F  k q1 q2 / r 2
(Equation 18.1). Substituting Coulomb’s law into Equation 1 yields
 k q q  k q1 q2
KE  12 r Fc  12 r F  12 r  12 2  
2r
 r

Setting q1  e and q2  3e , we have
KE 

k e 3e
2r
8.99 109 N  m 2 / C2  1.60  1019 C
21.76 1011 m 
3 1.60  10 19 C
 1.96 1017 J
37- REASONING The electric field is given by Equation 18.2 as the force F that acts on
a test charge q0, divided by q0. Although the force is not known, the acceleration and
mass of the charged object are given. Therefore, we can use Newton’s second law to
determine the force as the mass times the acceleration and then determine the
magnitude of the field directly from Equation 18.2. The force has the same direction
as the acceleration. The direction of the field, however, is in the direction opposite to
that of the acceleration and force. This is because the object carries a negative
charge, while the field has the same direction as the force acting on a positive test
charge.
SOLUTION According to Equation 18.2, the magnitude of the electric field is
E
F
q0
According to Newton’s second law, the net force acting on an object of mass m and
acceleration a is F = ma. Here, the net force is the electrostatic force F, since that
force alone acts on the object. Thus, the magnitude of the electric field is



3.0 103 kg 2.5 103 m/s 2
F
ma
E


 2.2 105 N/C

6
q0
q0
34 10 C
The direction of this field is opposite to the direction of the acceleration. Thus, the
field points along the x axis .