Solutions to p677 #Q 2,7
P678 #Q23
p679 # P2, 5, 12, 15, 48, 49, 57, 63, 64
Q20.2. Reason: The statement of the question informs us that ball A is negative since charge
is transferred to it from a plastic rod that has been rubbed with wool. Recall that a plastic rod
rubbed with wool has a negative charge.
The first statement informs us that balls B, C, and D are either neutral or positive. If they are
neutral, their attraction to ball A is due to charge polarization. If they are positive, their attraction
to ball A is due to the fact that unlike charges attract.
The second statement informs us that balls B and D are neutral. If both are charged, there will be
attraction or repulsion. If either one is charged, there will be attraction due to charge polarization
of the other. Since there is neither attraction nor repulsion, neither is charged.
The last statement informs us that ball C must be charged. It is attracted to ball B due to
polarization of charges on the neutral ball B.
Knowing that C is charged and attracted to A and that unlike charges attract, we can conclude
that C has a positive charge.
To recap, ball A is negative, balls B and D are neutral, and ball C is positive.
Assess: By careful application of our knowledge of electrostatics we have been able to account
for all observations. In some cases we had to combine two bits of information in order to make a
conclusion regarding charge. All conclusions are consistent with the knowledge that like charges
repel, unlike charges attract and charge polarization can occur for a neutral object.
Q20.7. Reason: Each sphere ends up with 1 unit of negative charge. Once they touch, the
two spheres become essentially one conductor. The overall net charge is – 4  2  – 2. Charge is
spread
uniformly
over
the
surface
of
a conductor, so each sphere ends up with –1.
Assess: The arithmetic is easy to compute the total charge. One must also realize that charge
spreads out over
a conductor.
Q20.23. Reason: Because of Newton’s third law we shouldn’t choose choices D or E for any
of the situations. We ignore the gravitational attraction between the balls as it is many orders of
magnitude smaller than usual electrical forces.
(a) Like charges repel, so the correct choice is B.
(b) Opposite charges attract so the correct choice is C.
(c) Like charges repel, so the correct choice is B.
(d) The changed rod will polarize the neutral rod and they will attract each other, so the correct
choice is C.
r
Assess: It is important to remember that F12  F21.
P20.2.
1.6  10
Prepare: Use the charge model. An electron has a negative charge of magnitude
19
C.
Solve: (a) In the process of charging by rubbing, electrons are removed from one material and
transferred into the other because they are relatively free to move. Protons, on the other hand, are
tightly bound in the nuclei. So, electrons have been added to the plastic rod to make it negatively
charged.
(b) The number of electrons added is
20  109 C
11
 1.25  10
1.6  1019 C
which to two significant figures is 1.3  1011.
Assess: To create a modest charge of 20 nC, a large number of electrons need to be transferred
from one material to the other.
P20.5. Prepare: Use the charge model and the model of a conductor as a material through
which electrons move. An electron has a negative charge of magnitude 1.6  10 19 C.
Solve: (a) The charge of the glass rod decreases from 12 nC to 8 nC. Because it is the
electrons that are transferred, 4 nC of electrons has been added to the glass rod. Thus, electrons
are removed from the metal sphere and added to the glass rod.
(b) Because each electron has a charge of 1.6  10 19 C and a charge of 4 nC was transferred, the
number of electrons transferred from the metal sphere to the glass rod is
4 109 C
10
 2.5  10
1.6  10 19 C
Assess: 25 billion electrons constitute a charge of 4 nC.
P20.12. Prepare: This is a straightforward Coulomb’s law problem. We know the magnitude
of one of the charges (q1) and from the graph we can determine F and r at various points. This will
allow us to determine the magnitude of the second charge (q2 ).
Solve: Use the first point from the graph (r, F)  (1 cm, 0.8 mN) to determine the second
charge (q2 ). From Coulomb’s law we have F kq1q2 /r2 , which when solved for q2 gives:
q2  Fr2 /kq1  8.1 108 C.
Since the force is attractive and q1 is positive, q2 must be negative.
Assess: After carefully canceling units we note that our answer has the correct units and this
amount of charge is comparable to the given charge. We can further check our results by
performing the same calculation for the second given point on the graph (r, F)  (2 cm, 0.2 mN).
This calculation gives the same value for q2 allowing us to conclude that we have the correct
magnitude for q2 .
P20.15. Prepare: Please refer to Figure P20.15. The charged particles are point charges. The
charge q2 is in static equilibrium, so the net force on q2 is zero. If q2 is positive, q1 will have to be
positive to make the net force zero on q2. And, if q2 is negative, q1 will still have to be positive
for q2 to be in equilibrium. We will assume that the charge q2 is positive. For this situation, the
force on q2 by –2 nC charge is to the left and by q1 is to the right.
Solve: We have
r
r
 1
  1 (2  10 9 C)| q 2 |

|q1 ||q 2 |
Fne t on q 2  Fq1 on q 2  F2 nC on q 2  
,

x-direc
tion
, x -direction  0 N/C

2
2
(0.10 m)
 4 0 ( 0.2 m)
  4  0

Thus,
q1
2  109 C
2 
2  0 N/C 
(0.2 m)
(0.10 m)
q1  8 nC
Assess: If the charge q2 is assumed negative, the force on q2 by –2 nC charge is to the right and
by q1 is to the left. The magnitude of q1 remains unchanged.
P20.48. Prepare: The charges are point charges. Please refer to Figure P20.48. The electric
force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1, where q1 is the 1 nC charge, q2 is the
left 2 nC charge, and q3 is the right 2 nC charge.
Solve: We have
 K | q1|| q2 |

F2 on 1  
, awa y from q 2 
r2


 (9.0 0  10 9 N m 2 /C 2 )(1  10 9 C)(2  10 9 C )


, away from q2 
2
2
(1  10 m )


 (1.8  1 0 4 N, awa y f rom q2 )
r
( F2 on 1 ) x  (1.8  1 0 4 N)(cos 60)  (0.9  10 4 N)
r
( F2 on 1 ) y  (1.8  1 0 4 N)(sin 60)  (1.5 6  10 4 N)
r
 K | q1|| q3 |

F3 on 1  
, towa rd q 3   (1.8  10 4 N, towa rd q 3 )
r2


r
(F3 on 1 ) x 
r
(F3 on 1 ) y 
r
( Fon 1 ) x 
r
( Fon 1 ) y 
4
4
(1.8  1 0 N)(cos 60)  (0.9  10 N)
4
4
 (1.8  10 N)(sin6 0)  (1.56  10 N)
r
r
( F2 on 1 )x  (F3 on 1 ) x  1.8  10 4 N
r
r
( F2 on 1 )y  (F3 on 1 ) y  0 N
So the force on the 1 nC charge is 1.8  10–4 N directed horizontally to the right.
Assess: The magnitudes and symmetry of q2 and q3 ensure that their y-component of the net
force is zero.
P20.49. Prepare: The charges are point charges. Please refer to Figure P20.49. Placing the 1
nC charge at the origin and calling it q1, the q2 charge is in the first quadrant, the q3 charge is in
the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second
quadrant. The electric force on q1 is the vector sum of the forces F2 on 1, F3 on 1, F4 on 1, and F5 on 1.
Solve: The magnitude of the four forces is the same because all four charges are equal and
equidistant from q1. So,
F2 on 1  F3 on 1  F4 on 1  F5 on 1 
(9.0  10 9 N m2/C2 )(2 109 C)(110 9 C)
 3.6 104 N
2
2
2
2
(0.5  10 m)  (0.5 10 m)
Thus, Fon 1  (3.6  104 N, toward q2)  (3.6  104 N, toward q3)  (3.6  104 N, toward q4) 
(3.6  104 N, toward q5). It is now easy to see that the net force on q1 is zero.
Assess: Look at the symmetry of the charges. It is no surprise that the net force on charge q1 is
zero.
P20.51. Prepare: The charges are point charges. Please refer to Figure P20.51. Placing the 1
nC charge at the origin and calling it q1, the 6 nC is q3, the q2 charge is in the first quadrant, and
the q4 charge is in the second quadrant. The net electric force on q1 is the vector sum of the electric
forces from the other three charges q2, q3, and q4.
Solve: We have
 K |q1||q2|

F2 on 1  
, away from q2 
 r2

 (9.0  109 N m2/C2)(110 9 C)(2 109 C)


, away from q2
2
2
(5.0 10 m)


 (0.72  10 5 N, away from q2 )
r
 K |q1||q3|

F3 on 1  
, toward q3
2
r


 (9.0  109 N m2/C2)(110 9 C)(6 109 C)


, toward q3 
2
2
(5.0 10 m)


 (2.16 10 5 N, away from q3 )
r
 K |q1||q4|

F4 on 1  
, away from q4   (0.72  10 5 N, away from q4 )
2
r


r
( F2 on 1 )x   (0 .72  10 5 N)( cos4 5)  ( 0.50 9  10 5 N)
r
( F2 on 1 )y   (0.72  10 5 N)(sin 45)   (0.509  10 5 N)
r
( F3 on 1 ) x  0 N
r
( F3 on 1 ) y  (2.16  10 5 N)
r
5
5
( F4 on 1 )x  (0.72  10 N)(cos 45)  ( 0.509  10 N)
r
5
5
( F4 on 1 )y   (0.72  10 N)( sin 45)   ( 0.509  10 N)
r
r
r
r
( Fo n 1 ) x  ( F2 on 1 )x  (F3 on 1 ) x  ( F4 on 1 )x  0 N
r
r
r
r
( Fo n 1 ) y  ( F2 on 1 )y  (F3 on 1 ) y  (F4 on 1 ) y  1.14  10 5 N
So the force on the 1 nC charge is 1.14  10–5 N directed vertically up.
P20.57. Prepare: The electron and the proton are point charges. The electric Coulomb force
between the electron and the proton provides the centripetal acceleration for the electron’s
circular motion.   2πf.
Solve:
K(e)(e) mv2
2

 mr 2  mr(2  f )
r2
r
 f 
1
2
Ke2
mr
3

(9.0  10 9 N m 2 /C 2 )(1.60  10 19 C) 2
(9.11  10 31 k g)(5.3  10 11 )3

1
( 4.12  10 16 rev/s)  6.6  1015 rev/s
2
Assess: This is an extremely fast motion, nearly 7  1015 rpm!
P20.63. Prepare: The charged plastic beads are point charges and the spring is an ideal spring
that obeys Hooke’s law. Let q be the charge on each plastic bead. The repulsive force between the
beads pushes the beads apart. The spring is stretched until the restoring spring force on either bead
is equal to the repulsive Coulomb force. The spring constant k is obtained by noting that the
weight of a 1.0 g mass stretches the spring 1.0 cm. Thus
mg  k(1.0  102 m)
k
(1.0  103 kg)(9.8 N/kg)
 0.98 N/m
2
1.0  10 m
Solve: We have
Kq 2
r2
 kx  q 
kx r 2
K
q
(0.98 N/m)( 4.5  10 2 m  4.0  10 2 m)( 4.5  10 2 m) 2
9.0  10 9 N m 2/C2
 33 nC
P20.64. Model: The charged spheres are point charges. Each sphere is in static equilibrium
when the string makes an angle of 20 with the vertical. The three forces acting on each sphere
are
the
electric
force,
the
weight
of the sphere, and the tension force. A visual overview of the problem is shown below.
Solve:
r
r
r
r
In the static equilibrium, Newton’s first law is Fnet  T  w  Fe  0. In component form,
( Fne t )x  Tx  wx  (Fe ) x  0 N
 T sin   0 N 
 T sin  
K q2
d
2

Kq
( Fnet ) y  Ty  wy  (Fe ) y  0 N
2
0N
d2
 K q2
( 2L sin  )
2
T c os   mg  0 N  0 N
T cos    mg
Dividing the two equations and solving for q,
q
4 sin 2  tan  L2 mg
K

4(sin 2 20 tan 20)(1.0 m)2 (3.0  10 3 kg)(9.8 N/kg)
9.0  10 9 N m 2/C 2
 750 nC